다변수 연쇄법칙 연습문제

61.APEX · 12.5절 · 28번대학 수준우리말 번역

연습문제 25–28에서 함수 $w=F(x,y,z)$ , 벡터 $\vec v$ , 점 $P$ 가 주어져 있다.

(a) $\nabla F(x,y,z)$ 를 구하세요. (b) $\vec u$ 가 $\vec v$ 의 방향의 단위벡터일 때, $P$ 에서 $D_{\vec u\,}F$ 를 구하세요.

$\displaystyle F(x,y,z) = \frac{2}{x^2+y^2+z^2}$ , $\vec v = \left\langle 1,1,-2\right\rangle$ , $P = (1,1,1)$

영어 원문 보기

In Exercises 25–28, a function $w=F(x,y,z)$ , a vector $\vec v$ and a point $P$ are given.

(a) Find $\nabla F(x,y,z)$ . (b) Find $D_{\vec u\,}F$ at $P$ , where $\vec u$ is the unit vector in the direction of $\vec v$ .

$\displaystyle F(x,y,z) = \frac{2}{x^2+y^2+z^2}$ , $\vec v = \left\langle 1,1,-2\right\rangle$ , $P = (1,1,1)$

정답·풀이 보기

(a) $\nabla F(x,y,z) = \left\langle -\frac{4x}{(x^2+y^2+z^2)^2},-\frac{4y}{(x^2+y^2+z^2)^2},-\frac{4z}{(x^2+y^2+z^2)^2}\right\rangle$ (b) $0$

원본 교재 풀이(영어)

(a) $\nabla F(x,y,z) = \left\langle -\frac{4x}{(x^2+y^2+z^2)^2},-\frac{4y}{(x^2+y^2+z^2)^2},-\frac{4z}{(x^2+y^2+z^2)^2}\right\rangle$ (b) $0$

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.5절 · 17번

# 다변수 연쇄법칙계산자동 분류

62.CLP-3 · 2.4절 · 4번대학 수준우리말 번역

열역학 교재에서는 다음 관계를 사용한다.

$\left(\frac{\partial y}{\partial x}\right) \left(\frac{\partial z}{\partial y}\right) \left(\frac{\partial x}{\partial z}\right)=-1$

이 등식의 의미를 설명하고, 이것이 성립함을 증명하세요.

영어 원문 보기

Thermodynamics texts use the relationship

$\left(\frac{\partial y}{\partial x}\right) \left(\frac{\partial z}{\partial y}\right) \left(\frac{\partial x}{\partial z}\right)=-1$

Explain the meaning of this equation and prove that it is true.

정답·풀이 보기

기본 가정은 세 양 $x$ , $y$ , $z$ 가 서로 독립이 아니라는 것이다. 이 중 어느 두 개가 주어지면 나머지 하나가 유일하게 정해진다. 이들이 관계식 $F(x,y,z)=0$ 을 만족한다고 가정하면, 이 식을 풀어서

$x$ 를 $y$ 와 $z$ 의 함수로 정할 수 있고(예를 들어 $x=f(y,z)$ ), 또한
$y$ 를 $x$ 와 $z$ 의 함수로 정할 수 있고(예를 들어 $y=g(x,z)$ ), 또한
$z$ 를 $x$ 와 $y$ 의 함수로 정할 수 있다 (예를 들어 $z=h(x,y)$ ).

예를 들어 $F(x,y,z) = xyz-1$ 이면

$F(x,y,z)=xyz-1=0$ 으로부터 $x=\frac{1}{yz}=f(y,z)$ 이고,
$F(x,y,z)=xyz-1=0$ 으로부터 $y=\frac{1}{xz}=g(x,z)$ 이고,
$F(x,y,z)=xyz-1=0$ 으로부터 $z=\frac{1}{xy}=h(x,y)$ 이다.

