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다변수 연쇄법칙

변수가 여러 단계로 연결될 때 나무 그림으로 편도함수를 구한다

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19문제 · 1 / 1쪽

1.CLP-3 · 2.4절 · 6번대학 수준우리말 번역

연쇄 법칙을 이용하는 방법을 포함하여 두 가지 방법으로 ∂w∂s\frac{\partial w}{\partial s} 와 ∂w∂t\frac{\partial w}{\partial t} 의 값을 구하세요. 단, 함수 w=x2+y2+z2w=x^2+y^2+z^2 이고 x=st, y=scos⁡tx=st,\ y=s\cos t, z=ssin⁡tz=s\sin t 입니다.

영어 원문 보기

Use two methods (one using the chain rule) to evaluate ∂w∂s\frac{\partial w}{\partial s} and ∂w∂t\frac{\partial w}{\partial t} given that the function w=x2+y2+z2w=x^2+y^2+z^2, with x=st, y=scos⁡tx=st,\ y=s\cos t and z=ssin⁡tz=s\sin t.

정답·풀이 보기

방법 1: w(s,t)=x(s,t)2+y(s,t)2+z(s,t)2w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2 이고 x(s,t)=stx(s,t)=st, y(s,t)=scos⁡ty(s,t)=s\cos t, z(s,t)=ssin⁡tz(s,t)=s\sin t 이므로 w(s,t)w(s,t) 를 명시적으로 나타낼 수 있다:

w(s,t)=(st)2+(scos⁡t)2+(ssin⁡t)2=s2(t2+1)  ⟹  ws(s,t)=2s(t2+1)그리고wt(s,t)=s2(2t)\begin{aligned} w(s,t)&=(st)^2+(s\cos t)^2+(s\sin t)^2 =s^2(t^2+1)\\ \implies w_s(s,t)&=2s(t^2+1)\qquad\text{그리고}\qquad w_t(s,t)=s^2(2t) \end{aligned}

  • 방법 2:* 문제에서 w(s,t)=x(s,t)2+y(s,t)2+z(s,t)2w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2 이고 x(s,t)=stx(s,t)=st, y(s,t)=scos⁡ty(s,t)=s\cos t, z(s,t)=ssin⁡tz(s,t)=s\sin t 로 주어졌다. 즉, w(s,t)=W(x(s,t),y(s,t),z(s,t))w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big) 이며 W(x,y,z)=x2+y2+z2W(x,y,z)= x^2+y^2+z^2 이다. w(s,t)=W(x(s,t),y(s,t),z(s,t))w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big) 에 연쇄 법칙을 적용하고 ∂W∂x=2x\frac{\partial W}{\partial x}=2x, ∂W∂y=2y\frac{\partial W}{\partial y}=2y, ∂W∂z=2z\frac{\partial W}{\partial z}=2z 임을 이용하면 다음을 얻는다.

∂w∂s(s,t)=∂W∂x(x(s,t),y(s,t),z(s,t))∂x∂s(s,t)+∂W∂y(x(s,t),y(s,t),z(s,t))∂y∂s(s,t)+∂W∂z(x(s,t),y(s,t),z(s,t))∂z∂s(s,t)=2x(s,t) xs(s,t)+2y(s,t) ys(s,t)+2z(s,t) zs(s,t)=2(st) t+2(scos⁡t) cos⁡t+2(ssin⁡t) sin⁡t=2st2+2s∂w∂t(s,t)=∂W∂x(x(s,t),y(s,t),z(s,t))∂x∂t(s,t)+∂W∂y(x(s,t),y(s,t),z(s,t))∂y∂t(s,t)+∂W∂z(x(s,t),y(s,t),z(s,t))∂z∂t(s,t)=2x(s,t) xt(s,t)+2y(s,t) yt(s,t)+2z(s,t) zt(s,t)=2(st) s+2(scos⁡t) (−ssin⁡t)+2(ssin⁡t) (scos⁡t)=2s2t\begin{aligned} \frac{\partial w}{\partial s}(s,t) &= \frac{\partial W}{\partial x}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial x}{\partial s}(s,t) + \frac{\partial W}{\partial y}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial y}{\partial s}(s,t)\\& + \frac{\partial W}{\partial z}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial z}{\partial s}(s,t) \\ &=2 x(s,t)\ x_s(s,t)+2 y(s,t)\ y_s(s,t)+2 z(s,t)\ z_s(s,t) \\ &=2 (st)\ t+2 (s\cos t)\ \cos t+2 (s\sin t)\ \sin t\\ &=2 st^2+2 s\\ \frac{\partial w}{\partial t}(s,t) &= \frac{\partial W}{\partial x}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial x}{\partial t}(s,t) + \frac{\partial W}{\partial y}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial y}{\partial t}(s,t)\\& + \frac{\partial W}{\partial z}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial z}{\partial t}(s,t) \\ &=2 x(s,t)\ x_t(s,t)+2 y(s,t)\ y_t(s,t)+2 z(s,t)\ z_t(s,t)\\ &=2 (st)\ s+2 (s\cos t)\ (-s\sin t)+2 (s\sin t)\ (s\cos t)\\ &=2 s^2t \end{aligned}

원본 교재 풀이(영어)

Method 1: Since w(s,t)=x(s,t)2+y(s,t)2+z(s,t)2w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2 with x(s,t)=stx(s,t)=st, y(s,t)=scos⁡ty(s,t)=s\cos t and z(s,t)=ssin⁡tz(s,t)=s\sin t we can write out w(s,t)w(s,t) explicitly:

w(s,t)=(st)2+(scos⁡t)2+(ssin⁡t)2=s2(t2+1)  ⟹  ws(s,t)=2s(t2+1)andwt(s,t)=s2(2t)\begin{aligned} w(s,t)&=(st)^2+(s\cos t)^2+(s\sin t)^2 =s^2(t^2+1)\\ \implies w_s(s,t)&=2s(t^2+1)\qquad\text{and}\qquad w_t(s,t)=s^2(2t) \end{aligned}

  • Method 2:* The question specifies that w(s,t)=x(s,t)2+y(s,t)2+z(s,t)2w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2 with x(s,t)=stx(s,t)=st, y(s,t)=scos⁡ty(s,t)=s\cos t and z(s,t)=ssin⁡tz(s,t)=s\sin t. That is, w(s,t)=W(x(s,t),y(s,t),z(s,t))w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big) with W(x,y,z)=x2+y2+z2W(x,y,z)= x^2+y^2+z^2. Applying the chain rule to w(s,t)=W(x(s,t),y(s,t),z(s,t))w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big) and noting that ∂W∂x=2x\frac{\partial W}{\partial x}=2x, ∂W∂y=2y\frac{\partial W}{\partial y}=2y, ∂W∂z=2z\frac{\partial W}{\partial z}=2z, gives

∂w∂s(s,t)=∂W∂x(x(s,t),y(s,t),z(s,t))∂x∂s(s,t)+∂W∂y(x(s,t),y(s,t),z(s,t))∂y∂s(s,t)+∂W∂z(x(s,t),y(s,t),z(s,t))∂z∂s(s,t)=2x(s,t) xs(s,t)+2y(s,t) ys(s,t)+2z(s,t) zs(s,t)=2(st) t+2(scos⁡t) cos⁡t+2(ssin⁡t) sin⁡t=2st2+2s∂w∂t(s,t)=∂W∂x(x(s,t),y(s,t),z(s,t))∂x∂t(s,t)+∂W∂y(x(s,t),y(s,t),z(s,t))∂y∂t(s,t)+∂W∂z(x(s,t),y(s,t),z(s,t))∂z∂t(s,t)=2x(s,t) xt(s,t)+2y(s,t) yt(s,t)+2z(s,t) zt(s,t)=2(st) s+2(scos⁡t) (−ssin⁡t)+2(ssin⁡t) (scos⁡t)=2s2t\begin{aligned} \frac{\partial w}{\partial s}(s,t) &= \frac{\partial W}{\partial x}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial x}{\partial s}(s,t) + \frac{\partial W}{\partial y}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial y}{\partial s}(s,t)\\& + \frac{\partial W}{\partial z}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial z}{\partial s}(s,t) \\ &=2 x(s,t)\ x_s(s,t)+2 y(s,t)\ y_s(s,t)+2 z(s,t)\ z_s(s,t) \\ &=2 (st)\ t+2 (s\cos t)\ \cos t+2 (s\sin t)\ \sin t\\ &=2 st^2+2 s\\ \frac{\partial w}{\partial t}(s,t) &= \frac{\partial W}{\partial x}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial x}{\partial t}(s,t) + \frac{\partial W}{\partial y}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial y}{\partial t}(s,t)\\& + \frac{\partial W}{\partial z}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial z}{\partial t}(s,t) \\ &=2 x(s,t)\ x_t(s,t)+2 y(s,t)\ y_t(s,t)+2 z(s,t)\ z_t(s,t)\\ &=2 (st)\ s+2 (s\cos t)\ (-s\sin t)+2 (s\sin t)\ (s\cos t)\\ &=2 s^2t \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 7번
# 다변수 연쇄법칙계산보통자동 분류
2.CLP-3 · 2.4절 · 7번대학 수준우리말 번역

∂3∂x∂y2f(2x+3y,xy)\frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) 의 값을 ff 의 편도함수로 나타내어 구하세요. 연쇄 법칙과, 혼합 편도함수가 서로 같음을 나타내는 클레로의 정리를 적용할 수 있도록 ff 는 매끄러운 함수라고 가정해도 됩니다.

영어 원문 보기

Evaluate ∂3∂x∂y2f(2x+3y,xy)\frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) in terms of partial derivatives of ff. You may assume that ff is a smooth function so that the Chain Rule and Clairaut's Theorem on the equality of the mixed partial derivatives apply.

정답·풀이 보기

정의에 따라

∂3∂x∂y2f(2x+3y,xy)=∂∂x[∂∂y(∂∂yf(2x+3y,xy))]\frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) =\frac{\partial }{\partial x}\left[\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f(2x+3y,xy)\right) \right]

이다.

안쪽부터 차례로 편도함수를 계산한다. F(x,y)=f(2x+3y,xy)F(x,y)=f(2x+3y,xy) 라 하면 가장 안쪽의 편도함수는 G(x,y)=∂∂yf(2x+3y,xy)=∂∂yF(x,y)G(x,y)=\frac{\partial }{\partial y}f(2x+3y,xy)=\frac{\partial }{\partial y}F(x,y) 이다. 연쇄 법칙에 의해

G(x,y)=∂∂yF(x,y)=f1(2x+3y,xy)∂∂y(2x+3y)+f2(2x+3y,xy)∂∂y(xy)=3f1(2x+3y,xy)+xf2(2x+3y,xy)\begin{aligned} G(x,y)=\frac{\partial }{\partial y}F(x,y) &=f_1(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_2(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_1(2x+3y,xy)+xf_2(2x+3y,xy) \end{aligned}

이다.

여기서 아래첨자 11 은 두 번째 변수를 고정한 채 ff 를 첫 번째 변수에 대해 편미분한다는 뜻이고, 아래첨자 22 는 첫 번째 변수를 고정한 채 ff 를 두 번째 변수에 대해 편미분한다는 뜻이다. 다음으로 가운데 편도함수를 H(x,y)=∂∂y(∂∂yf(2x+3y,xy))H(x,y)=\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f(2x+3y,xy)\right) 라 하면

H(x,y)=∂∂yG(x,y)=∂∂y(3f1(2x+3y,xy)+xf2(2x+3y,xy))=3∂∂y(f1(2x+3y,xy))+x∂∂y(f2(2x+3y,xy))\begin{aligned} H(x,y) &= \frac{\partial }{\partial y} G(x,y) \\ &=\frac{\partial }{\partial y}\Big(3f_1(2x+3y,xy)+xf_2(2x+3y,xy)\Big) \\ &=3\frac{\partial }{\partial y}\Big(f_1(2x+3y,xy)\Big) +x\frac{\partial }{\partial y}\Big(f_2(2x+3y,xy)\Big) \end{aligned}

이다.

연쇄 법칙을 (두 번) 쓰면

∂∂y(f1(2x+3y,xy))=f11(2x+3y,xy)∂∂y(2x+3y)+f12(2x+3y,xy)∂∂y(xy)=3f11(2x+3y,xy)+xf12(2x+3y,xy)∂∂y(f2(2x+3y,xy))=f21(2x+3y,xy)∂∂y(2x+3y)+f22(2x+3y,xy)∂∂y(xy)=3f21(2x+3y,xy)+xf22(2x+3y,xy)\begin{aligned} \frac{\partial }{\partial y}\Big(f_1(2x+3y,xy)\Big) &=f_{11}(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_{12}(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_{11}(2x+3y,xy)+xf_{12}(2x+3y,xy) \\ \frac{\partial }{\partial y}\Big(f_2(2x+3y,xy)\Big) &=f_{21}(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_{22}(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_{21}(2x+3y,xy)+xf_{22}(2x+3y,xy) \end{aligned}

이므로

H(x,y)=3(3f11(2x+3y,xy)+xf12(2x+3y,xy))+x(3f21(2x+3y,xy)+xf22(2x+3y,xy))=9f11(2x+3y,xy)+6xf12(2x+3y,xy)+x2f22(2x+3y,xy)\begin{aligned} H(x,y)&=3\Big(3f_{11}(2x+3y,xy)+xf_{12}(2x+3y,xy) \Big) \\[-0.05in] & +x\Big(3f_{21}(2x+3y,xy)+xf_{22}(2x+3y,xy)\Big) \\ &=9f_{11}(2x+3y,xy)+6xf_{12}(2x+3y,xy)+x^2f_{22}(2x+3y,xy) \end{aligned}

이다.

마지막 등호에서는 f21(2x+3y,xy)=f12(2x+3y,xy)f_{21}(2x+3y,xy)=f_{12}(2x+3y,xy) 임을 이용했다. 기호 f21f_{21} 은 먼저 두 번째 변수에 대해 미분한 다음 첫 번째 변수에 대해 미분한다는 뜻이다. 예를 들어 f(x,y)=e2ysin⁡xf(x,y)=e^{2y}\sin x 이면

f21(x,y)=∂∂x[∂∂y(e2ysin⁡x)]=∂∂x[2e2ysin⁡x]=2eycos⁡x\begin{aligned} f_{21}(x,y) =\frac{\partial }{\partial x} \Big[\frac{\partial }{\partial y}\big(e^{2y}\sin x\big)\Big] =\frac{\partial }{\partial x} \Big[2e^{2y}\sin x\Big] =2e^y\cos x \end{aligned}

이다.

마지막으로

∂3∂x∂y2f(2x+3y,xy)=∂∂xH(x,y)=∂∂x(9f11(2x+3y,xy)+6xf12(2x+3y,xy)+x2f22(2x+3y,xy))=9∂∂x(f11(2x+3y,xy))+6f12(2x+3y,xy)+6x∂∂x(f12(2x+3y,xy))+2xf22(2x+3y,xy)+x2∂∂x(f22(2x+3y,xy))\begin{aligned} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=\frac{\partial }{\partial x} H(x,y) \\ &=\frac{\partial }{\partial x}\Big(9f_{11}(2x+3y,xy)+6xf_{12}(2x+3y,xy)+x^2f_{22}(2x+3y,xy)\Big)\\ &=9\frac{\partial }{\partial x}\Big(f_{11}(2x+3y,xy)\Big)\\& +6f_{12}(2x+3y,xy) +6x\frac{\partial }{\partial x}\Big(f_{12}(2x+3y,xy)\Big) \\& +2xf_{22}(2x+3y,xy) +x^2\frac{\partial }{\partial x}\Big(f_{22}(2x+3y,xy)\Big) \end{aligned}

이다.

연쇄 법칙을 세 번 적용하면

∂3∂x∂y2f(2x+3y,xy)=9(2f111+yf112)+6f12+6x(2f121+yf122)+2xf22+x2(2f221+yf222)=6 f12+2x f22+18 f111+(9y+12x) f112+(6xy+2x2) f122+x2y f222\begin{aligned} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=9\Big(2f_{111}+yf_{112}\Big)\\& +6f_{12}+6x\Big(2f_{121}+yf_{122}\Big)\\& +2xf_{22}+x^2\Big(2f_{221}+yf_{222}\Big)\\ &=6\,f_{12}+2x\,f_{22}+18\,f_{111}+(9y+12x)\,f_{112} +(6xy+2x^2)\,f_{122}+x^2y\,f_{222} \end{aligned}

이다.

우변의 모든 함수의 인수는 (2x+3y,xy)(2x+3y,xy) 이다.

원본 교재 풀이(영어)

By definition,

∂3∂x∂y2f(2x+3y,xy)=∂∂x[∂∂y(∂∂yf(2x+3y,xy))]\frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) =\frac{\partial }{\partial x}\left[\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f(2x+3y,xy)\right) \right]

We'll compute the derivatives from the inside out. Let's call F(x,y)=f(2x+3y,xy)F(x,y)=f(2x+3y,xy) so that the innermost derivative is G(x,y)=∂∂yf(2x+3y,xy)=∂∂yF(x,y)G(x,y)=\frac{\partial }{\partial y}f(2x+3y,xy)=\frac{\partial }{\partial y}F(x,y). By the chain rule

G(x,y)=∂∂yF(x,y)=f1(2x+3y,xy)∂∂y(2x+3y)+f2(2x+3y,xy)∂∂y(xy)=3f1(2x+3y,xy)+xf2(2x+3y,xy)\begin{aligned} G(x,y)=\frac{\partial }{\partial y}F(x,y) &=f_1(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_2(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_1(2x+3y,xy)+xf_2(2x+3y,xy) \end{aligned}

Here the subscript 11 means take the partial derivative of ff with respect to the first argument while holding the second argument fixed, and the subscript 22 means take the partial derivative of ff with respect to the second argument while holding the first argument fixed. Next call the middle derivative H(x,y)=∂∂y(∂∂yf(2x+3y,xy))H(x,y)=\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f(2x+3y,xy)\right) so that

H(x,y)=∂∂yG(x,y)=∂∂y(3f1(2x+3y,xy)+xf2(2x+3y,xy))=3∂∂y(f1(2x+3y,xy))+x∂∂y(f2(2x+3y,xy))\begin{aligned} H(x,y) &= \frac{\partial }{\partial y} G(x,y) \\ &=\frac{\partial }{\partial y}\Big(3f_1(2x+3y,xy)+xf_2(2x+3y,xy)\Big) \\ &=3\frac{\partial }{\partial y}\Big(f_1(2x+3y,xy)\Big) +x\frac{\partial }{\partial y}\Big(f_2(2x+3y,xy)\Big) \end{aligned}

By the chain rule (twice),

∂∂y(f1(2x+3y,xy))=f11(2x+3y,xy)∂∂y(2x+3y)+f12(2x+3y,xy)∂∂y(xy)=3f11(2x+3y,xy)+xf12(2x+3y,xy)∂∂y(f2(2x+3y,xy))=f21(2x+3y,xy)∂∂y(2x+3y)+f22(2x+3y,xy)∂∂y(xy)=3f21(2x+3y,xy)+xf22(2x+3y,xy)\begin{aligned} \frac{\partial }{\partial y}\Big(f_1(2x+3y,xy)\Big) &=f_{11}(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_{12}(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_{11}(2x+3y,xy)+xf_{12}(2x+3y,xy) \\ \frac{\partial }{\partial y}\Big(f_2(2x+3y,xy)\Big) &=f_{21}(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_{22}(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_{21}(2x+3y,xy)+xf_{22}(2x+3y,xy) \end{aligned}

so that

H(x,y)=3(3f11(2x+3y,xy)+xf12(2x+3y,xy))+x(3f21(2x+3y,xy)+xf22(2x+3y,xy))=9f11(2x+3y,xy)+6xf12(2x+3y,xy)+x2f22(2x+3y,xy)\begin{aligned} H(x,y)&=3\Big(3f_{11}(2x+3y,xy)+xf_{12}(2x+3y,xy) \Big) \\[-0.05in] & +x\Big(3f_{21}(2x+3y,xy)+xf_{22}(2x+3y,xy)\Big) \\ &=9f_{11}(2x+3y,xy)+6xf_{12}(2x+3y,xy)+x^2f_{22}(2x+3y,xy) \end{aligned}

In the last equality we used that f21(2x+3y,xy)=f12(2x+3y,xy)f_{21}(2x+3y,xy)=f_{12}(2x+3y,xy). The notation f21f_{21} means first differentiate with respect to the second argument and then differentiate with respect to the first argument. For example, if f(x,y)=e2ysin⁡xf(x,y)=e^{2y}\sin x, then

f21(x,y)=∂∂x[∂∂y(e2ysin⁡x)]=∂∂x[2e2ysin⁡x]=2eycos⁡x\begin{aligned} f_{21}(x,y) =\frac{\partial }{\partial x} \Big[\frac{\partial }{\partial y}\big(e^{2y}\sin x\big)\Big] =\frac{\partial }{\partial x} \Big[2e^{2y}\sin x\Big] =2e^y\cos x \end{aligned}

Finally, we get to

∂3∂x∂y2f(2x+3y,xy)=∂∂xH(x,y)=∂∂x(9f11(2x+3y,xy)+6xf12(2x+3y,xy)+x2f22(2x+3y,xy))=9∂∂x(f11(2x+3y,xy))+6f12(2x+3y,xy)+6x∂∂x(f12(2x+3y,xy))+2xf22(2x+3y,xy)+x2∂∂x(f22(2x+3y,xy))\begin{aligned} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=\frac{\partial }{\partial x} H(x,y) \\ &=\frac{\partial }{\partial x}\Big(9f_{11}(2x+3y,xy)+6xf_{12}(2x+3y,xy)+x^2f_{22}(2x+3y,xy)\Big)\\ &=9\frac{\partial }{\partial x}\Big(f_{11}(2x+3y,xy)\Big)\\& +6f_{12}(2x+3y,xy) +6x\frac{\partial }{\partial x}\Big(f_{12}(2x+3y,xy)\Big) \\& +2xf_{22}(2x+3y,xy) +x^2\frac{\partial }{\partial x}\Big(f_{22}(2x+3y,xy)\Big) \end{aligned}

By three applications of the chain rule

∂3∂x∂y2f(2x+3y,xy)=9(2f111+yf112)+6f12+6x(2f121+yf122)+2xf22+x2(2f221+yf222)=6 f12+2x f22+18 f111+(9y+12x) f112+(6xy+2x2) f122+x2y f222\begin{aligned} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=9\Big(2f_{111}+yf_{112}\Big)\\& +6f_{12}+6x\Big(2f_{121}+yf_{122}\Big)\\& +2xf_{22}+x^2\Big(2f_{221}+yf_{222}\Big)\\ &=6\,f_{12}+2x\,f_{22}+18\,f_{111}+(9y+12x)\,f_{112} +(6xy+2x^2)\,f_{122}+x^2y\,f_{222} \end{aligned}

All functions on the right hand side have arguments (2x+3y,xy)(2x+3y,xy).

