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문제 은행 · 미적분 · 대학 미적분 · 다변수 미분

다변수 연쇄법칙

먼저 혼자 풀어 보고, 풀이로 확인한 다음, 아래 「한 번 더」 문제로 이어서 연습하세요.

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 문제CLP-3 · 2.4절 · 16번우리말 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.

답 쓰고 채점
다 풀었으면 정답 확인

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.

이 문제로 이어서 풀기
# 다변수 연쇄법칙계산보통자동 분류

「다변수 연쇄법칙」 문제 목록으로