일반적으로 $F(x,y,z)=0$ 이 $x=f(y,z)$ 를 정한다는 것은

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

이 모든 $y$ 와 $z$ 에 대해 성립한다는 뜻이다. $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)$ 라 하자. $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)$ 에 ( $y$ 와 $z$ 를 독립변수로 보고) 연쇄 법칙을 적용하면

$\begin{aligned} \frac{\partial \mathcal{F}}{\partial z}(y,z) &= \frac{\partial F}{\partial x}\big(f(y,z),y,z\big) \frac{\partial f}{\partial z}(y,z) +\frac{\partial F}{\partial z}\big(f(y,z),y,z\big) \end{aligned}$

이다.

등식 $(*)$ 는 모든 $y$ 와 $z$ 에 대해 $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)=0$ 임을 뜻한다. 따라서 등식 $(*)$ 를 $z$ 에 대해 미분하면

$\begin{aligned} \frac{\partial \mathcal{F}}{\partial z}(y,z)&=\frac{\partial F}{\partial x}\big(f(y,z),y,z\big) \frac{\partial f}{\partial z}(y,z) +\frac{\partial F}{\partial z}\big(f(y,z),y,z\big)=0 \\ &\implies \frac{\partial f}{\partial z}(y,z) =-\frac{\frac{\partial F}{\partial z}\big(f(y,z),y,z\big)} {\frac{\partial F}{\partial x}\big(f(y,z),y,z\big)} \end{aligned}$

이 모든 $y$ 와 $z$ 에 대해 성립한다. 마찬가지로 $F\big(x,g(x,z),z\big)=0$ 을 $x$ 에 대해 미분하고 $F\big(x,y,h(x,y)\big)=0$ 을 $y$ 에 대해 미분하면

$\begin{aligned} \frac{\partial g}{\partial x}(x,z) =-\frac{\frac{\partial F}{\partial x}\big(x,g(x,z),z\big)} {\frac{\partial F}{\partial y}\big(x,g(x,z),z\big)}\qquad \frac{\partial h}{\partial y}(x,y) =-\frac{\frac{\partial F}{\partial y}\big(x,y,h(x,y)\big)} {\frac{\partial F}{\partial z}\big(x,y,h(x,y)\big)} \end{aligned}$

를 얻는다.

$(x,y,z)$ 가 $F(x,y,z)=0$ 을 만족하는 임의의 점이면 (따라서 $x=f(y,z)$ , $y=g(x,z)$ , $z=h(x,y)$ 이다),

$\begin{aligned} \frac{\partial f}{\partial z}(y,z) =-\frac{\frac{\partial F}{\partial z}\big(x,y,z\big)} {\frac{\partial F}{\partial x}\big(x,y,z\big)}\qquad \frac{\partial g}{\partial x}(x,z) =-\frac{\frac{\partial F}{\partial x}\big(x,y,z\big)} {\frac{\partial F}{\partial y}\big(x,y,z\big)}\qquad \frac{\partial h}{\partial y}(x,y) =-\frac{\frac{\partial F}{\partial y}\big(x,y,z\big)} {\frac{\partial F}{\partial z}\big(x,y,z\big)} \end{aligned}$

이고,

$\begin{aligned} \frac{\partial f}{\partial z}(y,z)\ \frac{\partial g}{\partial x}(x,z)\ \frac{\partial h}{\partial y}(x,y) &=-\frac{\frac{\partial F}{\partial z}\big(x,y,z\big)} {\frac{\partial F}{\partial x}\big(x,y,z\big)}\ \frac{\frac{\partial F}{\partial x}\big(x,y,z\big)} {\frac{\partial F}{\partial y}\big(x,y,z\big)}\ \frac{\frac{\partial F}{\partial y}\big(x,y,z\big)} {\frac{\partial F}{\partial z}\big(x,y,z\big)} \\ &=-1 \end{aligned}$

이다.