한 번 더 — 비슷한 문제 풀기APEX · 12.5절 · 2번
# 다변수 연쇄법칙계산보통자동 분류
3.CLP-3 · 2.4절 · 8번대학 수준우리말 번역

g(s,t)=f(2s+3t,3s−2t)g(s,t)=f(2s+3t,3s-2t) 의 모든 이계도함수를 구하세요. f(x,y)f(x,y) 는 매끄러운 함수이므로 연쇄 법칙과 혼합 편도함수의 같음에 관한 클레로의 정리가 적용된다고 가정해도 됩니다.

영어 원문 보기

Find all second order derivatives of g(s,t)=f(2s+3t,3s−2t)g(s,t)=f(2s+3t,3s-2t). You may assume that f(x,y)f(x,y) is a smooth function so that the Chain Rule and Clairaut's Theorem on the equality of the mixed partial derivatives apply.

정답·풀이 보기

주어진 함수는

g(s,t)=f(2s+3t,3s−2t)g(s,t)=f(2s+3t,3s-2t)

이다.

일계도함수는

gs(s,t)=2f1(2s+3t,3s−2t)+3f2(2s+3t,3s−2t)gt(s,t)=3f1(2s+3t,3s−2t)−2f2(2s+3t,3s−2t)\begin{aligned} g_s(s,t)&=2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\\ g_t(s,t)&=3f_1(2s+3t,3s-2t)-2f_2(2s+3t,3s-2t) \end{aligned}

이다.

이계도함수는

gss(s,t)=∂∂s(2f1(2s+3t,3s−2t)+3f2(2s+3t,3s−2t))=2(2f11+3f12)+3(2f21+3f22)=4f11+6f12+6f21+9f22=4f11(2s+3t,3s−2t)+12f12(2s+3t,3s−2t)+9f22(2s+3t,3s−2t)gst(s,t)=∂∂t(2f1(2s+3t,3s−2t)+3f2(2s+3t,3s−2t))=2(3f11−2f12)+3(3f21−2f22)=6f11(2s+3t,3s−2t)+5f12(2s+3t,3s−2t)−6f22(2s+3t,3s−2t)gtt(s,t)=∂∂t(3f1(2s+3t,3s−2t)−2f2(2s+3t,3s−2t))=3(3f11−2f12)−2(3f21−2f22)=9f11(2s+3t,3s−2t)−12f12(2s+3t,3s−2t)+4f22(2s+3t,3s−2t)\begin{aligned} g_{ss}(s,t) &=\frac{\partial }{\partial s}\Big(2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\Big)\\ &=2\Big(2f_{11}+3f_{12}\Big) +3\Big(2f_{21}+3f_{22}\Big)\\ &=4f_{11}+6f_{12}+6f_{21}+9f_{22}\\ &=4f_{11}(2s+3t,3s-2t)+12f_{12}(2s+3t,3s-2t)+9f_{22}(2s+3t,3s-2t)\\ g_{st}(s,t)&=\frac{\partial }{\partial t}\Big(2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\Big)\\ &=2\Big(3f_{11}-2f_{12}\Big) +3\Big(3f_{21}-2f_{22}\Big)\\ &=6f_{11}(2s+3t,3s-2t)+5f_{12}(2s+3t,3s-2t)-6f_{22}(2s+3t,3s-2t)\\ g_{tt}(s,t)&=\frac{\partial }{\partial t}\Big(3f_1(2s+3t,3s-2t)-2f_2(2s+3t,3s-2t)\Big)\\ &=3\Big(3f_{11}-2f_{12}\Big) -2\Big(3f_{21}-2f_{22}\Big)\\ &=9f_{11}(2s+3t,3s-2t)-12f_{12}(2s+3t,3s-2t)+4f_{22}(2s+3t,3s-2t) \end{aligned}

이다.

여기서 f1f_1 은 ff 를 첫 번째 변수에 대해 편미분한 편도함수를 뜻하고, f12f_{12} 는 ff 를 먼저 첫 번째 변수에 대해 편미분한 다음 그 결과를 두 번째 변수에 대해 편미분한 것이며, 나머지도 같은 방식이다.

원본 교재 풀이(영어)

The given function is

g(s,t)=f(2s+3t,3s−2t)g(s,t)=f(2s+3t,3s-2t)

The first order derivatives are

gs(s,t)=2f1(2s+3t,3s−2t)+3f2(2s+3t,3s−2t)gt(s,t)=3f1(2s+3t,3s−2t)−2f2(2s+3t,3s−2t)\begin{aligned} g_s(s,t)&=2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\\ g_t(s,t)&=3f_1(2s+3t,3s-2t)-2f_2(2s+3t,3s-2t) \end{aligned}

The second order derivatives are

gss(s,t)=∂∂s(2f1(2s+3t,3s−2t)+3f2(2s+3t,3s−2t))=2(2f11+3f12)+3(2f21+3f22)=4f11+6f12+6f21+9f22=4f11(2s+3t,3s−2t)+12f12(2s+3t,3s−2t)+9f22(2s+3t,3s−2t)gst(s,t)=∂∂t(2f1(2s+3t,3s−2t)+3f2(2s+3t,3s−2t))=2(3f11−2f12)+3(3f21−2f22)=6f11(2s+3t,3s−2t)+5f12(2s+3t,3s−2t)−6f22(2s+3t,3s−2t)gtt(s,t)=∂∂t(3f1(2s+3t,3s−2t)−2f2(2s+3t,3s−2t))=3(3f11−2f12)−2(3f21−2f22)=9f11(2s+3t,3s−2t)−12f12(2s+3t,3s−2t)+4f22(2s+3t,3s−2t)\begin{aligned} g_{ss}(s,t) &=\frac{\partial }{\partial s}\Big(2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\Big)\\ &=2\Big(2f_{11}+3f_{12}\Big) +3\Big(2f_{21}+3f_{22}\Big)\\ &=4f_{11}+6f_{12}+6f_{21}+9f_{22}\\ &=4f_{11}(2s+3t,3s-2t)+12f_{12}(2s+3t,3s-2t)+9f_{22}(2s+3t,3s-2t)\\ g_{st}(s,t)&=\frac{\partial }{\partial t}\Big(2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\Big)\\ &=2\Big(3f_{11}-2f_{12}\Big) +3\Big(3f_{21}-2f_{22}\Big)\\ &=6f_{11}(2s+3t,3s-2t)+5f_{12}(2s+3t,3s-2t)-6f_{22}(2s+3t,3s-2t)\\ g_{tt}(s,t)&=\frac{\partial }{\partial t}\Big(3f_1(2s+3t,3s-2t)-2f_2(2s+3t,3s-2t)\Big)\\ &=3\Big(3f_{11}-2f_{12}\Big) -2\Big(3f_{21}-2f_{22}\Big)\\ &=9f_{11}(2s+3t,3s-2t)-12f_{12}(2s+3t,3s-2t)+4f_{22}(2s+3t,3s-2t) \end{aligned}

Here f1f_1 denotes the partial derivative of ff with respect to its first argument, f12f_{12} is the result of first taking one partial derivative of ff with respect to its first argument and then taking a partial derivative with respect to its second argument, and so on.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.5절 · 39번
# 다변수 연쇄법칙활용보통자동 분류
4.CLP-3 · 2.4절 · 9번대학 수준우리말 번역

f(x,y)f(x,y) 가 라플라스 방정식 ∂2f∂x2+∂2f∂y2=0\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0 을 만족시킨다고 가정하세요. 합성함수 g(s,t)=f(s−t,s+t)g(s,t) = f (s - t, s + t) 에 대해서도 같은 식이 성립함을 보이세요. 즉, ∂2g∂s2+∂2g∂t2=0\frac{\partial^2 g}{\partial s^2}+\frac{\partial^2 g}{\partial t^2}=0 임을 보이세요. f(x,y)f(x,y) 는 매끄러운 함수여서 연쇄 법칙과 혼합 편도함수의 같음에 관한 클레로의 정리를 적용할 수 있다고 가정해도 됩니다.

영어 원문 보기

Assume that f(x,y)f(x,y) satisfies Laplace's equation ∂2f∂x2+∂2f∂y2=0\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0. Show that this is also the case for the composite function g(s,t)=f(s−t,s+t)g(s,t) = f (s - t, s + t). That is, show that ∂2g∂s2+∂2g∂t2=0\frac{\partial^2 g}{\partial s^2}+\frac{\partial^2 g}{\partial t^2}=0. You may assume that f(x,y)f(x,y) is a smooth function so that the Chain Rule and Clairaut's Theorem on the equality of the mixed partial derivatives apply.

정답·풀이 보기

연쇄 법칙에 의해,

∂g∂s(s,t)=∂∂sf(s−t,s+t)=∂f∂x(s−t , s+t)∂∂s(s−t)+∂f∂y(s−t , s+t)∂∂s(s+t)=∂f∂x(s−t , s+t)+∂f∂y(s−t , s+t)∂2g∂s2(s,t)=∂∂s[∂f∂x(s−t , s+t)]+∂∂s[∂f∂y(s−t , s+t)]=∂2f∂x2(s−t , s+t)+∂2f∂y∂x(s−t , s+t)+∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)={∂2f∂x2(s−t , s+t)+2∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)}\begin{aligned} \frac{\partial g}{\partial s}(s,t) &=\frac{\partial }{\partial s} f (s - t, s + t) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial s}\big(s-t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial s}\big(s+t\big) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big) \\ \frac{\partial^2 g}{\partial s^2}(s,t) &= \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\right] + \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\right] \\ &= \frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y\partial x}\big(s-t\,,\,s+t\big) \\ & + \frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big) \\ &=\left\{\frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) + 2\frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big)\right\} \end{aligned}

이고

∂g∂t(s,t)=∂∂tf(s−t,s+t)=∂f∂x(s−t , s+t)∂∂t(s−t)+∂f∂y(s−t , s+t)∂∂t(s+t)=−∂f∂x(s−t , s+t)+∂f∂y(s−t , s+t)∂2g∂t2(s,t)=−∂∂t[∂f∂x(s−t , s+t)]+∂∂t[∂f∂y(s−t , s+t)]=−[−∂2f∂x2(s−t , s+t)+∂2f∂y∂x(s−t , s+t)]+[−∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)]={∂2f∂x2(s−t , s+t)−2∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)}\begin{aligned} \frac{\partial g}{\partial t}(s,t)&=\frac{\partial }{\partial t} f (s - t, s + t) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial t}\big(s-t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial t}\big(s+t\big) \\ &= -\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big) \\ \frac{\partial^2 g}{\partial t^2}(s,t) &=- \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\right] + \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\right] \\ &= - \Big[-\frac{\partial^2 f}{\partial x^2} \big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y\partial x}\big(s-t\,,\,s+t\big)\Big] \\ & + \Big[-\frac{\partial^2 f}{\partial x\partial y} \big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big) \Big] \\ &=\left\{\frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) - 2\frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big)\right\} \end{aligned}

이다. 인수를 생략하면

∂2g∂s2+∂2g∂t2={∂2f∂x2+2∂2f∂x∂y+∂2f∂y2}+{∂2f∂x2−2∂2f∂x∂y+∂2f∂y2}=2[∂2f∂x2+∂2f∂y2]=0\begin{aligned} \frac{\partial^2 g}{\partial s^2} + \frac{\partial^2 g}{\partial t^2} &=\left\{\frac{\partial^2 f}{\partial x^2} + 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} +\left\{\frac{\partial^2 f}{\partial x^2} - 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} \\ &=2\left[\frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2}\right] \\ &=0 \end{aligned}

이므로 원하는 결과가 얻어진다.

원본 교재 풀이(영어)

By the chain rule,

∂g∂s(s,t)=∂∂sf(s−t,s+t)=∂f∂x(s−t , s+t)∂∂s(s−t)+∂f∂y(s−t , s+t)∂∂s(s+t)=∂f∂x(s−t , s+t)+∂f∂y(s−t , s+t)∂2g∂s2(s,t)=∂∂s[∂f∂x(s−t , s+t)]+∂∂s[∂f∂y(s−t , s+t)]=∂2f∂x2(s−t , s+t)+∂2f∂y∂x(s−t , s+t)+∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)={∂2f∂x2(s−t , s+t)+2∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)}\begin{aligned} \frac{\partial g}{\partial s}(s,t) &=\frac{\partial }{\partial s} f (s - t, s + t) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial s}\big(s-t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial s}\big(s+t\big) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big) \\ \frac{\partial^2 g}{\partial s^2}(s,t) &= \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\right] + \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\right] \\ &= \frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y\partial x}\big(s-t\,,\,s+t\big) \\ & + \frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big) \\ &=\left\{\frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) + 2\frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big)\right\} \end{aligned}

and

∂g∂t(s,t)=∂∂tf(s−t,s+t)=∂f∂x(s−t , s+t)∂∂t(s−t)+∂f∂y(s−t , s+t)∂∂t(s+t)=−∂f∂x(s−t , s+t)+∂f∂y(s−t , s+t)∂2g∂t2(s,t)=−∂∂t[∂f∂x(s−t , s+t)]+∂∂t[∂f∂y(s−t , s+t)]=−[−∂2f∂x2(s−t , s+t)+∂2f∂y∂x(s−t , s+t)]+[−∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)]={∂2f∂x2(s−t , s+t)−2∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)}\begin{aligned} \frac{\partial g}{\partial t}(s,t)&=\frac{\partial }{\partial t} f (s - t, s + t) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial t}\big(s-t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial t}\big(s+t\big) \\ &= -\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big) \\ \frac{\partial^2 g}{\partial t^2}(s,t) &=- \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\right] + \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\right] \\ &= - \Big[-\frac{\partial^2 f}{\partial x^2} \big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y\partial x}\big(s-t\,,\,s+t\big)\Big] \\ & + \Big[-\frac{\partial^2 f}{\partial x\partial y} \big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big) \Big] \\ &=\left\{\frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) - 2\frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big)\right\} \end{aligned}

Suppressing the arguments

∂2g∂s2+∂2g∂t2={∂2f∂x2+2∂2f∂x∂y+∂2f∂y2}+{∂2f∂x2−2∂2f∂x∂y+∂2f∂y2}=2[∂2f∂x2+∂2f∂y2]=0\begin{aligned} \frac{\partial^2 g}{\partial s^2} + \frac{\partial^2 g}{\partial t^2} &=\left\{\frac{\partial^2 f}{\partial x^2} + 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} +\left\{\frac{\partial^2 f}{\partial x^2} - 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} \\ &=2\left[\frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2}\right] \\ &=0 \end{aligned}

as desired.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.5절 · 37번
# 다변수 연쇄법칙증명·논술보통자동 분류
5.CLP-3 · 2.4절 · 10번대학 수준우리말 번역

z=f(x,y)z = f(x,y) 이고 x=2s+tx = 2s + t, y=s−ty = s - t 라 하자. 다음 등식이 성립하도록 하는 상수 aa, bb, cc 의 값을 구하세요.

a∂2z∂x2+b∂2z∂x ∂y+c∂2z∂y2=∂2z∂s2+∂2z∂t2a\frac{\partial^2 z}{\partial x^2} +b\frac{\partial^2 z}{\partial x\,\partial y} +c\frac{\partial^2 z}{\partial y^2} =\frac{\partial^2 z}{\partial s^2} +\frac{\partial^2 z}{\partial t^2}

z=f(x,y)z = f(x,y) 는 매끄러운 함수이므로 연쇄 법칙과 혼합 편도함수의 같음에 대한 클레로의 정리를 적용할 수 있다고 가정해도 됩니다.

영어 원문 보기

Let z=f(x,y)z = f(x,y) where x=2s+tx = 2s + t and y=s−ty = s - t. Find the values of the constants aa, bb and cc such that

a∂2z∂x2+b∂2z∂x ∂y+c∂2z∂y2=∂2z∂s2+∂2z∂t2a\frac{\partial^2 z}{\partial x^2} +b\frac{\partial^2 z}{\partial x\,\partial y} +c\frac{\partial^2 z}{\partial y^2} =\frac{\partial^2 z}{\partial s^2} +\frac{\partial^2 z}{\partial t^2}

You may assume that z=f(x,y)z = f(x,y) is a smooth function so that the Chain Rule and Clairaut's Theorem on the equality of the mixed partial derivatives apply.

정답·풀이 보기

이 문제의 표기는 매우 좋지 않다. 기호 zz 가 한 등식 안에서 두 가지 다른 뜻으로 쓰였기 때문이다. 좌변에서는 xx 와 yy 의 함수이고, 우변에서는 ss 와 tt 의 함수이다. 안타깝게도 이러한 표기의 남용은 매우 흔하다. ss 와 tt 의 함수를 F(s,t)F(s,t) 로 바꾸어 표기의 충돌을 없애자. 즉,

F(s,t)=f(2s+t , s−t)F(s,t) = f\big(2s+t\,,\,s-t\big)

이 새 표기에서는 다음이 성립하도록 aa, bb, cc 를 구하면 된다.

a∂2f∂x2+b∂2f∂x∂y+c∂2f∂y2=∂2F∂s2+∂2F∂t2\begin{aligned} a\frac{\partial^2 f}{\partial x^2} +b\frac{\partial^2 f}{\partial x\partial y} +c\frac{\partial^2 f}{\partial y^2} &=\frac{\partial^2 F}{\partial s^2} + \frac{\partial^2 F}{\partial t^2} \end{aligned}

여기서 우변의 인수는 (s,t)(s,t) 이고 좌변의 인수는 (2s+t , s−t)\big(2s+t\,,\,s-t\big) 이다.

연쇄 법칙에 의해

∂F∂s(s,t)=∂f∂x(2s+t , s−t)∂∂s(2s+t)+∂f∂y(2s+t , s−t)∂∂s(s−t)=2∂f∂x(2s+t , s−t)+∂f∂y(2s+t , s−t)∂2F∂s2(s,t)=2∂∂s[∂f∂x(2s+t , s−t)]+∂∂s[∂f∂y(2s+t , s−t)]=4∂2f∂x2(2s+t , s−t)+2∂2f∂y∂x(2s+t , s−t)+2∂2f∂x∂y(2s+t , s−t)+∂2f∂y2(2s+t , s−t)\begin{aligned} \frac{\partial F}{\partial s}(s,t)&= \frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial s}(2s+t) +\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial s}(s-t) \\ &= 2\frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big) +\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big) \\ \frac{\partial^2 F}{\partial s^2}(s,t) &= 2\frac{\partial }{\partial s}\left[\frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\right] + \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\right]\\ &= 4\frac{\partial^2 f}{\partial x^2}\big(2s+t\,,\,s-t\big) + 2\frac{\partial^2 f}{\partial y\partial x} \big(2s+t\,,\,s-t\big) \\ & + 2\frac{\partial^2 f}{\partial x\partial y} \big(2s+t\,,\,s-t\big) +\frac{\partial^2 f}{\partial y^2}\big(2s+t\,,\,s-t\big) \end{aligned}

이고

∂F∂t(s,t)=∂f∂x(2s+t , s−t)∂∂t(2s+t)+∂f∂y(2s+t , s−t)∂∂t(s−t)=∂f∂x(2s+t , s−t)−∂f∂y(2s+t , s−t)∂2F∂t2(s,t)=∂∂t[∂f∂x(2s+t , s−t)]−∂∂t[∂f∂y(2s+t , s−t)]=∂2f∂x2(2s+t , s−t)−∂2f∂y∂x(2s+t , s−t)−∂2f∂x∂y(2s+t , s−t)+∂2f∂y2(2s+t , s−t)\begin{aligned} \frac{\partial F}{\partial t}(s,t)&= \frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial t}(2s+t) +\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial t}(s-t)\\ &= \frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big) -\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big) \\ \frac{\partial^2 F}{\partial t^2}(s,t) &= \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\right] -\frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\right] \\ &= \frac{\partial^2 f}{\partial x^2}\big(2s+t\,,\,s-t\big) -\frac{\partial^2 f}{\partial y\partial x}\big(2s+t\,,\,s-t\big) \\ & -\frac{\partial^2 f}{\partial x\partial y} \big(2s+t\,,\,s-t\big) +\frac{\partial^2 f}{\partial y^2}\big(2s+t\,,\,s-t\big) \end{aligned}

이다. 인수를 생략하면

∂2F∂s2+∂2F∂t2=5∂2f∂x2+2∂2f∂x∂y+2∂2f∂y2\begin{aligned} \frac{\partial^2 F}{\partial s^2} + \frac{\partial^2 F}{\partial t^2} &=5\frac{\partial^2 f}{\partial x^2} +2\frac{\partial^2 f}{\partial x\partial y} +2\frac{\partial^2 f}{\partial y^2} \end{aligned}

이다. 마지막으로 문제의 (좋지 않은) 표기로 되돌리면

∂2z∂s2+∂2z∂t2=5∂2z∂x2+2∂2z∂x∂y+2∂2z∂y2\begin{aligned} \frac{\partial^2 z}{\partial s^2} + \frac{\partial^2 z}{\partial t^2} &=5\frac{\partial^2 z}{\partial x^2} +2\frac{\partial^2 z}{\partial x\partial y} +2\frac{\partial^2 z}{\partial y^2} \end{aligned}

이므로 a=5a=5 이고 b=c=2b=c=2 이다.

