다변수 연쇄법칙 연습문제

1.CLP-3 · 2.4절 · 9번대학 수준우리말 번역

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

영어 원문 보기

Assume that $f(x,y)$ satisfies Laplace's equation $\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0$ . Show that this is also the case for the composite function $g(s,t) = f (s - t, s + t)$ . That is, show that $\frac{\partial^2 g}{\partial s^2}+\frac{\partial^2 g}{\partial t^2}=0$ . You may assume that $f(x,y)$ 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,

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

이고

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

이다. 인수를 생략하면

$\begin{aligned} \frac{\partial^2 g}{\partial s^2} + \frac{\partial^2 g}{\partial t^2} &=\left\{\frac{\partial^2 f}{\partial x^2} + 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} +\left\{\frac{\partial^2 f}{\partial x^2} - 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} \\ &=2\left[\frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2}\right] \\ &=0 \end{aligned}$

이므로 원하는 결과가 얻어진다.

원본 교재 풀이(영어)

By the chain rule,

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

and

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

Suppressing the arguments

$\begin{aligned} \frac{\partial^2 g}{\partial s^2} + \frac{\partial^2 g}{\partial t^2} &=\left\{\frac{\partial^2 f}{\partial x^2} + 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} +\left\{\frac{\partial^2 f}{\partial x^2} - 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} \\ &=2\left[\frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2}\right] \\ &=0 \end{aligned}$

as desired.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.5절 · 37번

# 다변수 연쇄법칙증명·논술보통자동 분류

2.CLP-3 · 2.4절 · 12번대학 수준우리말 번역

$u(x,y)$ 가 다음과 같이 정의된다.

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

여기서 $F(z)$ 는 임의의 함수이다.

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

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

영어 원문 보기

$u(x,y)$ is defined as

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

for an arbitrary function $F(z)$ .

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

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

정답·풀이 보기

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

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

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

$\begin{aligned} \frac{\partial u}{\partial x}(x,y)&= e^y\, \frac{1}{xe^{-y^2}}\ e^{-y^2} = \frac{e^y}{x}\\ \frac{\partial u}{\partial y}(x,y)&= e^y\, \ln\big(xe^{-y^2}\big) + e^y\, \frac{1}{xe^{-y^2}}\ (-2xy)e^{-y^2} = e^y\, \ln\big(xe^{-y^2}\big) -2y e^y \\ &= e^y\,\ln (x) -y^2\,e^y -2y e^y \end{aligned}$

(b) 일반적으로

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

원본 교재 풀이(영어)

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

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

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

$\begin{aligned} \frac{\partial u}{\partial x}(x,y)&= e^y\, \frac{1}{xe^{-y^2}}\ e^{-y^2} = \frac{e^y}{x}\\ \frac{\partial u}{\partial y}(x,y)&= e^y\, \ln\big(xe^{-y^2}\big) + e^y\, \frac{1}{xe^{-y^2}}\ (-2xy)e^{-y^2} = e^y\, \ln\big(xe^{-y^2}\big) -2y e^y \\ &= e^y\,\ln (x) -y^2\,e^y -2y e^y \end{aligned}$

(b) In general

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

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 11번

# 다변수 연쇄법칙증명·논술보통자동 분류

3.CLP-3 · 2.4절 · 14번대학 수준우리말 번역

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

$x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} = z\frac{\partial w}{\partial z}$

영어 원문 보기

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

$x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} = z\frac{\partial w}{\partial z}$

정답·풀이 보기

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

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

따라서

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

이므로 주어진 등식이 성립한다.

원본 교재 풀이(영어)

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

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

So

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

as desired.

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 22번

# 다변수 연쇄법칙증명·논술보통자동 분류

4.CLP-3 · 2.4절 · 19번대학 수준우리말 번역

$f(u,v)$ 를 두 변수의 미분가능한 함수라 하고, $z$ 를 $f(xz,yz) = 0$ 으로 음함수로 정의된 $x$ 와 $y$ 의 미분가능한 함수라 하자. 다음을 보이세요.

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -z$

영어 원문 보기

Let $f(u,v)$ be a differentiable function of two variables, and let $z$ be a differentiable function of $x$ and $y$ defined implicitly by $f(xz,yz) = 0$ . Show that

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -z$

정답·풀이 보기

함수 $z(x,y)$ 는 모든 $x$ 와 $y$ 에 대해

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

을 만족한다. 연쇄 법칙에 의해

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

이므로 $(*)$ 를 $x$ 에 대해, 그리고 $y$ 에 대해 미분하면

$\begin{aligned} f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,y z_x(x,y) &=0 \\ f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, x\,z_y(x,y) + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] &=0 \end{aligned}$

이고, 인수를 생략하여 나타내면

$\begin{aligned} f_u\,\big[z+x\,z_x\big] + f_v\,y z_x &=0 \\ f_u\, x\,z_y + f_v\,\big[z+y\,z_y\big] &=0 \end{aligned}$

이다. 첫 번째 식을 $z_x$ 에 대해, 두 번째 식을 $z_y$ 에 대해 풀면

$\begin{aligned} z_x & = -\frac{z\,f_u}{x\,f_u+y\,f_v} \\ z_y & = -\frac{z\,f_v}{x\,f_u+y\,f_v} \end{aligned}$

이므로

$\begin{aligned} x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -\frac{xz\,f_u}{x\,f_u+y\,f_v} -\frac{yz\,f_v}{x\,f_u+y\,f_v} =-\frac{z\,(x\,f_u+y\,f_v)}{x\,f_u+y\,f_v} = -z \end{aligned}$

이다. 따라서 주어진 등식이 성립한다.