원본 교재 풀이(영어)

The basic assumption is that the three quantites $x$ , $y$ and $z$ are not independent. Given any two of them, the third is uniquely determined. They are assumed to satisfy a relationship $F(x,y,z)=0$ , which can be solved to

determine $x$ as a function of $y$ and $z$ (say $x=f(y,z)$ ) and can alternatively be solved to
determine $y$ as a function of $x$ and $z$ (say $y=g(x,z)$ ) and can alternatively be solved to
determine $z$ as a function of $x$ and $y$ (say $z=h(x,y)$ ).

As an example, if $F(x,y,z) = xyz-1$ , then

$F(x,y,z)=xyz-1=0$ implies that $x=\frac{1}{yz}=f(y,z)$ and
$F(x,y,z)=xyz-1=0$ implies that $y=\frac{1}{xz}=g(x,z)$ and
$F(x,y,z)=xyz-1=0$ implies that $z=\frac{1}{xy}=h(x,y)$

In general, saying that $F(x,y,z)=0$ determines $x=f(y,z)$ means that

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

for all $y$ and $z$ . Set $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)$ . Applying the chain rule to $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)$ (with $y$ and $z$ independent variables) gives

$\begin{aligned} \frac{\partial \mathcal{F}}{\partial z}(y,z) &= \frac{\partial F}{\partial x}\big(f(y,z),y,z\big) \frac{\partial f}{\partial z}(y,z) +\frac{\partial F}{\partial z}\big(f(y,z),y,z\big) \end{aligned}$

The equation $(*)$ says that $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)=0$ for all $y$ and $z$ . So differentiating the equation $(*)$ with respect to $z$ gives

$\begin{aligned} \frac{\partial \mathcal{F}}{\partial z}(y,z)&=\frac{\partial F}{\partial x}\big(f(y,z),y,z\big) \frac{\partial f}{\partial z}(y,z) +\frac{\partial F}{\partial z}\big(f(y,z),y,z\big)=0 \\ &\implies \frac{\partial f}{\partial z}(y,z) =-\frac{\frac{\partial F}{\partial z}\big(f(y,z),y,z\big)} {\frac{\partial F}{\partial x}\big(f(y,z),y,z\big)} \end{aligned}$

for all $y$ and $z$ . Similarly, differentiating $F\big(x,g(x,z),z\big)=0$ with respect to $x$ and $F\big(x,y,h(x,y)\big)=0$ with respect to $y$ gives

$\begin{aligned} \frac{\partial g}{\partial x}(x,z) =-\frac{\frac{\partial F}{\partial x}\big(x,g(x,z),z\big)} {\frac{\partial F}{\partial y}\big(x,g(x,z),z\big)}\qquad \frac{\partial h}{\partial y}(x,y) =-\frac{\frac{\partial F}{\partial y}\big(x,y,h(x,y)\big)} {\frac{\partial F}{\partial z}\big(x,y,h(x,y)\big)} \end{aligned}$

If $(x,y,z)$ is any point satisfying $F(x,y,z)=0$ (so that $x=f(y,z)$ and $y=g(x,z)$ and $z=h(x,y)$ ), then

$\begin{aligned} \frac{\partial f}{\partial z}(y,z) =-\frac{\frac{\partial F}{\partial z}\big(x,y,z\big)} {\frac{\partial F}{\partial x}\big(x,y,z\big)}\qquad \frac{\partial g}{\partial x}(x,z) =-\frac{\frac{\partial F}{\partial x}\big(x,y,z\big)} {\frac{\partial F}{\partial y}\big(x,y,z\big)}\qquad \frac{\partial h}{\partial y}(x,y) =-\frac{\frac{\partial F}{\partial y}\big(x,y,z\big)} {\frac{\partial F}{\partial z}\big(x,y,z\big)} \end{aligned}$

and

$\begin{aligned} \frac{\partial f}{\partial z}(y,z)\ \frac{\partial g}{\partial x}(x,z)\ \frac{\partial h}{\partial y}(x,y) &=-\frac{\frac{\partial F}{\partial z}\big(x,y,z\big)} {\frac{\partial F}{\partial x}\big(x,y,z\big)}\ \frac{\frac{\partial F}{\partial x}\big(x,y,z\big)} {\frac{\partial F}{\partial y}\big(x,y,z\big)}\ \frac{\frac{\partial F}{\partial y}\big(x,y,z\big)} {\frac{\partial F}{\partial z}\big(x,y,z\big)} \\ &=-1 \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 5번