원본 교재 풀이(영어)

The notation in the statement of this question is horrendous — the symbol zz is used with two different meanings in one equation. On the left hand side, it is a function of xx and yy, and on the right hand side, it is a function of ss and tt. Unfortunately that abuse of notation is also very common. Let us undo the notation conflict by renaming the function of ss and tt to F(s,t)F(s,t). That is,

F(s,t)=f(2s+t , s−t)F(s,t) = f\big(2s+t\,,\,s-t\big)

In this new notation, we are being asked to find aa, bb and cc so that

a∂2f∂x2+b∂2f∂x∂y+c∂2f∂y2=∂2F∂s2+∂2F∂t2\begin{aligned} a\frac{\partial^2 f}{\partial x^2} +b\frac{\partial^2 f}{\partial x\partial y} +c\frac{\partial^2 f}{\partial y^2} &=\frac{\partial^2 F}{\partial s^2} + \frac{\partial^2 F}{\partial t^2} \end{aligned}

with the arguments on the right hand side being (s,t)(s,t) and the arguments on the left hand side being (2s+t , s−t)\big(2s+t\,,\,s-t\big).

By the chain rule,

∂F∂s(s,t)=∂f∂x(2s+t , s−t)∂∂s(2s+t)+∂f∂y(2s+t , s−t)∂∂s(s−t)=2∂f∂x(2s+t , s−t)+∂f∂y(2s+t , s−t)∂2F∂s2(s,t)=2∂∂s[∂f∂x(2s+t , s−t)]+∂∂s[∂f∂y(2s+t , s−t)]=4∂2f∂x2(2s+t , s−t)+2∂2f∂y∂x(2s+t , s−t)+2∂2f∂x∂y(2s+t , s−t)+∂2f∂y2(2s+t , s−t)\begin{aligned} \frac{\partial F}{\partial s}(s,t)&= \frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial s}(2s+t) +\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial s}(s-t) \\ &= 2\frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big) +\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big) \\ \frac{\partial^2 F}{\partial s^2}(s,t) &= 2\frac{\partial }{\partial s}\left[\frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\right] + \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\right]\\ &= 4\frac{\partial^2 f}{\partial x^2}\big(2s+t\,,\,s-t\big) + 2\frac{\partial^2 f}{\partial y\partial x} \big(2s+t\,,\,s-t\big) \\ & + 2\frac{\partial^2 f}{\partial x\partial y} \big(2s+t\,,\,s-t\big) +\frac{\partial^2 f}{\partial y^2}\big(2s+t\,,\,s-t\big) \end{aligned}

and

∂F∂t(s,t)=∂f∂x(2s+t , s−t)∂∂t(2s+t)+∂f∂y(2s+t , s−t)∂∂t(s−t)=∂f∂x(2s+t , s−t)−∂f∂y(2s+t , s−t)∂2F∂t2(s,t)=∂∂t[∂f∂x(2s+t , s−t)]−∂∂t[∂f∂y(2s+t , s−t)]=∂2f∂x2(2s+t , s−t)−∂2f∂y∂x(2s+t , s−t)−∂2f∂x∂y(2s+t , s−t)+∂2f∂y2(2s+t , s−t)\begin{aligned} \frac{\partial F}{\partial t}(s,t)&= \frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial t}(2s+t) +\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial t}(s-t)\\ &= \frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big) -\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big) \\ \frac{\partial^2 F}{\partial t^2}(s,t) &= \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\right] -\frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\right] \\ &= \frac{\partial^2 f}{\partial x^2}\big(2s+t\,,\,s-t\big) -\frac{\partial^2 f}{\partial y\partial x}\big(2s+t\,,\,s-t\big) \\ & -\frac{\partial^2 f}{\partial x\partial y} \big(2s+t\,,\,s-t\big) +\frac{\partial^2 f}{\partial y^2}\big(2s+t\,,\,s-t\big) \end{aligned}

Suppressing the arguments

∂2F∂s2+∂2F∂t2=5∂2f∂x2+2∂2f∂x∂y+2∂2f∂y2\begin{aligned} \frac{\partial^2 F}{\partial s^2} + \frac{\partial^2 F}{\partial t^2} &=5\frac{\partial^2 f}{\partial x^2} +2\frac{\partial^2 f}{\partial x\partial y} +2\frac{\partial^2 f}{\partial y^2} \end{aligned}

Finally, translating back into the (horrendous) notation of the question

∂2z∂s2+∂2z∂t2=5∂2z∂x2+2∂2z∂x∂y+2∂2z∂y2\begin{aligned} \frac{\partial^2 z}{\partial s^2} + \frac{\partial^2 z}{\partial t^2} &=5\frac{\partial^2 z}{\partial x^2} +2\frac{\partial^2 z}{\partial x\partial y} +2\frac{\partial^2 z}{\partial y^2} \end{aligned}

so that a=5a=5 and b=c=2b=c=2.

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 26번
# 다변수 연쇄법칙계산보통자동 분류
6.CLP-3 · 2.4절 · 11번대학 수준우리말 번역

FF 를 R2\mathbb{R}^2 위의 함수라 하자. R2\mathbb{R}^2 의 점을 (u,v)(u, v) 로 나타내고, FF 의 대응하는 편도함수를 Fu(u,v)F_u(u, v), Fv(u,v)F_v (u, v), Fuu(u,v)F_{uu}(u, v), Fuv(u,v)F_{uv}(u, v) 등으로 나타낸다. 이 편도함수들이 모두 연속이라고 가정한다.

∂2∂x ∂yF(x2−y2,2xy)\begin{aligned} \frac{\partial^2}{\partial x\, \partial y} F(x^2 - y^2 , 2xy) \end{aligned}

를 함수 FF 의 편도함수를 사용하여 나타내세요.

영어 원문 보기

Let FF be a function on R2\mathbb{R}^2. Denote points in R2\mathbb{R}^2 by (u,v)(u, v) and the corresponding partial derivatives of FF by Fu(u,v)F_u(u, v), Fv(u,v)F_v (u, v), Fuu(u,v)F_{uu}(u, v), Fuv(u,v)F_{uv}(u, v), etc.. Assume those derivatives are all continuous. Express

∂2∂x ∂yF(x2−y2,2xy)\begin{aligned} \frac{\partial^2}{\partial x\, \partial y} F(x^2 - y^2 , 2xy) \end{aligned}

in terms of partial derivatives of the function FF.

정답·풀이 보기

u(x,y)=x2−y2u(x,y) = x^2 - y^2, v(x,y)=2xyv(x,y) = 2xy 라 하자. 그러면 F(x2−y2,2xy)=F(u(x,y),v(x,y))F(x^2 - y^2 , 2xy) =F\big(u(x,y),v(x,y)\big) 이다. 연쇄 법칙에 의해

∂∂yF(x2−y2,2xy)=∂∂yF(u(x,y),v(xy))=Fu(u(x,y),v(xy))∂u∂y(x,y)+Fv(u(x,y),v(xy))∂v∂y(x,y)=Fu(x2−y2,2xy) (−2y)+Fv(x2−y2,2xy) (2x)∂2∂x ∂yF(x2−y2,2xy)=∂∂x{−2yFu(x2−y2,2xy)+2xFv(x2−y2,2xy)}=−2y∂∂x[Fu(x2−y2,2xy)]+2Fv(x2−y2,2xy)+2x∂∂x[Fv(x2−y2,2xy)]=−4xy Fuu(x2−y2,2xy)−4y2Fuv(x2−y2,2xy)+2Fv(x2−y2,2xy)+4x2Fvu(x2−y2,2xy)+4xy Fvv(x2−y2,2xy)=2 Fv(x2−y2,2xy)−4xy Fuu(x2−y2,2xy)+4(x2−y2) Fuv(x2−y2,2xy)+4xy Fvv(x2−y2,2xy)\begin{aligned} \frac{\partial }{\partial y}F(x^2 - y^2 , 2xy) &=\frac{\partial }{\partial y}F(u(x,y) , v(xy)) \\ &=F_u(u(x,y) , v(xy))\frac{\partial u}{\partial y}(x,y) + F_v(u(x,y) , v(xy))\frac{\partial v}{\partial y}(x,y) \\ &= F_u(x^2 - y^2 , 2xy)\ (-2y) + F_v(x^2 - y^2 , 2xy)\, (2x) \\ \frac{\partial^2}{\partial x\, \partial y} F(x^2 - y^2 , 2xy) &=\frac{\partial }{\partial x}\left\{-2y F_u(x^2 - y^2 , 2xy) + 2x F_v(x^2 - y^2 , 2xy) \right\} \\ &= -2y\frac{\partial }{\partial x}\left[ F_u(x^2 - y^2 , 2xy)\right] +2 F_v(x^2 - y^2 , 2xy) \\& + 2x \frac{\partial }{\partial x}\left[F_v(x^2 - y^2 , 2xy)\right] \\ &= -4xy\,F_{uu}(x^2 - y^2 , 2xy) -4y^2 F_{uv}(x^2 - y^2 , 2xy) +2 F_v(x^2 - y^2 , 2xy) \\& + 4x^2 F_{vu}(x^2 - y^2 , 2xy) +4xy\,F_{vv}(x^2 - y^2 , 2xy) \\ &= 2 \, F_v(x^2 - y^2 , 2xy) -4xy\, F_{uu}(x^2 - y^2 , 2xy) \\& +4(x^2-y^2)\, F_{uv}(x^2 - y^2 , 2xy) \\& +4xy\, F_{vv}(x^2 - y^2 , 2xy) \end{aligned}

원본 교재 풀이(영어)

Let u(x,y)=x2−y2u(x,y) = x^2 - y^2 , and v(x,y)=2xyv(x,y) = 2xy. Then F(x2−y2,2xy)=F(u(x,y),v(x,y))F(x^2 - y^2 , 2xy) =F\big(u(x,y),v(x,y)\big). By the chain rule

∂∂yF(x2−y2,2xy)=∂∂yF(u(x,y),v(xy))=Fu(u(x,y),v(xy))∂u∂y(x,y)+Fv(u(x,y),v(xy))∂v∂y(x,y)=Fu(x2−y2,2xy) (−2y)+Fv(x2−y2,2xy) (2x)∂2∂x ∂yF(x2−y2,2xy)=∂∂x{−2yFu(x2−y2,2xy)+2xFv(x2−y2,2xy)}=−2y∂∂x[Fu(x2−y2,2xy)]+2Fv(x2−y2,2xy)+2x∂∂x[Fv(x2−y2,2xy)]=−4xy Fuu(x2−y2,2xy)−4y2Fuv(x2−y2,2xy)+2Fv(x2−y2,2xy)+4x2Fvu(x2−y2,2xy)+4xy Fvv(x2−y2,2xy)=2 Fv(x2−y2,2xy)−4xy Fuu(x2−y2,2xy)+4(x2−y2) Fuv(x2−y2,2xy)+4xy Fvv(x2−y2,2xy)\begin{aligned} \frac{\partial }{\partial y}F(x^2 - y^2 , 2xy) &=\frac{\partial }{\partial y}F(u(x,y) , v(xy)) \\ &=F_u(u(x,y) , v(xy))\frac{\partial u}{\partial y}(x,y) + F_v(u(x,y) , v(xy))\frac{\partial v}{\partial y}(x,y) \\ &= F_u(x^2 - y^2 , 2xy)\ (-2y) + F_v(x^2 - y^2 , 2xy)\, (2x) \\ \frac{\partial^2}{\partial x\, \partial y} F(x^2 - y^2 , 2xy) &=\frac{\partial }{\partial x}\left\{-2y F_u(x^2 - y^2 , 2xy) + 2x F_v(x^2 - y^2 , 2xy) \right\} \\ &= -2y\frac{\partial }{\partial x}\left[ F_u(x^2 - y^2 , 2xy)\right] +2 F_v(x^2 - y^2 , 2xy) \\& + 2x \frac{\partial }{\partial x}\left[F_v(x^2 - y^2 , 2xy)\right] \\ &= -4xy\,F_{uu}(x^2 - y^2 , 2xy) -4y^2 F_{uv}(x^2 - y^2 , 2xy) +2 F_v(x^2 - y^2 , 2xy) \\& + 4x^2 F_{vu}(x^2 - y^2 , 2xy) +4xy\,F_{vv}(x^2 - y^2 , 2xy) \\ &= 2 \, F_v(x^2 - y^2 , 2xy) -4xy\, F_{uu}(x^2 - y^2 , 2xy) \\& +4(x^2-y^2)\, F_{uv}(x^2 - y^2 , 2xy) \\& +4xy\, F_{vv}(x^2 - y^2 , 2xy) \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 10번
# 다변수 연쇄법칙계산보통자동 분류
7.CLP-3 · 2.4절 · 12번대학 수준우리말 번역

u(x,y)u(x,y) 가 다음과 같이 정의된다.

u(x,y)=ey F(xe−y2)u(x,y) = e^y\, F\big(xe^{-y^2}\big)

여기서 F(z)F(z) 는 임의의 함수이다.

(a) F(z)=ln⁡(z)F(z) = \ln(z) 일 때, ∂u∂x\frac{\partial u}{\partial x} 와 ∂u∂y\frac{\partial u}{\partial y} 를 구하세요. (b) 임의의 함수 F(z)F(z) 에 대해 u(x,y)u(x,y) 가 다음을 만족시킴을 보이세요.

2xy∂u∂x+∂u∂y=u2xy\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = u

영어 원문 보기

u(x,y)u(x,y) is defined as

u(x,y)=ey F(xe−y2)u(x,y) = e^y\, F\big(xe^{-y^2}\big)

for an arbitrary function F(z)F(z).

(a) If F(z)=ln⁡(z)F(z) = \ln(z), find ∂u∂x\frac{\partial u}{\partial x} and ∂u∂y\frac{\partial u}{\partial y}. (b) For an arbitrary F(z)F(z) show that u(x,y)u(x,y) satisfies

2xy∂u∂x+∂u∂y=u2xy\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = u

정답·풀이 보기

임의의 (미분가능한) 함수 FF 에 대해, 연쇄법칙과 곱의 미분법에 의해 다음이 성립한다.

∂u∂x(x,y)=∂∂x[ey F(xe−y2)]=ey ∂∂x[F(xe−y2)]=ey F′(xe−y2)∂∂x(xe−y2)=ey F′(xe−y2) e−y2∂u∂y(x,y)=∂∂y[ey F(xe−y2)]=ey F(xe−y2)+ey ∂∂y[F(xe−y2)]=ey F(xe−y2)+ey F′(xe−y2) ∂∂y(xe−y2)=ey F(xe−y2)+ey F′(xe−y2) (−2xy)e−y2\begin{aligned} \frac{\partial u}{\partial x}(x,y)&= \frac{\partial }{\partial x}\Big[e^y\,F\big(xe^{-y^2}\big)\Big] =e^y\,\frac{\partial }{\partial x}\Big[F\big(xe^{-y^2}\big)\Big] \\ &=e^y\,F'\big(xe^{-y^2}\big)\frac{\partial }{\partial x}\Big(xe^{-y^2}\Big) \\ &= e^y\, F'\big(xe^{-y^2}\big)\ e^{-y^2} \\ \frac{\partial u}{\partial y}(x,y)&= \frac{\partial }{\partial y}\Big[e^y\,F\big(xe^{-y^2}\big)\Big]\\ &=e^y\,F\big(xe^{-y^2}\big) +e^y\,\frac{\partial }{\partial y}\Big[F\big(xe^{-y^2}\big)\Big] \\ &= e^y\, F\big(xe^{-y^2}\big) + e^y\,F'\big(xe^{-y^2}\big)\ \frac{\partial }{\partial y}\Big(xe^{-y^2}\Big) \\ &= e^y\, F\big(xe^{-y^2}\big) + e^y\, F'\big(xe^{-y^2}\big)\ (-2xy)e^{-y^2} \end{aligned}

(a) 특히 F(z)=ln⁡(z)F(z)=\ln(z) 일 때 F′(z)=1zF'(z)=\frac{1}{z} 이므로

∂u∂x(x,y)=ey 1xe−y2 e−y2=eyx∂u∂y(x,y)=ey ln⁡(xe−y2)+ey 1xe−y2 (−2xy)e−y2=ey ln⁡(xe−y2)−2yey=ey ln⁡(x)−y2 ey−2yey\begin{aligned} \frac{\partial u}{\partial x}(x,y)&= e^y\, \frac{1}{xe^{-y^2}}\ e^{-y^2} = \frac{e^y}{x}\\ \frac{\partial u}{\partial y}(x,y)&= e^y\, \ln\big(xe^{-y^2}\big) + e^y\, \frac{1}{xe^{-y^2}}\ (-2xy)e^{-y^2} = e^y\, \ln\big(xe^{-y^2}\big) -2y e^y \\ &= e^y\,\ln (x) -y^2\,e^y -2y e^y \end{aligned}

(b) 일반적으로

2xy∂u∂x+∂u∂y=2xy ey F′(xe−y2) e−y2+ey F(xe−y2)+ey F′(xe−y2) (−2xy)e−y2=ey F(xe−y2)=u\begin{aligned} 2xy\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} &=2xy\ e^y\, F'\big(xe^{-y^2}\big)\ e^{-y^2} +e^y\, F\big(xe^{-y^2}\big) + e^y\, F'\big(xe^{-y^2}\big)\ (-2xy)e^{-y^2} \\ &=e^y\, F\big(xe^{-y^2}\big) \\ &=u \end{aligned}

원본 교재 풀이(영어)

For any (differentiable) function FF, we have, by the chain and product rules,

∂u∂x(x,y)=∂∂x[ey F(xe−y2)]=ey ∂∂x[F(xe−y2)]=ey F′(xe−y2)∂∂x(xe−y2)=ey F′(xe−y2) e−y2∂u∂y(x,y)=∂∂y[ey F(xe−y2)]=ey F(xe−y2)+ey ∂∂y[F(xe−y2)]=ey F(xe−y2)+ey F′(xe−y2) ∂∂y(xe−y2)=ey F(xe−y2)+ey F′(xe−y2) (−2xy)e−y2\begin{aligned} \frac{\partial u}{\partial x}(x,y)&= \frac{\partial }{\partial x}\Big[e^y\,F\big(xe^{-y^2}\big)\Big] =e^y\,\frac{\partial }{\partial x}\Big[F\big(xe^{-y^2}\big)\Big] \\ &=e^y\,F'\big(xe^{-y^2}\big)\frac{\partial }{\partial x}\Big(xe^{-y^2}\Big) \\ &= e^y\, F'\big(xe^{-y^2}\big)\ e^{-y^2} \\ \frac{\partial u}{\partial y}(x,y)&= \frac{\partial }{\partial y}\Big[e^y\,F\big(xe^{-y^2}\big)\Big]\\ &=e^y\,F\big(xe^{-y^2}\big) +e^y\,\frac{\partial }{\partial y}\Big[F\big(xe^{-y^2}\big)\Big] \\ &= e^y\, F\big(xe^{-y^2}\big) + e^y\,F'\big(xe^{-y^2}\big)\ \frac{\partial }{\partial y}\Big(xe^{-y^2}\Big) \\ &= e^y\, F\big(xe^{-y^2}\big) + e^y\, F'\big(xe^{-y^2}\big)\ (-2xy)e^{-y^2} \end{aligned}

(a) In particular, when F(z)=ln⁡(z)F(z)=\ln(z), F′(z)=1zF'(z)=\frac{1}{z} and

∂u∂x(x,y)=ey 1xe−y2 e−y2=eyx∂u∂y(x,y)=ey ln⁡(xe−y2)+ey 1xe−y2 (−2xy)e−y2=ey ln⁡(xe−y2)−2yey=ey ln⁡(x)−y2 ey−2yey\begin{aligned} \frac{\partial u}{\partial x}(x,y)&= e^y\, \frac{1}{xe^{-y^2}}\ e^{-y^2} = \frac{e^y}{x}\\ \frac{\partial u}{\partial y}(x,y)&= e^y\, \ln\big(xe^{-y^2}\big) + e^y\, \frac{1}{xe^{-y^2}}\ (-2xy)e^{-y^2} = e^y\, \ln\big(xe^{-y^2}\big) -2y e^y \\ &= e^y\,\ln (x) -y^2\,e^y -2y e^y \end{aligned}

(b) In general

2xy∂u∂x+∂u∂y=2xy ey F′(xe−y2) e−y2+ey F(xe−y2)+ey F′(xe−y2) (−2xy)e−y2=ey F(xe−y2)=u\begin{aligned} 2xy\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} &=2xy\ e^y\, F'\big(xe^{-y^2}\big)\ e^{-y^2} +e^y\, F\big(xe^{-y^2}\big) + e^y\, F'\big(xe^{-y^2}\big)\ (-2xy)e^{-y^2} \\ &=e^y\, F\big(xe^{-y^2}\big) \\ &=u \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 11번
# 다변수 연쇄법칙증명·논술보통자동 분류
8.CLP-3 · 2.4절 · 13번대학 수준우리말 번역

f(x)f(x) 와 g(x)g(x) 가 xx 에 대한 두 함수이고 f′′(7)=−2f''(7) = -2 이고 g′′(−4)=−1g''(-4) = -1 을 만족시킨다고 하자. z=h(s,t)=f(2s+3t)+g(s−6t)z = h(s,t) = f(2s + 3t) + g(s - 6t) 가 ss 와 tt 에 대한 함수일 때, s=2s = 2 이고 t=1t = 1 일 때 ∂2z∂t2\frac{\partial^2 z}{\partial t^2} 의 값을 구하세요.