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

원본 교재 풀이(영어)

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

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

for all $x$ and $y$ . By the chain rule,

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

so differentiating $(*)$ with respect to $x$ and with respect to $y$ gives

$\begin{aligned} f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,y z_x(x,y) &=0 \\ f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, x\,z_y(x,y) + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] &=0 \end{aligned}$

or, leaving out the arguments,

$\begin{aligned} f_u\,\big[z+x\,z_x\big] + f_v\,y z_x &=0 \\ f_u\, x\,z_y + f_v\,\big[z+y\,z_y\big] &=0 \end{aligned}$

Solving the first equation for $z_x$ and the second for $z_y$ gives

$\begin{aligned} z_x & = -\frac{z\,f_u}{x\,f_u+y\,f_v} \\ z_y & = -\frac{z\,f_v}{x\,f_u+y\,f_v} \end{aligned}$

so that

$\begin{aligned} x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -\frac{xz\,f_u}{x\,f_u+y\,f_v} -\frac{yz\,f_v}{x\,f_u+y\,f_v} =-\frac{z\,(x\,f_u+y\,f_v)}{x\,f_u+y\,f_v} = -z \end{aligned}$

as desired.

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

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 18번

# 다변수 연쇄법칙증명·논술보통자동 분류

5.CLP-3 · 2.4절 · 23번대학 수준우리말 번역

(a) $f$ 를 실수 전체에서 정의된 임의의 미분가능한 함수라 하자. 평면 전체에서 다음과 같이 정의된 함수 $w$ 가

$w(x,y)=e^{-y}f(x-y)$

편미분방정식

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

을 만족시킴을 보이세요. (b) 방정식 $x=u^3-3uv^2$ , $y=3u^2v-v^3$ , $z=u^2-v^2$ 은 $z$ 를 $x$ 와 $y$ 의 함수로 정의한다. 점 $(x,y)=(2,11)$ 에 대응하는 점 $(u,v)=(2,1)$ 에서 $\frac{\partial z}{\partial x}$ 의 값을 구하세요.

영어 원문 보기

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

$w(x,y)=e^{-y}f(x-y)$

satisfies the partial differential equation:

$w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}=0$ (b) The equations $x=u^3-3uv^2$ , $y=3u^2v-v^3$ and $z=u^2-v^2$ define $z$ as a function of $x$ and $y$ . Determine $\frac{\partial z}{\partial x}$ at the point $(u,v)=(2,1)$ which corresponds to the point $(x,y)=(2,11)$ .

정답·풀이 보기

(a) 곱의 미분법과 연쇄 법칙에 의해

$\begin{aligned} w_x(x,y) &= \frac{\partial }{\partial x}\big[e^{-y}f(x-y)\big] = e^{-y} \frac{\partial }{\partial x}\big[f(x-y)\big] = e^{-y} f'(x-y)\frac{\partial }{\partial x}(x-y) \\ &= e^{-y}f'(x-y) \\% w_y(x,y) &= \frac{\partial }{\partial y}\big[e^{-y}f(x-y)\big] = -e^{-y}f(x-y) + e^{-y} \frac{\partial }{\partial y}\big[f(x-y)\big] \\ &= -e^{-y}f(x-y) + e^{-y} f'(x-y)\frac{\partial }{\partial y}(x-y) \\ &= -e^{-y}f(x-y)-e^{-y}f'(x-y) \end{aligned}$

이므로

$w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y} =e^{-y}f(x-y)+e^{-y}f'(x-y)-e^{-y}f(x-y)-e^{-y}f'(x-y) =0$

이다. 따라서 성립한다.

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

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

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

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

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

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

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

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

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

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

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

$\begin{aligned} &\frac{\partial z}{\partial x}(x,y) =2u(x,y)\frac{\partial u}{\partial x}(x,y) -2v(x,y)\frac{\partial v}{\partial x}(x,y)\\ \implies & \frac{\partial z}{\partial x}(2,11) =4\frac{\partial u}{\partial x}(2,11) -2\frac{\partial v}{\partial x}(2,11) =4\frac{1}{25}+2\frac{4}{75}=\frac{20}{75} =\frac{4}{15} \end{aligned}$

이다.

원본 교재 풀이(영어)

(a) By the product and chain rules

$\begin{aligned} w_x(x,y) &= \frac{\partial }{\partial x}\big[e^{-y}f(x-y)\big] = e^{-y} \frac{\partial }{\partial x}\big[f(x-y)\big] = e^{-y} f'(x-y)\frac{\partial }{\partial x}(x-y) \\ &= e^{-y}f'(x-y) \\% w_y(x,y) &= \frac{\partial }{\partial y}\big[e^{-y}f(x-y)\big] = -e^{-y}f(x-y) + e^{-y} \frac{\partial }{\partial y}\big[f(x-y)\big] \\ &= -e^{-y}f(x-y) + e^{-y} f'(x-y)\frac{\partial }{\partial y}(x-y) \\ &= -e^{-y}f(x-y)-e^{-y}f'(x-y) \end{aligned}$

Hence

$w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y} =e^{-y}f(x-y)+e^{-y}f'(x-y)-e^{-y}f(x-y)-e^{-y}f'(x-y) =0$

as desired.

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

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

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

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

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

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

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

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

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

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

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

$\begin{aligned} &\frac{\partial z}{\partial x}(x,y) =2u(x,y)\frac{\partial u}{\partial x}(x,y) -2v(x,y)\frac{\partial v}{\partial x}(x,y)\\ \implies & \frac{\partial z}{\partial x}(2,11) =4\frac{\partial u}{\partial x}(2,11) -2\frac{\partial v}{\partial x}(2,11) =4\frac{1}{25}+2\frac{4}{75}=\frac{20}{75} =\frac{4}{15} \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 20번

# 다변수 연쇄법칙증명·논술보통자동 분류