# 다변수 연쇄법칙증명·논술기본자동 분류

63.CLP-3 · 2.4절 · 5번대학 수준우리말 번역

다음 논증에서 잘못된 점은 무엇인가요? $w=f(x,y,z)$ 이고 $z=g(x,y)$ 라고 하자. 연쇄 법칙에 의해

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

이다. 따라서 $0=\frac{\partial w}{\partial z}\frac{\partial z}{\partial x}$ 이고, 그러므로 $\frac{\partial w}{\partial z}=0$ 또는 $\frac{\partial z}{\partial x}=0$ 이다.

영어 원문 보기

What is wrong with the following argument? Suppose that $w=f(x,y,z)$ and $z=g(x,y)$ . By the chain rule,

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

Hence $0=\frac{\partial w}{\partial z}\frac{\partial z}{\partial x}$ and so $\frac{\partial w}{\partial z}=0$ or $\frac{\partial z}{\partial x}=0$ .

정답·풀이 보기

문제는 같은 등식 안에서 $\frac{\partial w}{\partial x}$ 가 서로 완전히 다른 두 함수를 나타내는 데 쓰였다는 점이다. 이 식을 정확하게 쓰는 방법은 다음과 같다. $f(x,y,z)$ 와 $g(x,y)$ 를 연속적으로 미분가능한 함수라 하고 $w(x,y)=f\big(x,y,g(x,y)\big)$ 로 정의하자. 연쇄 법칙에 의해

$\begin{aligned} \frac{\partial w}{\partial x}(x,y) &=\frac{\partial f}{\partial x}\big(x,y,g(x,y)\big) \frac{\partial x}{\partial x} +\frac{\partial f}{\partial y}\big(x,y,g(x,y)\big) \frac{\partial y}{\partial x} +\frac{\partial f}{\partial z}\big(x,y,g(x,y)\big) \frac{\partial g}{\partial x}(x,y)\\ &=\frac{\partial f}{\partial x}\big(x,y,g(x,y)\big) +\frac{\partial f}{\partial z}\big(x,y,g(x,y)\big) \frac{\partial g}{\partial x}(x,y) \end{aligned}$

이다. $w(x,y)=f\big(x,y,g(x,y)\big)$ 이지만 $\frac{\partial w}{\partial x}(x,y) =\frac{\partial f}{\partial x}\big(x,y,g(x,y)\big)$ 는 성립하지 않는다. 예를 들어 $f(x,y,z)=x-z$ 이고 $g(x,y)=x$ 라 하자. 그러면 모든 $(x,y)$ 에 대해 $w(x,y)=f\big(x,y,g(x,y)\big)=x-g(x,y)=0$ 이므로 $\frac{\partial w}{\partial x}(x,y)=0$ 인 반면, 모든 $(x,y,z)$ 에 대해 $\frac{\partial f}{\partial x}(x,y,z)=1$ 이다.