영어 원문 보기

Let f(x)f(x) and g(x)g(x) be two functions of xx satisfying f′′(7)=−2f''(7) = -2 and g′′(−4)=−1g''(-4) = -1. If z=h(s,t)=f(2s+3t)+g(s−6t)z = h(s,t) = f(2s + 3t) + g(s - 6t) is a function of ss and tt, find the value of ∂2z∂t2\frac{\partial^2 z}{\partial t^2} when s=2s = 2 and t=1t = 1.

답 쓰고 채점
정답·풀이 보기

연쇄 법칙에 의해

∂h∂t(s,t)=∂∂t[f(2s+3t)]+∂∂t[g(s−6t)]=f′(2s+3t)∂∂t(2s+3t)+g(s−6t)∂∂t(s−6t)=3f′(2s+3t)−6g′(s−6t)∂2h∂t2(s,t)=3∂∂t[f′(2s+3t)]−6∂∂t[g′(s−6t)]=3 f′′(2s+3t)∂∂t(2s+3t)−6 g′′(s−6t)∂∂t(s−6t)=9f′′(2s+3t)+36g′′(s−6t)\begin{aligned} \frac{\partial h}{\partial t}(s,t) &=\frac{\partial }{\partial t}\big[f(2s + 3t)\big] + \frac{\partial }{\partial t}\big[g(s - 6t)\big] \\ &=f'(2s + 3t)\frac{\partial }{\partial t}(2s+3t) + g(s - 6t)\frac{\partial }{\partial t}(s-6t) \\ &= 3f'(2s + 3t) - 6g'(s - 6t) \\ \frac{\partial^2 h}{\partial t^2}(s,t) &= 3\frac{\partial }{\partial t}\big[f'(2s + 3t)\big] - 6\frac{\partial }{\partial t}\big[ g'(s - 6t) \big] \\ &= 3\,f''(2s + 3t)\frac{\partial }{\partial t}(2s+3t) - 6\,g''(s - 6t) \frac{\partial }{\partial t}(s-6t) \\ &= 9f''(2s+3t) + 36g''(s-6t) \end{aligned}

특히

∂2h∂t2(2,1)=9f′′(7)+36g′′(−4)=9(−2)+36(−1)=−54\begin{aligned} \frac{\partial^2 h}{\partial t^2}(2,1) =9f''(7) +36g''(-4) =9(-2) +36(-1) =-54 \end{aligned}

이다.

원본 교재 풀이(영어)

By the chain rule,

∂h∂t(s,t)=∂∂t[f(2s+3t)]+∂∂t[g(s−6t)]=f′(2s+3t)∂∂t(2s+3t)+g(s−6t)∂∂t(s−6t)=3f′(2s+3t)−6g′(s−6t)∂2h∂t2(s,t)=3∂∂t[f′(2s+3t)]−6∂∂t[g′(s−6t)]=3 f′′(2s+3t)∂∂t(2s+3t)−6 g′′(s−6t)∂∂t(s−6t)=9f′′(2s+3t)+36g′′(s−6t)\begin{aligned} \frac{\partial h}{\partial t}(s,t) &=\frac{\partial }{\partial t}\big[f(2s + 3t)\big] + \frac{\partial }{\partial t}\big[g(s - 6t)\big] \\ &=f'(2s + 3t)\frac{\partial }{\partial t}(2s+3t) + g(s - 6t)\frac{\partial }{\partial t}(s-6t) \\ &= 3f'(2s + 3t) - 6g'(s - 6t) \\ \frac{\partial^2 h}{\partial t^2}(s,t) &= 3\frac{\partial }{\partial t}\big[f'(2s + 3t)\big] - 6\frac{\partial }{\partial t}\big[ g'(s - 6t) \big] \\ &= 3\,f''(2s + 3t)\frac{\partial }{\partial t}(2s+3t) - 6\,g''(s - 6t) \frac{\partial }{\partial t}(s-6t) \\ &= 9f''(2s+3t) + 36g''(s-6t) \end{aligned}

In particular

∂2h∂t2(2,1)=9f′′(7)+36g′′(−4)=9(−2)+36(−1)=−54\begin{aligned} \frac{\partial^2 h}{\partial t^2}(2,1) =9f''(7) +36g''(-4) =9(-2) +36(-1) =-54 \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 12번
# 다변수 연쇄법칙계산보통자동 분류
9.CLP-3 · 2.4절 · 14번대학 수준우리말 번역

w=f(xz,yz)w = f (xz, yz) 라고 하자. 여기서 ff 는 미분가능한 함수이다. 다음을 보여라.

x∂w∂x+y∂w∂y=z∂w∂zx\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} = z\frac{\partial w}{\partial z}

영어 원문 보기

Suppose that w=f(xz,yz)w = f (xz, yz), where ff is a differentiable function. Show that

x∂w∂x+y∂w∂y=z∂w∂zx\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} = z\frac{\partial w}{\partial z}

정답·풀이 보기

먼저 w(x,y,z)w(x,y,z) 의 1계 편도함수를 구한다. u(x,y,z)=xzu(x,y,z)=xz, v(x,y,z)=yzv(x,y,z)=yz 라 하면 w(x,y,z)=f(u(x,y,z),v(x,y,z))w(x,y,z) = f\big(u(x,y,z), v(x,y,z)\big) 이다. 연쇄 법칙에 의해

∂w∂x(x,y,z)=∂∂x[f(u(x,y,z),v(x,y,z))]=∂f∂u(u(x,y,z),v(x,y,z))∂u∂x(x,y,z)+∂f∂v(u(x,y,z),v(x,y,z))∂v∂x(x,y,z)=z∂f∂u(xz,yz)∂w∂y(x,y,z)=∂∂y[f(u(x,y,z),v(x,y,z))]=∂f∂u(u(x,y,z),v(x,y,z))∂u∂y(x,y,z)+∂f∂v(u(x,y,z),v(x,y,z))∂v∂y(x,y,z)=z∂f∂v(xz,yz)∂w∂z(x,y,z)=∂∂z[f(u(x,y,z),v(x,y,z))]=∂f∂u(u(x,y,z),v(x,y,z))∂u∂z(x,y,z)+∂f∂v(u(x,y,z),v(x,y,z))∂v∂z(x,y,z)=x∂f∂u(xz,yz)+y∂f∂v(xz,yz)\begin{aligned} \frac{\partial w}{\partial x}(x,y,z) &= \frac{\partial }{\partial x}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial x}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial x}(x,y,z) \\ &=z\frac{\partial f}{\partial u}(xz, yz) \\ \frac{\partial w}{\partial y}(x,y,z) &= \frac{\partial }{\partial y}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial y}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial y}(x,y,z) \\ &=z\frac{\partial f}{\partial v}(xz, yz) \\ \frac{\partial w}{\partial z}(x,y,z) &= \frac{\partial }{\partial z}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial z}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial z}(x,y,z) \\ &=x\frac{\partial f}{\partial u}(xz, yz) + y\frac{\partial f}{\partial v}(xz, yz) \end{aligned}

따라서

x∂w∂x+y∂w∂y=xz∂f∂u(xz,yz)+yz∂f∂v(xz,yz)=z[x∂f∂u(xz,yz)+y∂f∂v(xz,yz)]=z∂w∂z\begin{aligned} x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} =xz\frac{\partial f}{\partial u}(xz, yz) + yz\frac{\partial f}{\partial v}(xz, yz) =z\left[x\frac{\partial f}{\partial u}(xz, yz) + y\frac{\partial f}{\partial v}(xz, yz)\right] =z\frac{\partial w}{\partial z} \end{aligned}

이므로 주어진 등식이 성립한다.

원본 교재 풀이(영어)

We'll first compute the first order partial derivatives of w(x,y,z)w(x,y,z). Write u(x,y,z)=xzu(x,y,z)=xz and v(x,y,z)=yzv(x,y,z)=yz so that w(x,y,z)=f(u(x,y,z),v(x,y,z))w(x,y,z) = f\big(u(x,y,z), v(x,y,z)\big). By the chain rule,

∂w∂x(x,y,z)=∂∂x[f(u(x,y,z),v(x,y,z))]=∂f∂u(u(x,y,z),v(x,y,z))∂u∂x(x,y,z)+∂f∂v(u(x,y,z),v(x,y,z))∂v∂x(x,y,z)=z∂f∂u(xz,yz)∂w∂y(x,y,z)=∂∂y[f(u(x,y,z),v(x,y,z))]=∂f∂u(u(x,y,z),v(x,y,z))∂u∂y(x,y,z)+∂f∂v(u(x,y,z),v(x,y,z))∂v∂y(x,y,z)=z∂f∂v(xz,yz)∂w∂z(x,y,z)=∂∂z[f(u(x,y,z),v(x,y,z))]=∂f∂u(u(x,y,z),v(x,y,z))∂u∂z(x,y,z)+∂f∂v(u(x,y,z),v(x,y,z))∂v∂z(x,y,z)=x∂f∂u(xz,yz)+y∂f∂v(xz,yz)\begin{aligned} \frac{\partial w}{\partial x}(x,y,z) &= \frac{\partial }{\partial x}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial x}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial x}(x,y,z) \\ &=z\frac{\partial f}{\partial u}(xz, yz) \\ \frac{\partial w}{\partial y}(x,y,z) &= \frac{\partial }{\partial y}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial y}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial y}(x,y,z) \\ &=z\frac{\partial f}{\partial v}(xz, yz) \\ \frac{\partial w}{\partial z}(x,y,z) &= \frac{\partial }{\partial z}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial z}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial z}(x,y,z) \\ &=x\frac{\partial f}{\partial u}(xz, yz) + y\frac{\partial f}{\partial v}(xz, yz) \end{aligned}

So

x∂w∂x+y∂w∂y=xz∂f∂u(xz,yz)+yz∂f∂v(xz,yz)=z[x∂f∂u(xz,yz)+y∂f∂v(xz,yz)]=z∂w∂z\begin{aligned} x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} =xz\frac{\partial f}{\partial u}(xz, yz) + yz\frac{\partial f}{\partial v}(xz, yz) =z\left[x\frac{\partial f}{\partial u}(xz, yz) + y\frac{\partial f}{\partial v}(xz, yz)\right] =z\frac{\partial w}{\partial z} \end{aligned}

as desired.

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 22번
# 다변수 연쇄법칙증명·논술보통자동 분류
10.CLP-3 · 2.4절 · 15번대학 수준우리말 번역

z=f(x,y)z = f (x, y) 가 연속인 이계 편도함수를 갖고, x=rcos⁡tx = r \cos t, y=rsin⁡ty = r \sin t 라 하자. 다음 편도함수를 rr, tt, 그리고 ff 의 편도함수로 나타내세요.

(a) ∂z∂t\frac{\partial z}{\partial t} (b) ∂2z∂t2\frac{\partial^2 z}{\partial t^2}

영어 원문 보기

Suppose z=f(x,y)z = f (x, y) has continuous second order partial derivatives, and x=rcos⁡tx = r \cos t, y=rsin⁡ty = r \sin t. Express the following partial derivatives in terms rr, tt, and partial derivatives of ff.

(a) ∂z∂t\frac{\partial z}{\partial t} (b) ∂2z∂t2\frac{\partial^2 z}{\partial t^2}

정답·풀이 보기

정의에 의해 z(r,t)=f(rcos⁡t , rsin⁡t)z(r,t) = f(r\cos t\,,\,r\sin t) 이다.

(a) 연쇄 법칙에 의해

∂z∂t(r,t)=∂∂t[f(rcos⁡t , rsin⁡t)]=∂f∂x(rcos⁡t , rsin⁡t)∂∂t(rcos⁡t)+∂f∂y(rcos⁡t , rsin⁡t)∂∂t(rsin⁡t)=−rsin⁡t ∂f∂x(rcos⁡t , rsin⁡t)+rcos⁡t ∂f∂y(rcos⁡t , rsin⁡t)\begin{aligned} \frac{\partial z}{\partial t}(r,t) &= \frac{\partial }{\partial t}\Big[f(r\cos t\,,\,r\sin t)\Big] \\ &= \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\frac{\partial }{\partial t}(r\cos t) +\frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\frac{\partial }{\partial t}(r\sin t) \\ &= -r\sin t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) +r\cos t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) \end{aligned}

(b) 선형성, 곱의 미분법, 연쇄 법칙에 의해

∂2z∂t2(r,t)=−∂∂t[rsin⁡t ∂f∂x(rcos⁡t , rsin⁡t)]+∂∂t[rcos⁡t ∂f∂y(rcos⁡t , rsin⁡t)]=−rcos⁡t ∂f∂x(rcos⁡t , rsin⁡t)−rsin⁡t ∂∂t[∂f∂x(rcos⁡t , rsin⁡t)]−rsin⁡t ∂f∂y(rcos⁡t , rsin⁡t)+rcos⁡t ∂∂t[∂f∂y(rcos⁡t , rsin⁡t)]=−rcos⁡t ∂f∂x(rcos⁡t , rsin⁡t)+r2sin⁡2t ∂2f∂x2(rcos⁡t , rsin⁡t)−r2sin⁡tcos⁡t ∂2 f∂y∂x(rcos⁡t , rsin⁡t)=−rsin⁡t ∂f∂y(rcos⁡t , rsin⁡t)−r2sin⁡tcos⁡t ∂2 f∂x∂y(rcos⁡t , rsin⁡t)+r2cos⁡2t ∂2f∂y2(rcos⁡t , rsin⁡t)=−rcos⁡t ∂f∂x−rsin⁡t ∂f∂y+r2sin⁡2t ∂2f∂x2−2r2sin⁡tcos⁡t ∂2 f∂x∂y+r2cos⁡2t ∂2f∂y2\begin{aligned} \frac{\partial^2 z}{\partial t^2}(r,t) &= -\frac{\partial }{\partial t}\left[r\sin t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\right] +\frac{\partial }{\partial t}\left[r\cos t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\right] \\ &= -r\cos t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) -r\sin t\ \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\right] \\ & -r\sin t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) +r\cos t\ \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\right] \\ &= -r\cos t \ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) \\ & +r^2\sin^2 t\ \frac{\partial^2 f}{\partial x^2}(r\cos t\,,\,r\sin t) -r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial y\partial x}(r\cos t\,,\,r\sin t) \\ &\phantom{=}-r\sin t \ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) \\ & -r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial x\partial y}(r\cos t\,,\,r\sin t) +r^2\cos^2 t\ \frac{\partial^2 f}{\partial y^2}(r\cos t\,,\,r\sin t) \\ &= -r\cos t \ \frac{\partial f}{\partial x} -r\sin t \ \frac{\partial f}{\partial y} \\ & +r^2\sin^2 t\ \frac{\partial^2 f}{\partial x^2} -2r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial x\partial y} +r^2\cos^2 t\ \frac{\partial^2 f}{\partial y^2} \end{aligned}

여기서 ff 의 모든 편도함수는 (rcos⁡t , rsin⁡t)(r\cos t\,,\,r\sin t) 에서의 값이다.

원본 교재 풀이(영어)

By definition z(r,t)=f(rcos⁡t , rsin⁡t)z(r,t) = f(r\cos t\,,\,r\sin t).

(a) By the chain rule

∂z∂t(r,t)=∂∂t[f(rcos⁡t , rsin⁡t)]=∂f∂x(rcos⁡t , rsin⁡t)∂∂t(rcos⁡t)+∂f∂y(rcos⁡t , rsin⁡t)∂∂t(rsin⁡t)=−rsin⁡t ∂f∂x(rcos⁡t , rsin⁡t)+rcos⁡t ∂f∂y(rcos⁡t , rsin⁡t)\begin{aligned} \frac{\partial z}{\partial t}(r,t) &= \frac{\partial }{\partial t}\Big[f(r\cos t\,,\,r\sin t)\Big] \\ &= \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\frac{\partial }{\partial t}(r\cos t) +\frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\frac{\partial }{\partial t}(r\sin t) \\ &= -r\sin t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) +r\cos t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) \end{aligned}

(b) By linearity, the product rule and the chain rule

∂2z∂t2(r,t)=−∂∂t[rsin⁡t ∂f∂x(rcos⁡t , rsin⁡t)]+∂∂t[rcos⁡t ∂f∂y(rcos⁡t , rsin⁡t)]=−rcos⁡t ∂f∂x(rcos⁡t , rsin⁡t)−rsin⁡t ∂∂t[∂f∂x(rcos⁡t , rsin⁡t)]−rsin⁡t ∂f∂y(rcos⁡t , rsin⁡t)+rcos⁡t ∂∂t[∂f∂y(rcos⁡t , rsin⁡t)]=−rcos⁡t ∂f∂x(rcos⁡t , rsin⁡t)+r2sin⁡2t ∂2f∂x2(rcos⁡t , rsin⁡t)−r2sin⁡tcos⁡t ∂2 f∂y∂x(rcos⁡t , rsin⁡t)=−rsin⁡t ∂f∂y(rcos⁡t , rsin⁡t)−r2sin⁡tcos⁡t ∂2 f∂x∂y(rcos⁡t , rsin⁡t)+r2cos⁡2t ∂2f∂y2(rcos⁡t , rsin⁡t)=−rcos⁡t ∂f∂x−rsin⁡t ∂f∂y+r2sin⁡2t ∂2f∂x2−2r2sin⁡tcos⁡t ∂2 f∂x∂y+r2cos⁡2t ∂2f∂y2\begin{aligned} \frac{\partial^2 z}{\partial t^2}(r,t) &= -\frac{\partial }{\partial t}\left[r\sin t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\right] +\frac{\partial }{\partial t}\left[r\cos t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\right] \\ &= -r\cos t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) -r\sin t\ \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\right] \\ & -r\sin t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) +r\cos t\ \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\right] \\ &= -r\cos t \ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) \\ & +r^2\sin^2 t\ \frac{\partial^2 f}{\partial x^2}(r\cos t\,,\,r\sin t) -r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial y\partial x}(r\cos t\,,\,r\sin t) \\ &\phantom{=}-r\sin t \ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) \\ & -r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial x\partial y}(r\cos t\,,\,r\sin t) +r^2\cos^2 t\ \frac{\partial^2 f}{\partial y^2}(r\cos t\,,\,r\sin t) \\ &= -r\cos t \ \frac{\partial f}{\partial x} -r\sin t \ \frac{\partial f}{\partial y} \\ & +r^2\sin^2 t\ \frac{\partial^2 f}{\partial x^2} -2r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial x\partial y} +r^2\cos^2 t\ \frac{\partial^2 f}{\partial y^2} \end{aligned}

with all of the partial derivatives of ff evaluated at (rcos⁡t , rsin⁡t)(r\cos t\,,\,r\sin t).

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 25번
# 다변수 연쇄법칙활용보통자동 분류
11.CLP-3 · 2.4절 · 16번대학 수준우리말 번역

z=f(x,y)z = f(x, y) 라 하자. 여기서 f(x,y)f(x, y) 는 연속인 이계 편도함수를 가지며,

fx(2,1)=5,fy(2,1)=−2,fxx(2,1)=2,fxy(2,1)=1,fyy(2,1)=−4f_x (2, 1) = 5, \qquad f_y(2, 1) =-2, \qquad f_{xx}(2, 1) = 2,\qquad f_{xy}(2, 1) = 1, \qquad f_{yy}(2, 1) = -4

이다.

x(t)=2t2x(t)=2t^2, y(t)=t3y(t)=t^3 이고 t=1t=1 일 때, d2dt2z(x(t),y(t)) \frac{\mathrm{d^2}}{\mathrm{d}{t}^2} z\big(x(t),y(t)\big) 의 값을 구하세요.

영어 원문 보기

Let z=f(x,y)z = f(x, y), where f(x,y)f(x, y) has continuous second-order partial derivatives, and

fx(2,1)=5,fy(2,1)=−2,fxx(2,1)=2,fxy(2,1)=1,fyy(2,1)=−4f_x (2, 1) = 5, \qquad f_y(2, 1) =-2, \qquad f_{xx}(2, 1) = 2,\qquad f_{xy}(2, 1) = 1, \qquad f_{yy}(2, 1) = -4

Find d2dt2z(x(t),y(t)) \frac{\mathrm{d^2}}{\mathrm{d}{t}^2} z\big(x(t),y(t)\big) when x(t)=2t2x(t)=2t^2, y(t)=t3y(t)=t^3 and t=1t=1.

답 쓰고 채점
정답·풀이 보기

x(t)=2t2x(t)=2t^2, y(t)=t3y(t)=t^3 에 대해 w(t)=z(x(t),y(t))=f(x(t),y(t))w(t) = z\big(x(t),y(t)\big) = f\big(x(t),y(t)\big) 라 하자. d2wdt2(1)\frac{\mathrm{d^2}w}{\mathrm{d}{t}^2}(1) 을 계산하면 된다. 연쇄 법칙에 의해

dwdt(t)=ddtf(x(t),y(t))=fx(x(t) , y(t)) dxdt(t)+fy(x(t) , y(t)) dydt(t)=4t fx(x(t) , y(t))+3t2 fy(x(t) , y(t))\begin{aligned} \frac{\mathrm{d}w}{\mathrm{d}t}(t)&= \frac{\mathrm{d}}{\mathrm{d}t} f\big(x(t),y(t)\big) \\ &= f_x(x(t)\,,\,y(t))\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) +f_y(x(t)\,,\,y(t))\,\frac{\mathrm{d}y}{\mathrm{d}t}(t) \\ &= 4t\,f_x(x(t)\,,\,y(t)) +3t^2\,f_y(x(t)\,,\,y(t)) \end{aligned}

이다. 선형성, 곱의 미분법, 연쇄 법칙에 의해

d2dt2f(x(t),y(t))=ddt[4t fx(x(t),y(t))]+ddt[3t2 fy(x(t),y(t))]=4 fx(x(t),y(t))+4tddt[fx(x(t),y(t))]+6t fy(x(t),y(t))+3t2ddt[fy(x(t),y(t))]=4 fx(2t2 , t3)+4t[fxx(x(t),y(t)) dxdt(t)+fxy(x(t),y(t)) dydt(t)]=+6t fy(2t2 , t3)+3t2[fyx(x(t),y(t)) dxdt(t)+fyy(x(t),y(t)) dydt(t)]=4 fx(2t2 , t3)+16t2 fxx(2t2 , t3)+12t3 fxy(2t2 , t3)=+6t fy(2t2 , t3)+12t3 fyx(2t2 , t3)+9t4 fyy(2t2 , t3)\begin{aligned} \frac{\mathrm{d^2}}{\mathrm{d}{t}^2} f\big(x(t),y(t)\big) &= \frac{\mathrm{d}}{\mathrm{d}t}\left[4t\,f_x\big(x(t),y(t)\big) \right] +\frac{\mathrm{d}}{\mathrm{d}t}\left[3t^2\,f_y\big(x(t),y(t)\big) \right] \\ &= 4\,f_x\big(x(t),y(t)\big) + 4t \frac{\mathrm{d}}{\mathrm{d}t}\left[f_x\big(x(t),y(t)\big) \right] \\& +6t\,f_y\big(x(t),y(t)\big) + 3t^2\frac{\mathrm{d}}{\mathrm{d}t}\left[f_y\big(x(t),y(t)\big) \right] \\ &= 4\,f_x(2t^2\,,\,t^3) + 4t\Big[f_{xx}\big(x(t),y(t)\big)\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) + f_{xy}\big(x(t),y(t)\big)\,\frac{\mathrm{d}y}{\mathrm{d}t}(t)\Big] \\ &\phantom{=} + 6t\,f_y(2t^2\,,\,t^3) + 3t^2\Big[f_{yx}\big(x(t),y(t)\big)\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) + f_{yy}\big(x(t),y(t)\big)\,\frac{\mathrm{d}y}{\mathrm{d}t}(t)\Big] \\ &= 4\,f_x(2t^2\,,\,t^3) + 16t^2\,f_{xx}(2t^2\,,\,t^3) + 12t^3\,f_{xy}(2t^2\,,\,t^3) \\ &\phantom{=} + 6t\,f_y(2t^2\,,\,t^3) + 12t^3\,f_{yx}(2t^2\,,\,t^3) + 9t^4\,f_{yy}(2t^2\,,\,t^3) \end{aligned}

이다. 특히 t=1t=1 일 때, fxy(2,1)=fyx(2,1)f_{xy}(2, 1)=f_{yx}(2, 1) 이므로

d2dt2f(x(t),y(t))∣t=1=4 (5)+16 (2)+12 (1)=+6 (−2)+12 (1)+9 (−4)=28\begin{aligned} \left.\frac{\mathrm{d^2}}{\mathrm{d}{t}^2} f\big(x(t),y(t)\big) \right|_{t=1} &= 4\,(5) + 16\,(2) + 12\,(1) \\ &\phantom{=} + 6\,(-2) + 12\,(1) + 9\,(-4) \\ &=28 \end{aligned}

이다.

원본 교재 풀이(영어)

Write w(t)=z(x(t),y(t))=f(x(t),y(t))w(t) = z\big(x(t),y(t)\big) = f\big(x(t),y(t)\big) with x(t)=2t2x(t)=2t^2, y(t)=t3y(t)=t^3. We are to compute d2wdt2(1)\frac{\mathrm{d^2}w}{\mathrm{d}{t}^2}(1). By the chain rule

dwdt(t)=ddtf(x(t),y(t))=fx(x(t) , y(t)) dxdt(t)+fy(x(t) , y(t)) dydt(t)=4t fx(x(t) , y(t))+3t2 fy(x(t) , y(t))\begin{aligned} \frac{\mathrm{d}w}{\mathrm{d}t}(t)&= \frac{\mathrm{d}}{\mathrm{d}t} f\big(x(t),y(t)\big) \\ &= f_x(x(t)\,,\,y(t))\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) +f_y(x(t)\,,\,y(t))\,\frac{\mathrm{d}y}{\mathrm{d}t}(t) \\ &= 4t\,f_x(x(t)\,,\,y(t)) +3t^2\,f_y(x(t)\,,\,y(t)) \end{aligned}

By linearity, the product rule, and the chain rule,

d2dt2f(x(t),y(t))=ddt[4t fx(x(t),y(t))]+ddt[3t2 fy(x(t),y(t))]=4 fx(x(t),y(t))+4tddt[fx(x(t),y(t))]+6t fy(x(t),y(t))+3t2ddt[fy(x(t),y(t))]=4 fx(2t2 , t3)+4t[fxx(x(t),y(t)) dxdt(t)+fxy(x(t),y(t)) dydt(t)]=+6t fy(2t2 , t3)+3t2[fyx(x(t),y(t)) dxdt(t)+fyy(x(t),y(t)) dydt(t)]=4 fx(2t2 , t3)+16t2 fxx(2t2 , t3)+12t3 fxy(2t2 , t3)=+6t fy(2t2 , t3)+12t3 fyx(2t2 , t3)+9t4 fyy(2t2 , t3)\begin{aligned} \frac{\mathrm{d^2}}{\mathrm{d}{t}^2} f\big(x(t),y(t)\big) &= \frac{\mathrm{d}}{\mathrm{d}t}\left[4t\,f_x\big(x(t),y(t)\big) \right] +\frac{\mathrm{d}}{\mathrm{d}t}\left[3t^2\,f_y\big(x(t),y(t)\big) \right] \\ &= 4\,f_x\big(x(t),y(t)\big) + 4t \frac{\mathrm{d}}{\mathrm{d}t}\left[f_x\big(x(t),y(t)\big) \right] \\& +6t\,f_y\big(x(t),y(t)\big) + 3t^2\frac{\mathrm{d}}{\mathrm{d}t}\left[f_y\big(x(t),y(t)\big) \right] \\ &= 4\,f_x(2t^2\,,\,t^3) + 4t\Big[f_{xx}\big(x(t),y(t)\big)\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) + f_{xy}\big(x(t),y(t)\big)\,\frac{\mathrm{d}y}{\mathrm{d}t}(t)\Big] \\ &\phantom{=} + 6t\,f_y(2t^2\,,\,t^3) + 3t^2\Big[f_{yx}\big(x(t),y(t)\big)\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) + f_{yy}\big(x(t),y(t)\big)\,\frac{\mathrm{d}y}{\mathrm{d}t}(t)\Big] \\ &= 4\,f_x(2t^2\,,\,t^3) + 16t^2\,f_{xx}(2t^2\,,\,t^3) + 12t^3\,f_{xy}(2t^2\,,\,t^3) \\ &\phantom{=} + 6t\,f_y(2t^2\,,\,t^3) + 12t^3\,f_{yx}(2t^2\,,\,t^3) + 9t^4\,f_{yy}(2t^2\,,\,t^3) \end{aligned}

In particular, when t=1t=1, and since fxy(2,1)=fyx(2,1)f_{xy}(2, 1)=f_{yx}(2, 1),

d2dt2f(x(t),y(t))∣t=1=4 (5)+16 (2)+12 (1)=+6 (−2)+12 (1)+9 (−4)=28\begin{aligned} \left.\frac{\mathrm{d^2}}{\mathrm{d}{t}^2} f\big(x(t),y(t)\big) \right|_{t=1} &= 4\,(5) + 16\,(2) + 12\,(1) \\ &\phantom{=} + 6\,(-2) + 12\,(1) + 9\,(-4) \\ &=28 \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 24번
# 다변수 연쇄법칙활용보통자동 분류
12.CLP-3 · 2.4절 · 17번대학 수준우리말 번역

함수 F(x,y,z)F(x,y,z) 가 등식 ∂F∂z=∂2F∂x2+∂2F∂y2\frac{\partial F}{\partial z} = \frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2} 을 만족시키고, 혼합 편도함수 ∂2F∂x∂y\frac{\partial^2 F}{\partial x \partial y} 와 ∂2F∂y∂x\frac{\partial^2 F}{\partial y \partial x} 가 서로 같다고 하자. AA 를 어떤 상수라 하고 G(γ,s,t)=F(γ+s,γ−s,At)G(\gamma, s, t) = F (\gamma + s, \gamma-s, At) 라 하자. ∂G∂t=∂2G∂γ2+∂2G∂s2\frac{\partial G}{\partial t} = \frac{\partial^2 G}{\partial \gamma^2} + \frac{\partial^2 G}{\partial s^2} 이 되도록 하는 AA 의 값을 구하세요.

영어 원문 보기

Assume that the function F(x,y,z)F(x,y,z) satisfies the equation ∂F∂z=∂2F∂x2+∂2F∂y2\frac{\partial F}{\partial z} = \frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2} and the mixed partial derivatives ∂2F∂x∂y\frac{\partial^2 F}{\partial x \partial y} and ∂2F∂y∂x\frac{\partial^2 F}{\partial y \partial x} are equal. Let AA be some constant and let G(γ,s,t)=F(γ+s,γ−s,At)G(\gamma, s, t) = F (\gamma + s, \gamma-s, At). Find the value of AA such that ∂G∂t=∂2G∂γ2+∂2G∂s2\frac{\partial G}{\partial t} = \frac{\partial^2 G}{\partial \gamma^2} + \frac{\partial^2 G}{\partial s^2}.

답 쓰고 채점
정답·풀이 보기

연쇄 법칙에 의해

∂G∂t(γ,s,t)=∂∂t[F(γ+s,γ−s,At)]=∂F∂x(γ+s,γ−s,At) ∂∂t(γ+s)+∂F∂y(γ+s,γ−s,At) ∂∂t(γ−s)+∂F∂z(γ+s,γ−s,At) ∂∂t(At)=A ∂F∂z(γ+s,γ−s,At)\begin{aligned} \frac{\partial G}{\partial t}(\gamma ,s,t) &= \frac{\partial }{\partial t} \Big[F (\gamma + s, \gamma-s, At)\Big] \\ &=\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial t}(\gamma + s) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial t}(\gamma - s) \\& +\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At)\,\frac{\partial }{\partial t}(At) \\ &= A\,\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At) \end{aligned}

이고

∂G∂γ(γ,s,t)=∂∂γ[F(γ+s,γ−s,At)]=∂F∂x(γ+s,γ−s,At) ∂∂γ(γ+s)+∂F∂y(γ+s,γ−s,At) ∂∂γ(γ−s)+∂F∂z(γ+s,γ−s,At) ∂∂γ(At)=∂F∂x(γ+s,γ−s,At)+∂F∂y(γ+s,γ−s,At)(E1)\begin{aligned} \frac{\partial G}{\partial \gamma }(\gamma ,s,t) &= \frac{\partial }{\partial \gamma } \Big[F (\gamma + s, \gamma-s, At)\Big] \\ &=\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial \gamma }(\gamma + s) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial \gamma }(\gamma - s) \\& +\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At)\,\frac{\partial }{\partial \gamma }(At) \\ &= \frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \tag{E1} \end{aligned}

이고

∂G∂s(γ,s,t)=∂∂s[F(γ+s,γ−s,At)]=∂F∂x(γ+s,γ−s,At) ∂∂s(γ+s)+∂F∂y(γ+s,γ−s,At) ∂∂s(γ−s)+∂F∂z(γ+s,γ−s,At) ∂∂s(At)=∂F∂x(γ+s,γ−s,At)−∂F∂y(γ+s,γ−s,At)(E2)\begin{aligned} \frac{\partial G}{\partial s}(\gamma ,s,t) &= \frac{\partial }{\partial s} \Big[F (\gamma + s, \gamma-s, At)\Big] \\ &=\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial s}(\gamma + s) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial s}(\gamma - s) \\& +\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At)\,\frac{\partial }{\partial s}(At) \\ &= \frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At) -\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \tag{E2} \end{aligned}

이다.

다음 식의 우변에 있는 네 항에 연쇄 법칙을 적용하면 이계도함수를 구할 수 있다.

∂2G∂γ2(γ,s,t)=∂∂γ[∂G∂γ(γ,s,t)]=∂∂γ[∂F∂x(γ+s,γ−s,At)]+∂∂γ[∂F∂y(γ+s,γ−s,At)]∂2G∂s2(γ,s,t)=∂∂s[∂G∂s(γ,s,t)]=∂∂s[∂F∂x(γ+s,γ−s,At)]−∂∂s[∂F∂y(γ+s,γ−s,At)]\begin{aligned} \frac{\partial^2G}{\partial\gamma ^2}(\gamma ,s,t) &=\frac{\partial }{\partial \gamma }\Big[\frac{\partial G}{\partial \gamma }(\gamma , s, t)\Big] = \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\Big] + \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \Big] \\ \frac{\partial^2G}{\partial s^2}(\gamma ,s,t) &=\frac{\partial }{\partial s}\Big[\frac{\partial G}{\partial s}(\gamma , s, t)\Big] = \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\Big] - \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \Big] \end{aligned}

다른 방법으로, (E1)과 (E2)에서 FF 를 ∂F∂x\frac{\partial F}{\partial x} 로 바꾸면 다음을 얻는다.

∂∂γ[∂F∂x(γ+s,γ−s,At)]=∂2F∂x2(γ+s,γ−s,At)+∂2F∂y∂x(γ+s,γ−s,At)∂∂s[∂F∂x(γ+s,γ−s,At)]=∂2F∂x2(γ+s,γ−s,At)−∂2F∂y∂x(γ+s,γ−s,At)\begin{aligned} \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\Big] &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \\ \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At) \Big] &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \end{aligned}

(E1)과 (E2)에서 FF 를 ∂F∂y\frac{\partial F}{\partial y} 로 바꾸면 다음을 얻는다.

∂∂γ[∂F∂y(γ+s,γ−s,At)]=∂2F∂x∂y(γ+s,γ−s,At)+∂2F∂y2(γ+s,γ−s,At)∂∂s[∂F∂y(γ+s,γ−s,At)]=∂2F∂x∂y(γ+s,γ−s,At)−∂2F∂y2(γ+s,γ−s,At)\begin{aligned} \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s,At)\Big] &= \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s,At) \\ \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \Big] &= \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{aligned}

따라서

∂2G∂γ2(γ,s,t)=∂2F∂x2(γ+s,γ−s,At)+∂2F∂y∂x(γ+s,γ−s,At)=+∂2F∂x∂y(γ+s,γ−s,At)+∂2F∂y2(γ+s,γ−s,At)=∂2F∂x2(γ+s,γ−s,At)+2∂2F∂y∂x(γ+s,γ−s,At)+∂2F∂y2(γ+s,γ−s,At)\begin{aligned} \frac{\partial^2G}{\partial\gamma ^2}(\gamma ,s,t) &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \\ &\phantom{=}+ \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \\ &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +2\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{aligned}

이고

∂2G∂s2(γ,s,t)=∂2F∂x2(γ+s,γ−s,At)−∂2F∂y∂x(γ+s,γ−s,At)=−[∂2F∂x∂y(γ+s,γ−s,At)−∂2F∂y2(γ+s,γ−s,At)]=∂2F∂x2(γ+s,γ−s,At)−2∂2F∂y∂x(γ+s,γ−s,At)+∂2F∂y2(γ+s,γ−s,At)\begin{aligned} \frac{\partial^2G}{\partial s^2}(\gamma ,s,t) &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \\ &\phantom{=}-\Big[ \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At)\Big] \\ &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -2\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{aligned}

이다.

그러므로 변수를 생략하고 나타내면,

∂2G∂γ2+∂2G∂s2−∂G∂t=2∂2F∂x2+2∂2F∂y2−A∂F∂z=2∂F∂z−A∂F∂z=0\begin{aligned} \frac{\partial^2 G}{\partial \gamma^2} + \frac{\partial^2 G}{\partial s^2} -\frac{\partial G}{\partial t} &=2\frac{\partial^2 F}{\partial x^2} + 2\frac{\partial^2 F}{\partial y^2} -A \frac{\partial F}{\partial z} =2\frac{\partial F}{\partial z}-A \frac{\partial F}{\partial z} =0 \end{aligned}

이다. 이는 A=2A=2 이면 성립한다.

원본 교재 풀이(영어)

By the chain rule

∂G∂t(γ,s,t)=∂∂t[F(γ+s,γ−s,At)]=∂F∂x(γ+s,γ−s,At) ∂∂t(γ+s)+∂F∂y(γ+s,γ−s,At) ∂∂t(γ−s)+∂F∂z(γ+s,γ−s,At) ∂∂t(At)=A ∂F∂z(γ+s,γ−s,At)\begin{aligned} \frac{\partial G}{\partial t}(\gamma ,s,t) &= \frac{\partial }{\partial t} \Big[F (\gamma + s, \gamma-s, At)\Big] \\ &=\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial t}(\gamma + s) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial t}(\gamma - s) \\& +\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At)\,\frac{\partial }{\partial t}(At) \\ &= A\,\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At) \end{aligned}

and

∂G∂γ(γ,s,t)=∂∂γ[F(γ+s,γ−s,At)]=∂F∂x(γ+s,γ−s,At) ∂∂γ(γ+s)+∂F∂y(γ+s,γ−s,At) ∂∂γ(γ−s)+∂F∂z(γ+s,γ−s,At) ∂∂γ(At)=∂F∂x(γ+s,γ−s,At)+∂F∂y(γ+s,γ−s,At)(E1)\begin{aligned} \frac{\partial G}{\partial \gamma }(\gamma ,s,t) &= \frac{\partial }{\partial \gamma } \Big[F (\gamma + s, \gamma-s, At)\Big] \\ &=\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial \gamma }(\gamma + s) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial \gamma }(\gamma - s) \\& +\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At)\,\frac{\partial }{\partial \gamma }(At) \\ &= \frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \tag{E1} \end{aligned}

and

∂G∂s(γ,s,t)=∂∂s[F(γ+s,γ−s,At)]=∂F∂x(γ+s,γ−s,At) ∂∂s(γ+s)+∂F∂y(γ+s,γ−s,At) ∂∂s(γ−s)+∂F∂z(γ+s,γ−s,At) ∂∂s(At)=∂F∂x(γ+s,γ−s,At)−∂F∂y(γ+s,γ−s,At)(E2)\begin{aligned} \frac{\partial G}{\partial s}(\gamma ,s,t) &= \frac{\partial }{\partial s} \Big[F (\gamma + s, \gamma-s, At)\Big] \\ &=\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial s}(\gamma + s) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial s}(\gamma - s) \\& +\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At)\,\frac{\partial }{\partial s}(At) \\ &= \frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At) -\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \tag{E2} \end{aligned}

We can evaluate the second derivatives by applying the chain rule to the four terms on the right hand sides of

∂2G∂γ2(γ,s,t)=∂∂γ[∂G∂γ(γ,s,t)]=∂∂γ[∂F∂x(γ+s,γ−s,At)]+∂∂γ[∂F∂y(γ+s,γ−s,At)]∂2G∂s2(γ,s,t)=∂∂s[∂G∂s(γ,s,t)]=∂∂s[∂F∂x(γ+s,γ−s,At)]−∂∂s[∂F∂y(γ+s,γ−s,At)]\begin{aligned} \frac{\partial^2G}{\partial\gamma ^2}(\gamma ,s,t) &=\frac{\partial }{\partial \gamma }\Big[\frac{\partial G}{\partial \gamma }(\gamma , s, t)\Big] = \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\Big] + \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \Big] \\ \frac{\partial^2G}{\partial s^2}(\gamma ,s,t) &=\frac{\partial }{\partial s}\Big[\frac{\partial G}{\partial s}(\gamma , s, t)\Big] = \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\Big] - \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \Big] \end{aligned}

Alternatively, we can observe that replacing FF by ∂F∂x\frac{\partial F}{\partial x} in (E1) and (E2) gives

∂∂γ[∂F∂x(γ+s,γ−s,At)]=∂2F∂x2(γ+s,γ−s,At)+∂2F∂y∂x(γ+s,γ−s,At)∂∂s[∂F∂x(γ+s,γ−s,At)]=∂2F∂x2(γ+s,γ−s,At)−∂2F∂y∂x(γ+s,γ−s,At)\begin{aligned} \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\Big] &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \\ \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At) \Big] &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \end{aligned}

replacing FF by ∂F∂y\frac{\partial F}{\partial y} in (E1) and (E2) gives

∂∂γ[∂F∂y(γ+s,γ−s,At)]=∂2F∂x∂y(γ+s,γ−s,At)+∂2F∂y2(γ+s,γ−s,At)∂∂s[∂F∂y(γ+s,γ−s,At)]=∂2F∂x∂y(γ+s,γ−s,At)−∂2F∂y2(γ+s,γ−s,At)\begin{aligned} \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s,At)\Big] &= \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s,At) \\ \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \Big] &= \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{aligned}

Consequently

∂2G∂γ2(γ,s,t)=∂2F∂x2(γ+s,γ−s,At)+∂2F∂y∂x(γ+s,γ−s,At)=+∂2F∂x∂y(γ+s,γ−s,At)+∂2F∂y2(γ+s,γ−s,At)=∂2F∂x2(γ+s,γ−s,At)+2∂2F∂y∂x(γ+s,γ−s,At)+∂2F∂y2(γ+s,γ−s,At)\begin{aligned} \frac{\partial^2G}{\partial\gamma ^2}(\gamma ,s,t) &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \\ &\phantom{=}+ \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \\ &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +2\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{aligned}

and

∂2G∂s2(γ,s,t)=∂2F∂x2(γ+s,γ−s,At)−∂2F∂y∂x(γ+s,γ−s,At)=−[∂2F∂x∂y(γ+s,γ−s,At)−∂2F∂y2(γ+s,γ−s,At)]=∂2F∂x2(γ+s,γ−s,At)−2∂2F∂y∂x(γ+s,γ−s,At)+∂2F∂y2(γ+s,γ−s,At)\begin{aligned} \frac{\partial^2G}{\partial s^2}(\gamma ,s,t) &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \\ &\phantom{=}-\Big[ \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At)\Big] \\ &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -2\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{aligned}

So, suppressing the arguments,

∂2G∂γ2+∂2G∂s2−∂G∂t=2∂2F∂x2+2∂2F∂y2−A∂F∂z=2∂F∂z−A∂F∂z=0\begin{aligned} \frac{\partial^2 G}{\partial \gamma^2} + \frac{\partial^2 G}{\partial s^2} -\frac{\partial G}{\partial t} &=2\frac{\partial^2 F}{\partial x^2} + 2\frac{\partial^2 F}{\partial y^2} -A \frac{\partial F}{\partial z} =2\frac{\partial F}{\partial z}-A \frac{\partial F}{\partial z} =0 \end{aligned}

if A=2A=2.