원본 교재 풀이(영어)

The problem is that $\frac{\partial w}{\partial x}$ is used to represent two completely different functions in the same equation. The careful way to write the equation is the following. Let $f(x,y,z)$ and $g(x,y)$ be continuously differentiable functions and define $w(x,y)=f\big(x,y,g(x,y)\big)$ . By the chain rule,

$\begin{aligned} \frac{\partial w}{\partial x}(x,y) &=\frac{\partial f}{\partial x}\big(x,y,g(x,y)\big) \frac{\partial x}{\partial x} +\frac{\partial f}{\partial y}\big(x,y,g(x,y)\big) \frac{\partial y}{\partial x} +\frac{\partial f}{\partial z}\big(x,y,g(x,y)\big) \frac{\partial g}{\partial x}(x,y)\\ &=\frac{\partial f}{\partial x}\big(x,y,g(x,y)\big) +\frac{\partial f}{\partial z}\big(x,y,g(x,y)\big) \frac{\partial g}{\partial x}(x,y) \end{aligned}$

While $w(x,y)=f\big(x,y,g(x,y)\big)$ , it is not true that $\frac{\partial w}{\partial x}(x,y) =\frac{\partial f}{\partial x}\big(x,y,g(x,y)\big)$ . For example, take $f(x,y,z)=x-z$ and $g(x,y)=x$ . Then $w(x,y)=f\big(x,y,g(x,y)\big)=x-g(x,y)=0$ for all $(x,y)$ , so that $\frac{\partial w}{\partial x}(x,y)=0$ while $\frac{\partial f}{\partial x}(x,y,z)=1$ for all $(x,y,z)$ .

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 6번

# 다변수 연쇄법칙계산기본자동 분류

64.CLP-3 · 2.4절 · 6번대학 수준우리말 번역

연쇄 법칙을 이용하는 방법을 포함하여 두 가지 방법으로 $\frac{\partial w}{\partial s}$ 와 $\frac{\partial w}{\partial t}$ 의 값을 구하세요. 단, 함수 $w=x^2+y^2+z^2$ 이고 $x=st,\ y=s\cos t$ , $z=s\sin t$ 입니다.

영어 원문 보기

Use two methods (one using the chain rule) to evaluate $\frac{\partial w}{\partial s}$ and $\frac{\partial w}{\partial t}$ given that the function $w=x^2+y^2+z^2$ , with $x=st,\ y=s\cos t$ and $z=s\sin t$ .

정답·풀이 보기

방법 1: $w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2$ 이고 $x(s,t)=st$ , $y(s,t)=s\cos t$ , $z(s,t)=s\sin t$ 이므로 $w(s,t)$ 를 명시적으로 나타낼 수 있다:

$\begin{aligned} w(s,t)&=(st)^2+(s\cos t)^2+(s\sin t)^2 =s^2(t^2+1)\\ \implies w_s(s,t)&=2s(t^2+1)\qquad\text{그리고}\qquad w_t(s,t)=s^2(2t) \end{aligned}$

방법 2:* 문제에서 $w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2$ 이고 $x(s,t)=st$ , $y(s,t)=s\cos t$ , $z(s,t)=s\sin t$ 로 주어졌다. 즉, $w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big)$ 이며 $W(x,y,z)= x^2+y^2+z^2$ 이다. $w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big)$ 에 연쇄 법칙을 적용하고 $\frac{\partial W}{\partial x}=2x$ , $\frac{\partial W}{\partial y}=2y$ , $\frac{\partial W}{\partial z}=2z$ 임을 이용하면 다음을 얻는다.

$\begin{aligned} \frac{\partial w}{\partial s}(s,t) &= \frac{\partial W}{\partial x}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial x}{\partial s}(s,t) + \frac{\partial W}{\partial y}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial y}{\partial s}(s,t)\\& + \frac{\partial W}{\partial z}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial z}{\partial s}(s,t) \\ &=2 x(s,t)\ x_s(s,t)+2 y(s,t)\ y_s(s,t)+2 z(s,t)\ z_s(s,t) \\ &=2 (st)\ t+2 (s\cos t)\ \cos t+2 (s\sin t)\ \sin t\\ &=2 st^2+2 s\\ \frac{\partial w}{\partial t}(s,t) &= \frac{\partial W}{\partial x}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial x}{\partial t}(s,t) + \frac{\partial W}{\partial y}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial y}{\partial t}(s,t)\\& + \frac{\partial W}{\partial z}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial z}{\partial t}(s,t) \\ &=2 x(s,t)\ x_t(s,t)+2 y(s,t)\ y_t(s,t)+2 z(s,t)\ z_t(s,t)\\ &=2 (st)\ s+2 (s\cos t)\ (-s\sin t)+2 (s\sin t)\ (s\cos t)\\ &=2 s^2t \end{aligned}$