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 16번
# 다변수 연쇄법칙계산보통자동 분류
13.CLP-3 · 2.4절 · 18번대학 수준우리말 번역

f(x)f(x) 를 미분가능한 함수라 하고, f′(0)=10f'(0) = 10 이 주어졌다고 하자. g(s,t)=f(as−bt)g(s,t) = f (as - bt) 라 하자. 여기서 aa 와 bb 는 상수이다. 점 (s,t)=(b,a)(s,t) = (b,a) 에서 ∂g∂s\frac{\partial g}{\partial s} 의 값을 구하세요. 즉, ∂g∂s∣(b,a)\frac{\partial g}{\partial s}\big|_{(b,a)} 를 구하세요.

영어 원문 보기

Let f(x)f(x) be a differentiable function, and suppose it is given that f′(0)=10f'(0) = 10. Let g(s,t)=f(as−bt)g(s,t) = f (as - bt), where aa and bb are constants. Evaluate ∂g∂s\frac{\partial g}{\partial s} at the point (s,t)=(b,a)(s,t) = (b,a), that is, find ∂g∂s∣(b,a)\frac{\partial g}{\partial s}\big|_{(b,a)}.

답 쓰고 채점
정답·풀이 보기

연쇄 법칙에 의해

∂g∂s(s,t)=∂∂s[f(as−bt)]=f′(as−bt)∂∂s(as−bt)=af′(as−bt)\begin{aligned} \frac{\partial g}{\partial s}(s,t) = \frac{\partial }{\partial s}\big[f(as-bt)\big] =f'(as-bt)\frac{\partial }{\partial s}(as-bt) = af'(as-bt) \end{aligned}

특히

∂g∂s(b,a)=af′(ab−ba)=af′(0)=10a\begin{aligned} \frac{\partial g}{\partial s}(b,a) = af'(ab-ba) =af'(0) =10 a \end{aligned}

원본 교재 풀이(영어)

By the chain rule

∂g∂s(s,t)=∂∂s[f(as−bt)]=f′(as−bt)∂∂s(as−bt)=af′(as−bt)\begin{aligned} \frac{\partial g}{\partial s}(s,t) = \frac{\partial }{\partial s}\big[f(as-bt)\big] =f'(as-bt)\frac{\partial }{\partial s}(as-bt) = af'(as-bt) \end{aligned}

In particular

∂g∂s(b,a)=af′(ab−ba)=af′(0)=10a\begin{aligned} \frac{\partial g}{\partial s}(b,a) = af'(ab-ba) =af'(0) =10 a \end{aligned}

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.5절 · 23번
# 다변수 연쇄법칙계산보통자동 분류
14.CLP-3 · 2.4절 · 19번대학 수준우리말 번역

f(u,v)f(u,v) 를 두 변수의 미분가능한 함수라 하고, zz 를 f(xz,yz)=0f(xz,yz) = 0 으로 음함수로 정의된 xx 와 yy 의 미분가능한 함수라 하자. 다음을 보이세요.

x∂z∂x+y∂z∂y=−zx\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -z

영어 원문 보기

Let f(u,v)f(u,v) be a differentiable function of two variables, and let zz be a differentiable function of xx and yy defined implicitly by f(xz,yz)=0f(xz,yz) = 0. Show that

x∂z∂x+y∂z∂y=−zx\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -z

정답·풀이 보기

함수 z(x,y)z(x,y) 는 모든 xx 와 yy 에 대해

f(x z(x,y) , y z(x,y))=0(∗)f\big(x\,z(x,y)\,,\, y\,z(x,y)\big) =0 \tag{$*$}

을 만족한다. 연쇄 법칙에 의해

∂∂x[f(x z(x,y) , y z(x,y))]=fu(x z(x,y) , y z(x,y))∂∂x[x z(x,y)]+fv(x z(x,y) , y z(x,y))∂∂x[y z(x,y)]=fu(x z(x,y) , y z(x,y)) [z(x,y)+x zx(x,y)]+fv(x z(x,y) , y z(x,y)) y zx(x,y)∂∂y[f(x z(x,y) , y z(x,y))]=fu(x z(x,y) , y z(x,y))∂∂y[x z(x,y)]+fv(x z(x,y) , y z(x,y))∂∂y[y z(x,y)]=fu(x z(x,y) , y z(x,y)) x zy(x,y)+fv(x z(x,y) , y z(x,y)) [z(x,y)+y zy(x,y)]\begin{aligned} &\frac{\partial }{\partial x}\big[f\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\big] \\[-0.05in] &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial x}\big[x\,z(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial x}\big[y\,z(x,y)\big] \\ &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,y\,z_x(x,y) \\ &\frac{\partial }{\partial y}\big[f\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\big] \\[-0.05in] &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial y}\big[x\,z(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial y}\big[y\,z(x,y)\big] \\ &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, x\,z_y(x,y) + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] \end{aligned}

이므로 (∗)(*) 를 xx 에 대해, 그리고 yy 에 대해 미분하면

fu(x z(x,y) , y z(x,y)) [z(x,y)+x zx(x,y)]+fv(x z(x,y) , y z(x,y)) yzx(x,y)=0fu(x z(x,y) , y z(x,y)) x zy(x,y)+fv(x z(x,y) , y z(x,y)) [z(x,y)+y zy(x,y)]=0\begin{aligned} f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,y z_x(x,y) &=0 \\ f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, x\,z_y(x,y) + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] &=0 \end{aligned}

이고, 인수를 생략하여 나타내면

fu [z+x zx]+fv yzx=0fu x zy+fv [z+y zy]=0\begin{aligned} f_u\,\big[z+x\,z_x\big] + f_v\,y z_x &=0 \\ f_u\, x\,z_y + f_v\,\big[z+y\,z_y\big] &=0 \end{aligned}

이다. 첫 번째 식을 zxz_x 에 대해, 두 번째 식을 zyz_y 에 대해 풀면

zx=−z fux fu+y fvzy=−z fvx fu+y fv\begin{aligned} z_x & = -\frac{z\,f_u}{x\,f_u+y\,f_v} \\ z_y & = -\frac{z\,f_v}{x\,f_u+y\,f_v} \end{aligned}

이므로

x∂z∂x+y∂z∂y=−xz fux fu+y fv−yz fvx fu+y fv=−z (x fu+y fv)x fu+y fv=−z\begin{aligned} x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -\frac{xz\,f_u}{x\,f_u+y\,f_v} -\frac{yz\,f_v}{x\,f_u+y\,f_v} =-\frac{z\,(x\,f_u+y\,f_v)}{x\,f_u+y\,f_v} = -z \end{aligned}

이다. 따라서 주어진 등식이 성립한다.

참고: 이는 물론 x fu+y fvx\,f_u+y\,f_v 가 0이 아니라는 가정 아래에서이다. 연쇄 법칙에 의해 이 가정은 ∂∂z[f(xz,yz)]\frac{\partial }{\partial z}\big[f(xz,yz)\big] 가 0이 아니라는 가정과 같다. 이는 다시 f(xz,yz)=0f(xz,yz)=0 을 xx 와 yy 의 함수로서 zz 에 대해 풀 수 있다는 진술과 거의 같지만, 완전히 같지는 않다.

원본 교재 풀이(영어)

We are told that the function z(x,y)z(x,y) obeys

f(x z(x,y) , y z(x,y))=0(∗)f\big(x\,z(x,y)\,,\, y\,z(x,y)\big) =0 \tag{$*$}

for all xx and yy. By the chain rule,

∂∂x[f(x z(x,y) , y z(x,y))]=fu(x z(x,y) , y z(x,y))∂∂x[x z(x,y)]+fv(x z(x,y) , y z(x,y))∂∂x[y z(x,y)]=fu(x z(x,y) , y z(x,y)) [z(x,y)+x zx(x,y)]+fv(x z(x,y) , y z(x,y)) y zx(x,y)∂∂y[f(x z(x,y) , y z(x,y))]=fu(x z(x,y) , y z(x,y))∂∂y[x z(x,y)]+fv(x z(x,y) , y z(x,y))∂∂y[y z(x,y)]=fu(x z(x,y) , y z(x,y)) x zy(x,y)+fv(x z(x,y) , y z(x,y)) [z(x,y)+y zy(x,y)]\begin{aligned} &\frac{\partial }{\partial x}\big[f\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\big] \\[-0.05in] &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial x}\big[x\,z(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial x}\big[y\,z(x,y)\big] \\ &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,y\,z_x(x,y) \\ &\frac{\partial }{\partial y}\big[f\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\big] \\[-0.05in] &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial y}\big[x\,z(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial y}\big[y\,z(x,y)\big] \\ &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, x\,z_y(x,y) + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] \end{aligned}

so differentiating (∗)(*) with respect to xx and with respect to yy gives

fu(x z(x,y) , y z(x,y)) [z(x,y)+x zx(x,y)]+fv(x z(x,y) , y z(x,y)) yzx(x,y)=0fu(x z(x,y) , y z(x,y)) x zy(x,y)+fv(x z(x,y) , y z(x,y)) [z(x,y)+y zy(x,y)]=0\begin{aligned} f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,y z_x(x,y) &=0 \\ f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, x\,z_y(x,y) + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] &=0 \end{aligned}

or, leaving out the arguments,

fu [z+x zx]+fv yzx=0fu x zy+fv [z+y zy]=0\begin{aligned} f_u\,\big[z+x\,z_x\big] + f_v\,y z_x &=0 \\ f_u\, x\,z_y + f_v\,\big[z+y\,z_y\big] &=0 \end{aligned}

Solving the first equation for zxz_x and the second for zyz_y gives

zx=−z fux fu+y fvzy=−z fvx fu+y fv\begin{aligned} z_x & = -\frac{z\,f_u}{x\,f_u+y\,f_v} \\ z_y & = -\frac{z\,f_v}{x\,f_u+y\,f_v} \end{aligned}

so that

x∂z∂x+y∂z∂y=−xz fux fu+y fv−yz fvx fu+y fv=−z (x fu+y fv)x fu+y fv=−z\begin{aligned} x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -\frac{xz\,f_u}{x\,f_u+y\,f_v} -\frac{yz\,f_v}{x\,f_u+y\,f_v} =-\frac{z\,(x\,f_u+y\,f_v)}{x\,f_u+y\,f_v} = -z \end{aligned}

as desired.

Remark: This is of course under the assumption that x fu+y fvx\,f_u+y\,f_v is nonzero. That is equivalent, by the chain rule, to the assumption that ∂∂z[f(xz,yz)]\frac{\partial }{\partial z}\big[f(xz,yz)\big] is non zero. That, in turn, is almost, but not quite, equivalent to the statement that f(xz,yz)=0f(xz,yz)=0 is can be solved for zz as a function of xx and yy.

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 18번
# 다변수 연쇄법칙증명·논술보통자동 분류
15.CLP-3 · 2.4절 · 20번대학 수준우리말 번역

어떤 두 번 미분가능한 함수 u=u(x,y)u = u(x, y) 에 대하여 w(s,t)=u(2s+3t,3s−2t)w(s,t) = u(2s + 3t, 3s - 2t) 라 하자.

(a) wssw_{ss} 를 uxxu_{xx}, uxyu_{xy}, uyyu_{yy} 로 나타내세요 (uxy=uyxu_{xy} = u_{yx} 라고 가정해도 됩니다). (b) uxx+uyy=0u_{xx} + u_{yy} = 0 이라 하자. 상수 AA 가 얼마일 때 wss=Awttw_{ss} = Aw_{tt} 가 되는지 구하세요.

영어 원문 보기

Let w(s,t)=u(2s+3t,3s−2t)w(s,t) = u(2s + 3t, 3s - 2t) for some twice differentiable function u=u(x,y)u = u(x, y).

(a) Find wssw_{ss} in terms of uxxu_{xx}, uxyu_{xy} , and uyyu_{yy} (you can assume that uxy=uyxu_{xy} = u_{yx}). (b) Suppose uxx+uyy=0u_{xx} + u_{yy} = 0. For what constant AA will wss=Awttw_{ss} = Aw_{tt}?

정답·풀이 보기

(a) 연쇄 법칙에 의해

ws(s,t)=∂∂s[u(2s+3t,3s−2t)]=ux(2s+3t,3s−2t)∂∂s(2s+3t)+uy(2s+3t,3s−2t)∂∂s(3s−2t)=2 ux(2s+3t,3s−2t)+3 uy(2s+3t,3s−2t)\begin{aligned} w_s(s,t) &=\frac{\partial }{\partial s}\big[u(2s + 3t, 3s - 2t)\big] \\ &= u_x(2s + 3t, 3s - 2t)\frac{\partial }{\partial s}\big(2s + 3t\big) +u_y(2s + 3t, 3s - 2t)\frac{\partial }{\partial s}\big(3s - 2t\big) \\ &= 2\,u_x(2s + 3t, 3s - 2t) +3\,u_y(2s + 3t, 3s - 2t) \end{aligned}

이고

wss(s,t)=2∂∂s[ux(2s+3t,3s−2t)]+3∂∂s[uy(2s+3t,3s−2t)]=[4uxx(2s+3t,3s−2t)+6uxy(2s+3t,3s−2t)]+[6uyx(2s+3t,3s−2t)+9uyy(2s+3t,3s−2t)]=4 uxx(2s+3t,3s−2t)+12 uxy(2s+3t,3s−2t)+9 uyy(2s+3t,3s−2t)\begin{aligned} w_{ss}(s,t) &= 2\frac{\partial }{\partial s}\big[u_x(2s + 3t, 3s - 2t)\big] + 3\frac{\partial }{\partial s}\big[u_y(2s + 3t, 3s - 2t)\big] \\ &= \big[4u_{xx}(2s + 3t, 3s - 2t) +6u_{xy}(2s + 3t, 3s - 2t) \big] \\& + \big[6u_{yx}(2s + 3t, 3s - 2t) +9u_{yy}(2s + 3t, 3s - 2t) \big] \\ &=4\,u_{xx}(2s + 3t, 3s - 2t) +12\,u_{xy}(2s + 3t, 3s - 2t) +9\,u_{yy}(2s + 3t, 3s - 2t) \end{aligned}

이다.

(b) 다시 연쇄 법칙에 의해

wt(s,t)=∂∂t[u(2s+3t,3s−2t)]=ux(2s+3t,3s−2t)∂∂t(2s+3t)+uy(2s+3t,3s−2t)∂∂t(3s−2t)=3 ux(2s+3t,3s−2t)−2 uy(2s+3t,3s−2t)\begin{aligned} w_t(s,t) &=\frac{\partial }{\partial t}\big[u(2s + 3t, 3s - 2t)\big] \\ &= u_x(2s + 3t, 3s - 2t)\frac{\partial }{\partial t}\big(2s + 3t\big) +u_y(2s + 3t, 3s - 2t)\frac{\partial }{\partial t}\big(3s - 2t\big) \\ &= 3\,u_x(2s + 3t, 3s - 2t) -2\, u_y(2s + 3t, 3s - 2t) \end{aligned}

이고

wtt(s,t)=3∂∂t[ux(2s+3t,3s−2t)]−2∂∂t[uy(2s+3t,3s−2t)]=[9uxx(2s+3t,3s−2t)−6uxy(2s+3t,3s−2t)]−[6uyx(2s+3t,3s−2t)−4uyy(2s+3t,3s−2t)]=9 uxx(2s+3t,3s−2t)−12 uxy(2s+3t,3s−2t)+4 uyy(2s+3t,3s−2t)\begin{aligned} w_{tt}(s,t) &= 3\frac{\partial }{\partial t}\big[u_x(2s + 3t, 3s - 2t)\big] - 2\frac{\partial }{\partial t}\big[u_y(2s + 3t, 3s - 2t)\big] \\ &= \big[9u_{xx}(2s + 3t, 3s - 2t) -6u_{xy}(2s + 3t, 3s - 2t) \big] \\& - \big[6u_{yx}(2s + 3t, 3s - 2t) -4u_{yy}(2s + 3t, 3s - 2t) \big] \\ &=9\,u_{xx}(2s + 3t, 3s - 2t) -12\,u_{xy}(2s + 3t, 3s - 2t) +4\,u_{yy}(2s + 3t, 3s - 2t) \end{aligned}

이다.

따라서 임의의 상수 AA 에 대해

wss−Awtt=(4−9A)uxx+(12+12A)uxy+(9−4A)uyy\begin{aligned} w_{ss} - Aw_{tt} &= (4-9A) u_{xx} +(12+12A) u_{xy} + (9-4A) u_{yy} \end{aligned}

이다.

uxx+uyy=0u_{xx} + u_{yy}=0 이 주어졌으므로, A=−1A=-1 이면 이 값은 원하는 대로 0이 된다. (이때 (4−9A)=(9−4A)=13(4-9A)=(9-4A)=13 이다.)

원본 교재 풀이(영어)

(a) By the chain rule

ws(s,t)=∂∂s[u(2s+3t,3s−2t)]=ux(2s+3t,3s−2t)∂∂s(2s+3t)+uy(2s+3t,3s−2t)∂∂s(3s−2t)=2 ux(2s+3t,3s−2t)+3 uy(2s+3t,3s−2t)\begin{aligned} w_s(s,t) &=\frac{\partial }{\partial s}\big[u(2s + 3t, 3s - 2t)\big] \\ &= u_x(2s + 3t, 3s - 2t)\frac{\partial }{\partial s}\big(2s + 3t\big) +u_y(2s + 3t, 3s - 2t)\frac{\partial }{\partial s}\big(3s - 2t\big) \\ &= 2\,u_x(2s + 3t, 3s - 2t) +3\,u_y(2s + 3t, 3s - 2t) \end{aligned}

and

wss(s,t)=2∂∂s[ux(2s+3t,3s−2t)]+3∂∂s[uy(2s+3t,3s−2t)]=[4uxx(2s+3t,3s−2t)+6uxy(2s+3t,3s−2t)]+[6uyx(2s+3t,3s−2t)+9uyy(2s+3t,3s−2t)]=4 uxx(2s+3t,3s−2t)+12 uxy(2s+3t,3s−2t)+9 uyy(2s+3t,3s−2t)\begin{aligned} w_{ss}(s,t) &= 2\frac{\partial }{\partial s}\big[u_x(2s + 3t, 3s - 2t)\big] + 3\frac{\partial }{\partial s}\big[u_y(2s + 3t, 3s - 2t)\big] \\ &= \big[4u_{xx}(2s + 3t, 3s - 2t) +6u_{xy}(2s + 3t, 3s - 2t) \big] \\& + \big[6u_{yx}(2s + 3t, 3s - 2t) +9u_{yy}(2s + 3t, 3s - 2t) \big] \\ &=4\,u_{xx}(2s + 3t, 3s - 2t) +12\,u_{xy}(2s + 3t, 3s - 2t) +9\,u_{yy}(2s + 3t, 3s - 2t) \end{aligned}

(b) Again by the chain rule

wt(s,t)=∂∂t[u(2s+3t,3s−2t)]=ux(2s+3t,3s−2t)∂∂t(2s+3t)+uy(2s+3t,3s−2t)∂∂t(3s−2t)=3 ux(2s+3t,3s−2t)−2 uy(2s+3t,3s−2t)\begin{aligned} w_t(s,t) &=\frac{\partial }{\partial t}\big[u(2s + 3t, 3s - 2t)\big] \\ &= u_x(2s + 3t, 3s - 2t)\frac{\partial }{\partial t}\big(2s + 3t\big) +u_y(2s + 3t, 3s - 2t)\frac{\partial }{\partial t}\big(3s - 2t\big) \\ &= 3\,u_x(2s + 3t, 3s - 2t) -2\, u_y(2s + 3t, 3s - 2t) \end{aligned}

and

wtt(s,t)=3∂∂t[ux(2s+3t,3s−2t)]−2∂∂t[uy(2s+3t,3s−2t)]=[9uxx(2s+3t,3s−2t)−6uxy(2s+3t,3s−2t)]−[6uyx(2s+3t,3s−2t)−4uyy(2s+3t,3s−2t)]=9 uxx(2s+3t,3s−2t)−12 uxy(2s+3t,3s−2t)+4 uyy(2s+3t,3s−2t)\begin{aligned} w_{tt}(s,t) &= 3\frac{\partial }{\partial t}\big[u_x(2s + 3t, 3s - 2t)\big] - 2\frac{\partial }{\partial t}\big[u_y(2s + 3t, 3s - 2t)\big] \\ &= \big[9u_{xx}(2s + 3t, 3s - 2t) -6u_{xy}(2s + 3t, 3s - 2t) \big] \\& - \big[6u_{yx}(2s + 3t, 3s - 2t) -4u_{yy}(2s + 3t, 3s - 2t) \big] \\ &=9\,u_{xx}(2s + 3t, 3s - 2t) -12\,u_{xy}(2s + 3t, 3s - 2t) +4\,u_{yy}(2s + 3t, 3s - 2t) \end{aligned}

Consquently, for any constant AA,

wss−Awtt=(4−9A)uxx+(12+12A)uxy+(9−4A)uyy\begin{aligned} w_{ss} - Aw_{tt} &= (4-9A) u_{xx} +(12+12A) u_{xy} + (9-4A) u_{yy} \end{aligned}

Given that uxx+uyy=0u_{xx} + u_{yy}=0, this will be zero, as desired, if A=−1A=-1. (Then (4−9A)=(9−4A)=13(4-9A)=(9-4A)=13.)

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.5절 · 15번
# 다변수 연쇄법칙계산보통자동 분류
16.CLP-3 · 2.4절 · 22번대학 수준우리말 번역

정의에 따르면, 미분가능한 함수 f(x,y)f(x,y) 의 점 (x0 , y0)\big( x_0\,,\,y_0\big) 에서의 기울기 벡터는 다음과 같다.

∇f(x0,y0)=<∂f∂x(x0 , y0) , ∂f∂y(x0 , y0)>{ \boldsymbol{\nabla} } f(x_0,y_0) =\left< \frac{\partial f}{\partial x}\big( x_0\,,\,y_0\big)\,,\, \frac{\partial f}{\partial y}\big( x_0\,,\,y_0\big)\right>

다음을 알고 있다고 하자.

∇f(3,6)=<7,8>{ \boldsymbol{\nabla} } f(3,6)=\left< 7,8\right>

또한

∇g(1,2)=<−1,4>,{ \boldsymbol{\nabla} } g(1,2)=\left< -1,4\right>,

이고

∇h(1,2)=<−5,10>.{ \boldsymbol{\nabla} } h(1,2)=\left< -5,10\right>.

이라고 하자.

g(1,2)=3g(1,2)=3, h(1,2)=6h(1,2)=6 이고 z(s,t)=f(g(s,t),h(s,t))z(s,t)=f\big(g(s,t),h(s,t)\big) 일 때, 다음을 구하세요.

∇z(1,2){ \boldsymbol{\nabla} } z(1,2)

영어 원문 보기

By definition, the gradient of the differentiable function f(x,y)f(x,y) at the point (x0 , y0)\big( x_0\,,\,y_0\big) is

∇f(x0,y0)=<∂f∂x(x0 , y0) , ∂f∂y(x0 , y0)>{ \boldsymbol{\nabla} } f(x_0,y_0) =\left< \frac{\partial f}{\partial x}\big( x_0\,,\,y_0\big)\,,\, \frac{\partial f}{\partial y}\big( x_0\,,\,y_0\big)\right>

Suppose that we know

∇f(3,6)=<7,8>{ \boldsymbol{\nabla} } f(3,6)=\left< 7,8\right>

Suppose also that

∇g(1,2)=<−1,4>,{ \boldsymbol{\nabla} } g(1,2)=\left< -1,4\right>,

and

∇h(1,2)=<−5,10>.{ \boldsymbol{\nabla} } h(1,2)=\left< -5,10\right>.

Assuming g(1,2)=3g(1,2)=3, h(1,2)=6h(1,2)=6, and z(s,t)=f(g(s,t),h(s,t))z(s,t)=f\big(g(s,t),h(s,t)\big), find

∇z(1,2){ \boldsymbol{\nabla} } z(1,2)

정답·풀이 보기

연쇄 법칙에 의해

∂z∂s(s,t)=∂∂sf(g(s,t),h(s,t))=∂f∂x(g(s,t),h(s,t))∂g∂s(s,t)+∂f∂y(g(s,t),h(s,t))∂h∂s(s,t)∂z∂t(s,t)=∂∂tf(g(s,t),h(s,t))=∂f∂x(g(s,t),h(s,t))∂g∂t(s,t)+∂f∂y(g(s,t),h(s,t))∂h∂t(s,t)\begin{aligned} \frac{\partial z}{\partial s}(s,t) &=\frac{\partial }{\partial s}f\big(g(s,t),h(s,t)\big) &&=\frac{\partial f}{\partial x}\big(g(s,t),h(s,t)\big) \frac{\partial g}{\partial s}(s,t) +\frac{\partial f}{\partial y}\big(g(s,t),h(s,t)\big) \frac{\partial h}{\partial s}(s,t) \\ \frac{\partial z}{\partial t}(s,t) &=\frac{\partial }{\partial t}f\big(g(s,t),h(s,t)\big) &&=\frac{\partial f}{\partial x}\big(g(s,t),h(s,t)\big) \frac{\partial g}{\partial t}(s,t) +\frac{\partial f}{\partial y}\big(g(s,t),h(s,t)\big) \frac{\partial h}{\partial t}(s,t) \end{aligned}

특히

∂z∂s(1,2)=∂f∂x(g(1,2),h(1,2))∂g∂s(1,2)+∂f∂y(g(1,2),h(1,2))∂h∂s(1,2)=∂f∂x(3,6)∂g∂s(1,2)+∂f∂y(3,6))∂h∂s(1,2)=7×(−1)+8×(−5)=−47∂z∂t(1,2)=∂f∂x(g(1,2),h(1,2))∂g∂t(1,2)+∂f∂y(g(1,2),h(1,2))∂h∂t(1,2)=7×4+8×10=108\begin{aligned} \frac{\partial z}{\partial s}(1,2) &=\frac{\partial f}{\partial x}\big(g(1,2),h(1,2)\big) \frac{\partial g}{\partial s}(1,2) +\frac{\partial f}{\partial y}\big(g(1,2),h(1,2)\big) \frac{\partial h}{\partial s}(1,2) \\ &=\frac{\partial f}{\partial x}(3,6) \frac{\partial g}{\partial s}(1,2) +\frac{\partial f}{\partial y}(3,6)) \frac{\partial h}{\partial s}(1,2) \\ &=7\times(-1)+8\times(-5) =-47\\ \frac{\partial z}{\partial t}(1,2) &=\frac{\partial f}{\partial x}\big(g(1,2),h(1,2)\big) \frac{\partial g}{\partial t}(1,2) +\frac{\partial f}{\partial y}\big(g(1,2),h(1,2)\big) \frac{\partial h}{\partial t}(1,2) \\ &=7\times4+8\times 10 =108 \end{aligned}

따라서 ∇z(1,2)=<−47,108>{ \boldsymbol{\nabla} } z(1,2)=\left< -47,108\right> 이다.

원본 교재 풀이(영어)

By the chain rule

∂z∂s(s,t)=∂∂sf(g(s,t),h(s,t))=∂f∂x(g(s,t),h(s,t))∂g∂s(s,t)+∂f∂y(g(s,t),h(s,t))∂h∂s(s,t)∂z∂t(s,t)=∂∂tf(g(s,t),h(s,t))=∂f∂x(g(s,t),h(s,t))∂g∂t(s,t)+∂f∂y(g(s,t),h(s,t))∂h∂t(s,t)\begin{aligned} \frac{\partial z}{\partial s}(s,t) &=\frac{\partial }{\partial s}f\big(g(s,t),h(s,t)\big) &&=\frac{\partial f}{\partial x}\big(g(s,t),h(s,t)\big) \frac{\partial g}{\partial s}(s,t) +\frac{\partial f}{\partial y}\big(g(s,t),h(s,t)\big) \frac{\partial h}{\partial s}(s,t) \\ \frac{\partial z}{\partial t}(s,t) &=\frac{\partial }{\partial t}f\big(g(s,t),h(s,t)\big) &&=\frac{\partial f}{\partial x}\big(g(s,t),h(s,t)\big) \frac{\partial g}{\partial t}(s,t) +\frac{\partial f}{\partial y}\big(g(s,t),h(s,t)\big) \frac{\partial h}{\partial t}(s,t) \end{aligned}

In particular

∂z∂s(1,2)=∂f∂x(g(1,2),h(1,2))∂g∂s(1,2)+∂f∂y(g(1,2),h(1,2))∂h∂s(1,2)=∂f∂x(3,6)∂g∂s(1,2)+∂f∂y(3,6))∂h∂s(1,2)=7×(−1)+8×(−5)=−47∂z∂t(1,2)=∂f∂x(g(1,2),h(1,2))∂g∂t(1,2)+∂f∂y(g(1,2),h(1,2))∂h∂t(1,2)=7×4+8×10=108\begin{aligned} \frac{\partial z}{\partial s}(1,2) &=\frac{\partial f}{\partial x}\big(g(1,2),h(1,2)\big) \frac{\partial g}{\partial s}(1,2) +\frac{\partial f}{\partial y}\big(g(1,2),h(1,2)\big) \frac{\partial h}{\partial s}(1,2) \\ &=\frac{\partial f}{\partial x}(3,6) \frac{\partial g}{\partial s}(1,2) +\frac{\partial f}{\partial y}(3,6)) \frac{\partial h}{\partial s}(1,2) \\ &=7\times(-1)+8\times(-5) =-47\\ \frac{\partial z}{\partial t}(1,2) &=\frac{\partial f}{\partial x}\big(g(1,2),h(1,2)\big) \frac{\partial g}{\partial t}(1,2) +\frac{\partial f}{\partial y}\big(g(1,2),h(1,2)\big) \frac{\partial h}{\partial t}(1,2) \\ &=7\times4+8\times 10 =108 \end{aligned}

Hence ∇z(1,2)=<−47,108>{ \boldsymbol{\nabla} } z(1,2)=\left< -47,108\right>.

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 23번
# 다변수 연쇄법칙계산보통자동 분류
17.CLP-3 · 2.4절 · 23번대학 수준우리말 번역

(a) ff 를 실수 전체에서 정의된 임의의 미분가능한 함수라 하자. 평면 전체에서 다음과 같이 정의된 함수 ww 가

w(x,y)=e−yf(x−y)w(x,y)=e^{-y}f(x-y)

편미분방정식

w+∂w∂x+∂w∂y=0w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}=0

을 만족시킴을 보이세요. (b) 방정식 x=u3−3uv2x=u^3-3uv^2, y=3u2v−v3y=3u^2v-v^3, z=u2−v2z=u^2-v^2 은 zz 를 xx 와 yy 의 함수로 정의한다. 점 (x,y)=(2,11)(x,y)=(2,11) 에 대응하는 점 (u,v)=(2,1)(u,v)=(2,1) 에서 ∂z∂x\frac{\partial z}{\partial x} 의 값을 구하세요.

영어 원문 보기

(a) Let ff be an arbitrary differentiable function defined on the entire real line. Show that the function ww defined on the entire plane as

w(x,y)=e−yf(x−y)w(x,y)=e^{-y}f(x-y)

satisfies the partial differential equation:

w+∂w∂x+∂w∂y=0w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}=0 (b) The equations x=u3−3uv2x=u^3-3uv^2, y=3u2v−v3y=3u^2v-v^3 and z=u2−v2z=u^2-v^2 define zz as a function of xx and yy. Determine ∂z∂x\frac{\partial z}{\partial x} at the point (u,v)=(2,1)(u,v)=(2,1) which corresponds to the point (x,y)=(2,11)(x,y)=(2,11).

정답·풀이 보기

(a) 곱의 미분법과 연쇄 법칙에 의해

wx(x,y)=∂∂x[e−yf(x−y)]=e−y∂∂x[f(x−y)]=e−yf′(x−y)∂∂x(x−y)=e−yf′(x−y)wy(x,y)=∂∂y[e−yf(x−y)]=−e−yf(x−y)+e−y∂∂y[f(x−y)]=−e−yf(x−y)+e−yf′(x−y)∂∂y(x−y)=−e−yf(x−y)−e−yf′(x−y)\begin{aligned} w_x(x,y) &= \frac{\partial }{\partial x}\big[e^{-y}f(x-y)\big] = e^{-y} \frac{\partial }{\partial x}\big[f(x-y)\big] = e^{-y} f'(x-y)\frac{\partial }{\partial x}(x-y) \\ &= e^{-y}f'(x-y) \\% w_y(x,y) &= \frac{\partial }{\partial y}\big[e^{-y}f(x-y)\big] = -e^{-y}f(x-y) + e^{-y} \frac{\partial }{\partial y}\big[f(x-y)\big] \\ &= -e^{-y}f(x-y) + e^{-y} f'(x-y)\frac{\partial }{\partial y}(x-y) \\ &= -e^{-y}f(x-y)-e^{-y}f'(x-y) \end{aligned}

이므로

w+∂w∂x+∂w∂y=e−yf(x−y)+e−yf′(x−y)−e−yf(x−y)−e−yf′(x−y)=0w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y} =e^{-y}f(x-y)+e^{-y}f'(x-y)-e^{-y}f(x-y)-e^{-y}f'(x-y) =0

이다. 따라서 성립한다.

(b) x=u3−3uv2x=u^3-3uv^2, y=3u2v−v3y=3u^2v-v^3 을 xx, yy 가 주어진 매개변수이고 uu, vv 가 미지수인 두 방정식으로 생각하자. 문제는 적어도 (u,v)=(2,1)(u,v)=(2,1), (x,y)=(2,11)(x,y)=(2,11) 근처에서 이 두 방정식을 uu, vv 에 대해 xx, yy 로 풀 수 있음을 암묵적으로 알려 준다. 즉, 함수 u(x,y)u(x,y) 와 v(x,y)v(x,y) 가 다음 식으로 정해진다는 것을 알려 준다.

x=u(x,y)3−3u(x,y) v(x,y)2y=3u(x,y)2v(x,y)−v(x,y)3x=u(x,y)^3-3u(x,y)\,v(x,y)^2\qquad y=3u(x,y)^2v(x,y)-v(x,y)^3

등식 x=u(x,y)3−3u(x,y)v(x,y)2x=u(x,y)^3-3u(x,y)v(x,y)^2 의 양변에 ∂∂x\frac{\partial }{\partial x} 를 취하면

1=3 u(x,y)2 ∂u∂x(x,y)−3∂u∂x(x,y) v(x,y)2−6 u(x,y) v(x,y) ∂v∂x(x,y)\begin{aligned} 1&=3\,u(x,y)^2\,\frac{\partial u}{\partial x}(x,y) -3\frac{\partial u}{\partial x}(x,y)\,v(x,y)^2 -6\,u(x,y)\,v(x,y)\,\frac{\partial v}{\partial x}(x,y) \\[0.05in] \end{aligned}

이다. 다음으로 y=3u(x,y)2v(x,y)−v(x,y)3y=3u(x,y)^2v(x,y)-v(x,y)^3 의 양변에 ∂∂x\frac{\partial }{\partial x} 를 취하면

0=6 u(x,y) ∂u∂x(x,y) v(x,y)+3 u(x,y)2 ∂v∂x(x,y)−3 v(x,y)2 ∂v∂x(x,y)\begin{aligned} 0&=6\,u(x,y)\,\frac{\partial u}{\partial x}(x,y)\,v(x,y) +3\,u(x,y)^2\,\frac{\partial v}{\partial x}(x,y) -3\,v(x,y)^2\,\frac{\partial v}{\partial x}(x,y) \end{aligned}

이다. x=2x=2, y=11y=11, u=2u=2, v=1v=1 을 대입하면

1=12∂u∂x(2,11)−3∂u∂x(2,11)−12∂v∂x(2,11)=9∂u∂x(2,11)−12∂v∂x(2,11)0=12∂u∂x(2,11)+12∂v∂x(2,11)−3∂v∂x(2,11)=12∂u∂x(2,11)+9∂v∂x(2,11)\begin{aligned} 1&=12\frac{\partial u}{\partial x}(2,11) -3\frac{\partial u}{\partial x}(2,11) -12\frac{\partial v}{\partial x}(2,11) &&=9\frac{\partial u}{\partial x}(2,11) -12\frac{\partial v}{\partial x}(2,11)\\[0.05in] 0&=12\frac{\partial u}{\partial x}(2,11) +12\frac{\partial v}{\partial x}(2,11) -3\frac{\partial v}{\partial x}(2,11) &&=12\frac{\partial u}{\partial x}(2,11) +9\frac{\partial v}{\partial x}(2,11) \end{aligned}

이다. 두 번째 식에서 ∂v∂x(2,11)=−43∂u∂x(2,11)\frac{\partial v}{\partial x}(2,11) =-\frac{4}{3}\frac{\partial u}{\partial x}(2,11) 이다. 이를 첫 번째 식에 대입하면

1=9∂u∂x(2,11)−12[−43∂u∂x(2,11)]=25∂u∂x(2,11)1=9\frac{\partial u}{\partial x}(2,11) -12\left[-\frac{4}{3}\frac{\partial u}{\partial x}(2,11)\right] = 25\frac{\partial u}{\partial x}(2,11)

이므로 ∂u∂x(2,11)=125\frac{\partial u}{\partial x}(2,11)=\frac{1}{25} 이고 ∂v∂x(2,11)=−475\frac{\partial v}{\partial x}(2,11)=-\frac{4}{75} 이다. 또한 문제에서 z(x,y)=u(x,y)2−v(x,y)2z(x,y)=u(x,y)^2-v(x,y)^2 임을 알 수 있다. 따라서

∂z∂x(x,y)=2u(x,y)∂u∂x(x,y)−2v(x,y)∂v∂x(x,y)  ⟹  ∂z∂x(2,11)=4∂u∂x(2,11)−2∂v∂x(2,11)=4125+2475=2075=415\begin{aligned} &\frac{\partial z}{\partial x}(x,y) =2u(x,y)\frac{\partial u}{\partial x}(x,y) -2v(x,y)\frac{\partial v}{\partial x}(x,y)\\ \implies & \frac{\partial z}{\partial x}(2,11) =4\frac{\partial u}{\partial x}(2,11) -2\frac{\partial v}{\partial x}(2,11) =4\frac{1}{25}+2\frac{4}{75}=\frac{20}{75} =\frac{4}{15} \end{aligned}

이다.

원본 교재 풀이(영어)

(a) By the product and chain rules

wx(x,y)=∂∂x[e−yf(x−y)]=e−y∂∂x[f(x−y)]=e−yf′(x−y)∂∂x(x−y)=e−yf′(x−y)wy(x,y)=∂∂y[e−yf(x−y)]=−e−yf(x−y)+e−y∂∂y[f(x−y)]=−e−yf(x−y)+e−yf′(x−y)∂∂y(x−y)=−e−yf(x−y)−e−yf′(x−y)\begin{aligned} w_x(x,y) &= \frac{\partial }{\partial x}\big[e^{-y}f(x-y)\big] = e^{-y} \frac{\partial }{\partial x}\big[f(x-y)\big] = e^{-y} f'(x-y)\frac{\partial }{\partial x}(x-y) \\ &= e^{-y}f'(x-y) \\% w_y(x,y) &= \frac{\partial }{\partial y}\big[e^{-y}f(x-y)\big] = -e^{-y}f(x-y) + e^{-y} \frac{\partial }{\partial y}\big[f(x-y)\big] \\ &= -e^{-y}f(x-y) + e^{-y} f'(x-y)\frac{\partial }{\partial y}(x-y) \\ &= -e^{-y}f(x-y)-e^{-y}f'(x-y) \end{aligned}

Hence

w+∂w∂x+∂w∂y=e−yf(x−y)+e−yf′(x−y)−e−yf(x−y)−e−yf′(x−y)=0w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y} =e^{-y}f(x-y)+e^{-y}f'(x-y)-e^{-y}f(x-y)-e^{-y}f'(x-y) =0

as desired.

(b) Think of x=u3−3uv2x=u^3-3uv^2, y=3u2v−v3y=3u^2v-v^3 as two equations in the two unknowns uu, vv with xx, yy just being given parameters. The question implicitly tells us that those two equations can be solved for uu, vv in terms of xx, yy, at least near (u,v)=(2,1)(u,v)=(2,1), (x,y)=(2,11)(x,y)=(2,11). That is, the question implicitly tells us that the functions u(x,y)u(x,y) and v(x,y)v(x,y) are determined by

x=u(x,y)3−3u(x,y) v(x,y)2y=3u(x,y)2v(x,y)−v(x,y)3x=u(x,y)^3-3u(x,y)\,v(x,y)^2\qquad y=3u(x,y)^2v(x,y)-v(x,y)^3

Applying ∂∂x\frac{\partial }{\partial x} to both sides of the equation x=u(x,y)3−3u(x,y)v(x,y)2x=u(x,y)^3-3u(x,y)v(x,y)^2 gives

1=3 u(x,y)2 ∂u∂x(x,y)−3∂u∂x(x,y) v(x,y)2−6 u(x,y) v(x,y) ∂v∂x(x,y)\begin{aligned} 1&=3\,u(x,y)^2\,\frac{\partial u}{\partial x}(x,y) -3\frac{\partial u}{\partial x}(x,y)\,v(x,y)^2 -6\,u(x,y)\,v(x,y)\,\frac{\partial v}{\partial x}(x,y) \\[0.05in] \end{aligned}

Then applying ∂∂x\frac{\partial }{\partial x} to both sides of y=3u(x,y)2v(x,y)−v(x,y)3y=3u(x,y)^2v(x,y)-v(x,y)^3 gives

0=6 u(x,y) ∂u∂x(x,y) v(x,y)+3 u(x,y)2 ∂v∂x(x,y)−3 v(x,y)2 ∂v∂x(x,y)\begin{aligned} 0&=6\,u(x,y)\,\frac{\partial u}{\partial x}(x,y)\,v(x,y) +3\,u(x,y)^2\,\frac{\partial v}{\partial x}(x,y) -3\,v(x,y)^2\,\frac{\partial v}{\partial x}(x,y) \end{aligned}

Substituting in x=2x=2, y=11y=11, u=2u=2, v=1v=1 gives

1=12∂u∂x(2,11)−3∂u∂x(2,11)−12∂v∂x(2,11)=9∂u∂x(2,11)−12∂v∂x(2,11)0=12∂u∂x(2,11)+12∂v∂x(2,11)−3∂v∂x(2,11)=12∂u∂x(2,11)+9∂v∂x(2,11)\begin{aligned} 1&=12\frac{\partial u}{\partial x}(2,11) -3\frac{\partial u}{\partial x}(2,11) -12\frac{\partial v}{\partial x}(2,11) &&=9\frac{\partial u}{\partial x}(2,11) -12\frac{\partial v}{\partial x}(2,11)\\[0.05in] 0&=12\frac{\partial u}{\partial x}(2,11) +12\frac{\partial v}{\partial x}(2,11) -3\frac{\partial v}{\partial x}(2,11) &&=12\frac{\partial u}{\partial x}(2,11) +9\frac{\partial v}{\partial x}(2,11) \end{aligned}

From the second equation ∂v∂x(2,11)=−43∂u∂x(2,11)\frac{\partial v}{\partial x}(2,11) =-\frac{4}{3}\frac{\partial u}{\partial x}(2,11). Substituting into the first equation gives

1=9∂u∂x(2,11)−12[−43∂u∂x(2,11)]=25∂u∂x(2,11)1=9\frac{\partial u}{\partial x}(2,11) -12\left[-\frac{4}{3}\frac{\partial u}{\partial x}(2,11)\right] = 25\frac{\partial u}{\partial x}(2,11)

so that ∂u∂x(2,11)=125\frac{\partial u}{\partial x}(2,11)=\frac{1}{25} and ∂v∂x(2,11)=−475\frac{\partial v}{\partial x}(2,11)=-\frac{4}{75}. The question also tells us that z(x,y)=u(x,y)2−v(x,y)2z(x,y)=u(x,y)^2-v(x,y)^2. Hence

∂z∂x(x,y)=2u(x,y)∂u∂x(x,y)−2v(x,y)∂v∂x(x,y)  ⟹  ∂z∂x(2,11)=4∂u∂x(2,11)−2∂v∂x(2,11)=4125+2475=2075=415\begin{aligned} &\frac{\partial z}{\partial x}(x,y) =2u(x,y)\frac{\partial u}{\partial x}(x,y) -2v(x,y)\frac{\partial v}{\partial x}(x,y)\\ \implies & \frac{\partial z}{\partial x}(2,11) =4\frac{\partial u}{\partial x}(2,11) -2\frac{\partial v}{\partial x}(2,11) =4\frac{1}{25}+2\frac{4}{75}=\frac{20}{75} =\frac{4}{15} \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 20번
# 다변수 연쇄법칙증명·논술보통자동 분류
18.CLP-3 · 2.4절 · 24번대학 수준우리말 번역

방정식

x2−ycos⁡(uv)=vx2+y2−sin⁡(uv)=4πu\begin{aligned} x^2-y\cos(uv)&=v\\ x^2+y^2-\sin(uv)&=\frac{4}{\pi}u \end{aligned}

은 (u,v)=(π2,0)(u,v)=\big(\frac{\pi}{2},0\big) 일 때의 점 (x,y)=(1,1)(x,y)=(1,1) 근처에서 xx 와 yy 를 uu 와 vv 의 함수로 암시적으로 정의한다(즉, x=x(u,v)x=x(u,v), y=y(u,v)y=y(u,v)).

(a) (u,v)=(π2,0)(u,v)=\big(\frac{\pi}{2},0\big) 에서

∂x∂u 와 ∂y∂u\frac{\partial x}{\partial u}\text{ 와 } \frac{\partial y}{\partial u}

의 값을 구하세요. (b) z=x4+y4z=x^4+y^4 일 때, 점 (u,v)=(π2,0)(u,v)=\big(\frac{\pi}{2},0\big) 에서 ∂z∂u\frac{\partial z}{\partial u} 의 값을 구하세요.

영어 원문 보기

The equations

x2−ycos⁡(uv)=vx2+y2−sin⁡(uv)=4πu\begin{aligned} x^2-y\cos(uv)&=v\\ x^2+y^2-\sin(uv)&=\frac{4}{\pi}u \end{aligned}

define xx and yy implicitly as functions of uu and vv (i.e. x=x(u,v)x=x(u,v), and y=y(u,v)y=y(u,v)) near the point (x,y)=(1,1)(x,y)=(1,1) at which (u,v)=(π2,0)(u,v)=\big(\frac{\pi}{2},0\big).

(a) Find

∂x∂u and ∂y∂u\frac{\partial x}{\partial u}\text{ and } \frac{\partial y}{\partial u}

at (u,v)=(π2,0)(u,v)=\big(\frac{\pi}{2},0\big). (b) If z=x4+y4z=x^4+y^4, determine ∂z∂u\frac{\partial z}{\partial u} at the point (u,v)=(π2,0)(u,v)=\big(\frac{\pi}{2},0\big).

정답·풀이 보기

(a) 주어진 식은

x(u,v)2−y(u,v)cos⁡(uv)=vx(u,v)2+y(u,v)2−sin⁡(uv)=4πux(u,v)^2-y(u,v)\cos(uv)=v\qquad x(u,v)^2+y(u,v)^2-\sin(uv)=\frac{4}{\pi}u

이다. 두 방정식에 ∂∂u\frac{\partial }{\partial u} 를 적용하면

2x(u,v)∂x∂u(u,v)−∂y∂u(u,v)cos⁡(uv)+v y(u,v)sin⁡(uv)=02x(u,v)∂x∂u(u,v)+2y(u,v)∂y∂u(u,v)−vcos⁡(uv)=4π\begin{aligned} 2x(u,v)\frac{\partial x}{\partial u}(u,v) -\frac{\partial y}{\partial u}(u,v)\cos(uv)+v\,y(u,v)\sin(uv)&=0 \\ 2x(u,v)\frac{\partial x}{\partial u}(u,v) +2y(u,v)\frac{\partial y}{\partial u}(u,v) -v\cos(uv)&=\frac{4}{\pi} \end{aligned}

이다. u=π2u=\frac{\pi}{2}, v=0v=0, x(π2,0)=1x\big(\frac{\pi}{2},0\big)=1, y(π2,0)=1y\big(\frac{\pi}{2},0\big)=1 을 대입하면

2∂x∂u(π2,0)−∂y∂u(π2,0)=02∂x∂u(π2,0)+2∂y∂u(π2,0)=4π\begin{aligned} 2\frac{\partial x}{\partial u}\left(\frac{\pi}{2},0\right) -\frac{\partial y}{\partial u}\left(\frac{\pi}{2},0\right)&=0 \\[0.1in] 2\frac{\partial x}{\partial u}\left(\frac{\pi}{2},0\right) +2\frac{\partial y}{\partial u}\left(\frac{\pi}{2},0\right) &=\frac{4}{\pi} \end{aligned}

이다. 첫째 방정식에서 얻은 ∂y∂u(π2,0)=2∂x∂u(π2,0)\frac{\partial y}{\partial u}\big(\frac{\pi}{2},0\big)=2\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big) 를 둘째 방정식에 대입하면 6∂x∂u(π2,0)=4π6\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{4}{\pi} 이므로 ∂x∂u(π2,0)=23π\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{2}{3\pi} 이고 ∂y∂u(π2,0)=43π\frac{\partial y}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{4}{3\pi} 이다.

(b) z(u,v)=x(u,v)4+y(u,v)4z(u,v)=x(u,v)^4+y(u,v)^4 으로 주어져 있다. 따라서

∂z∂u(u,v)=4x(u,v)3 ∂x∂u(u,v)+4y(u,v)3 ∂y∂u(u,v)\begin{aligned} \frac{\partial z}{\partial u}(u,v) &=4x(u,v)^3\ \frac{\partial x}{\partial u}(u,v) +4y(u,v)^3\ \frac{\partial y}{\partial u}(u,v) \end{aligned}

이다. u=π2u=\frac{\pi}{2}, v=0v=0, x(π2,0)=1x\big(\frac{\pi}{2},0\big)=1, y(π2,0)=1y\big(\frac{\pi}{2},0\big)=1 을 대입하고 (a)의 결과를 이용하면

∂z∂u(π2,0)=4 x ⁣(π2,0)3 ∂x∂u ⁣(π2,0)+4 y ⁣(π2,0)3 ∂y∂u ⁣(π2,0)=4(23π)+4(43π)=8π\begin{aligned} \frac{\partial z}{\partial u}\left(\frac{\pi}{2},0\right) &=4\,x\!\left(\frac{\pi}{2},0\right)^3\ \frac{\partial x}{\partial u}\!\left(\frac{\pi}{2},0\right) +4\,y\!\left(\frac{\pi}{2},0\right)^3\ \frac{\partial y}{\partial u}\!\left(\frac{\pi}{2},0\right) \\ &=4\left(\frac{2}{3\pi}\right) +4\left(\frac{4}{3\pi}\right) \\ &=\frac{8}{\pi} \end{aligned}

이다.

원본 교재 풀이(영어)

(a) We are told that

x(u,v)2−y(u,v)cos⁡(uv)=vx(u,v)2+y(u,v)2−sin⁡(uv)=4πux(u,v)^2-y(u,v)\cos(uv)=v\qquad x(u,v)^2+y(u,v)^2-\sin(uv)=\frac{4}{\pi}u

Applying ∂∂u\frac{\partial }{\partial u} to both equations gives

2x(u,v)∂x∂u(u,v)−∂y∂u(u,v)cos⁡(uv)+v y(u,v)sin⁡(uv)=02x(u,v)∂x∂u(u,v)+2y(u,v)∂y∂u(u,v)−vcos⁡(uv)=4π\begin{aligned} 2x(u,v)\frac{\partial x}{\partial u}(u,v) -\frac{\partial y}{\partial u}(u,v)\cos(uv)+v\,y(u,v)\sin(uv)&=0 \\ 2x(u,v)\frac{\partial x}{\partial u}(u,v) +2y(u,v)\frac{\partial y}{\partial u}(u,v) -v\cos(uv)&=\frac{4}{\pi} \end{aligned}

Setting u=π2u=\frac{\pi}{2}, v=0v=0, x(π2,0)=1x\big(\frac{\pi}{2},0\big)=1, y(π2,0)=1y\big(\frac{\pi}{2},0\big)=1 gives

2∂x∂u(π2,0)−∂y∂u(π2,0)=02∂x∂u(π2,0)+2∂y∂u(π2,0)=4π\begin{aligned} 2\frac{\partial x}{\partial u}\left(\frac{\pi}{2},0\right) -\frac{\partial y}{\partial u}\left(\frac{\pi}{2},0\right)&=0 \\[0.1in] 2\frac{\partial x}{\partial u}\left(\frac{\pi}{2},0\right) +2\frac{\partial y}{\partial u}\left(\frac{\pi}{2},0\right) &=\frac{4}{\pi} \end{aligned}

Substituting ∂y∂u(π2,0)=2∂x∂u(π2,0)\frac{\partial y}{\partial u}\big(\frac{\pi}{2},0\big)=2\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big), from the first equation, into the second equation gives 6∂x∂u(π2,0)=4π6\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{4}{\pi} so that ∂x∂u(π2,0)=23π\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{2}{3\pi} and ∂y∂u(π2,0)=43π\frac{\partial y}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{4}{3\pi}.

(b) We are told that z(u,v)=x(u,v)4+y(u,v)4z(u,v)=x(u,v)^4+y(u,v)^4. So

∂z∂u(u,v)=4x(u,v)3 ∂x∂u(u,v)+4y(u,v)3 ∂y∂u(u,v)\begin{aligned} \frac{\partial z}{\partial u}(u,v) &=4x(u,v)^3\ \frac{\partial x}{\partial u}(u,v) +4y(u,v)^3\ \frac{\partial y}{\partial u}(u,v) \end{aligned}

Substituting in u=π2u=\frac{\pi}{2}, v=0v=0, x(π2,0)=1x\big(\frac{\pi}{2},0\big)=1, y(π2,0)=1y\big(\frac{\pi}{2},0\big)=1 and using the results of part (a),

∂z∂u(π2,0)=4 x ⁣(π2,0)3 ∂x∂u ⁣(π2,0)+4 y ⁣(π2,0)3 ∂y∂u ⁣(π2,0)=4(23π)+4(43π)=8π\begin{aligned} \frac{\partial z}{\partial u}\left(\frac{\pi}{2},0\right) &=4\,x\!\left(\frac{\pi}{2},0\right)^3\ \frac{\partial x}{\partial u}\!\left(\frac{\pi}{2},0\right) +4\,y\!\left(\frac{\pi}{2},0\right)^3\ \frac{\partial y}{\partial u}\!\left(\frac{\pi}{2},0\right) \\ &=4\left(\frac{2}{3\pi}\right) +4\left(\frac{4}{3\pi}\right) \\ &=\frac{8}{\pi} \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 15번
# 다변수 연쇄법칙계산보통자동 분류
19.CLP-3 · 2.4절 · 25번대학 수준우리말 번역

f(u,v)f(u,v) 를 미분가능한 함수라 하고, u=x+yu=x+y , v=x−yv=x-y 라 하자. 다음을 만족하는 상수 α\alpha 를 구하세요.

(fx)2+(fy)2=α((fu)2+(fv)2)\begin{aligned} (f_x)^2+(f_y)^2=\alpha \big((f_u)^2+(f_v)^2\big) \end{aligned}

영어 원문 보기

Let f(u,v)f(u,v) be a differentiable function, and let u=x+yu=x+y and v=x−yv=x-y. Find a constant, α\alpha , such that

(fx)2+(fy)2=α((fu)2+(fv)2)\begin{aligned} (f_x)^2+(f_y)^2=\alpha \big((f_u)^2+(f_v)^2\big) \end{aligned}

정답·풀이 보기

이 문제는 나쁜(그러나 흔히 쓰는) 표기를 사용하고 있다. 하나의 기호 ff 가 서로 다른 두 함수, 즉 f(u,v)f(u,v) 와 f(x,y)=f(u,v)∣u=x+y,v=x−yf(x,y)=f(u,v)\big|_{u=x+y,v=x-y} 에 쓰였기 때문이다. 더 나은 표현은 다음과 같다.

  • f(u,v)f(u,v) 와 F(x,y)F(x,y) 를 F(x,y)=f(x+y,x−y)F(x,y)=f(x+y,x-y) 를 만족하는 미분가능한 함수라 하자. 다음을 만족하는 상수 α\alpha 를 구하세요.

Fx(x,y)2+Fy(x,y)2=α{fu(x+y,x−y)2+fv(x+y,x−y)2}F_x(x,y)^2+F_y(x,y)^2=\alpha \big\{f_u(x+y,x-y)^2+f_v(x+y,x-y)^2\big\}

연쇄 법칙에 의해

∂F∂x(x,y)=fu(x+y,x−y)∂∂x(x+y)+fv(x+y,x−y)∂∂x(x−y)=fu(x+y,x−y)+fv(x+y,x−y)∂F∂y(x,y)=fu(x+y,x−y)∂∂y(x+y)+fv(x+y,x−y)∂∂y(x−y)=fu(x+y,x−y)−fv(x+y,x−y)\begin{aligned} \frac{\partial F}{\partial x}(x,y) &=f_u(x+y,x-y)\frac{\partial }{\partial x}(x+y) +f_v(x+y,x-y)\frac{\partial }{\partial x}(x-y) \\ &=f_u(x+y,x-y)+f_v(x+y,x-y)\\ \frac{\partial F}{\partial y}(x,y) &=f_u(x+y,x-y)\frac{\partial }{\partial y}(x+y) +f_v(x+y,x-y)\frac{\partial }{\partial y}(x-y) \\ &=f_u(x+y,x-y)-f_v(x+y,x-y) \end{aligned}

따라서

Fx(x,y)2+Fy(x,y)2=[fu(x+y,x−y)+fv(x+y,x−y)]2+[fu(x+y,x−y)−fv(x+y,x−y)]2=2fu(x+y,x−y)2+2fv(x+y,x−y)2\begin{aligned} F_x(x,y)^2+F_y(x,y)^2 &=\Big[f_u(x+y,x-y)+f_v(x+y,x-y)\Big]^2 \\ & +\Big[f_u(x+y,x-y)-f_v(x+y,x-y)\Big]^2\\ &=2f_u(x+y,x-y)^2+2f_v(x+y,x-y)^2 \end{aligned}

그러므로 α=2\alpha =2 이면 된다.

원본 교재 풀이(영어)

This question uses bad (but standard) notation, in that the one symbol ff is used for two different functions, namely f(u,v)f(u,v) and f(x,y)=f(u,v)∣u=x+y,v=x−yf(x,y)=f(u,v)\big|_{u=x+y,v=x-y}. A better wording is

  • Let f(u,v)f(u,v) and F(x,y)F(x,y) be differentiable functions such that F(x,y)=f(x+y,x−y)F(x,y)=f(x+y,x-y). Find a constant, α\alpha , such that

Fx(x,y)2+Fy(x,y)2=α{fu(x+y,x−y)2+fv(x+y,x−y)2}F_x(x,y)^2+F_y(x,y)^2=\alpha \big\{f_u(x+y,x-y)^2+f_v(x+y,x-y)^2\big\}

By the chain rule

∂F∂x(x,y)=fu(x+y,x−y)∂∂x(x+y)+fv(x+y,x−y)∂∂x(x−y)=fu(x+y,x−y)+fv(x+y,x−y)∂F∂y(x,y)=fu(x+y,x−y)∂∂y(x+y)+fv(x+y,x−y)∂∂y(x−y)=fu(x+y,x−y)−fv(x+y,x−y)\begin{aligned} \frac{\partial F}{\partial x}(x,y) &=f_u(x+y,x-y)\frac{\partial }{\partial x}(x+y) +f_v(x+y,x-y)\frac{\partial }{\partial x}(x-y) \\ &=f_u(x+y,x-y)+f_v(x+y,x-y)\\ \frac{\partial F}{\partial y}(x,y) &=f_u(x+y,x-y)\frac{\partial }{\partial y}(x+y) +f_v(x+y,x-y)\frac{\partial }{\partial y}(x-y) \\ &=f_u(x+y,x-y)-f_v(x+y,x-y) \end{aligned}

Hence

Fx(x,y)2+Fy(x,y)2=[fu(x+y,x−y)+fv(x+y,x−y)]2+[fu(x+y,x−y)−fv(x+y,x−y)]2=2fu(x+y,x−y)2+2fv(x+y,x−y)2\begin{aligned} F_x(x,y)^2+F_y(x,y)^2 &=\Big[f_u(x+y,x-y)+f_v(x+y,x-y)\Big]^2 \\ & +\Big[f_u(x+y,x-y)-f_v(x+y,x-y)\Big]^2\\ &=2f_u(x+y,x-y)^2+2f_v(x+y,x-y)^2 \end{aligned}

So α=2\alpha =2 does the job.

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 14번
# 다변수 연쇄법칙계산보통자동 분류