원본 교재 풀이(영어)

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

$\begin{aligned} w(s,t)&=(st)^2+(s\cos t)^2+(s\sin t)^2 =s^2(t^2+1)\\ \implies w_s(s,t)&=2s(t^2+1)\qquad\text{and}\qquad w_t(s,t)=s^2(2t) \end{aligned}$

Method 2:* The question specifies that $w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2$ with $x(s,t)=st$ , $y(s,t)=s\cos t$ and $z(s,t)=s\sin t$ . That is, $w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big)$ with $W(x,y,z)= x^2+y^2+z^2$ . Applying the chain rule to $w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big)$ and noting that $\frac{\partial W}{\partial x}=2x$ , $\frac{\partial W}{\partial y}=2y$ , $\frac{\partial W}{\partial z}=2z$ , gives

$\begin{aligned} \frac{\partial w}{\partial s}(s,t) &= \frac{\partial W}{\partial x}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial x}{\partial s}(s,t) + \frac{\partial W}{\partial y}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial y}{\partial s}(s,t)\\& + \frac{\partial W}{\partial z}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial z}{\partial s}(s,t) \\ &=2 x(s,t)\ x_s(s,t)+2 y(s,t)\ y_s(s,t)+2 z(s,t)\ z_s(s,t) \\ &=2 (st)\ t+2 (s\cos t)\ \cos t+2 (s\sin t)\ \sin t\\ &=2 st^2+2 s\\ \frac{\partial w}{\partial t}(s,t) &= \frac{\partial W}{\partial x}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial x}{\partial t}(s,t) + \frac{\partial W}{\partial y}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial y}{\partial t}(s,t)\\& + \frac{\partial W}{\partial z}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial z}{\partial t}(s,t) \\ &=2 x(s,t)\ x_t(s,t)+2 y(s,t)\ y_t(s,t)+2 z(s,t)\ z_t(s,t)\\ &=2 (st)\ s+2 (s\cos t)\ (-s\sin t)+2 (s\sin t)\ (s\cos t)\\ &=2 s^2t \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 7번

# 다변수 연쇄법칙계산보통자동 분류

65.CLP-3 · 2.4절 · 7번대학 수준우리말 번역

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

영어 원문 보기

Evaluate $\frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy)$ in terms of partial derivatives of $f$ . You may assume that $f$ 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

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

이다.

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

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

이다.

여기서 아래첨자 $1$ 은 두 번째 변수를 고정한 채 $f$ 를 첫 번째 변수에 대해 편미분한다는 뜻이고, 아래첨자 $2$ 는 첫 번째 변수를 고정한 채 $f$ 를 두 번째 변수에 대해 편미분한다는 뜻이다. 다음으로 가운데 편도함수를 $H(x,y)=\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f(2x+3y,xy)\right)$ 라 하면

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

이다.

연쇄 법칙을 (두 번) 쓰면

$\begin{aligned} \frac{\partial }{\partial y}\Big(f_1(2x+3y,xy)\Big) &=f_{11}(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_{12}(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_{11}(2x+3y,xy)+xf_{12}(2x+3y,xy) \\ \frac{\partial }{\partial y}\Big(f_2(2x+3y,xy)\Big) &=f_{21}(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_{22}(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_{21}(2x+3y,xy)+xf_{22}(2x+3y,xy) \end{aligned}$

이므로

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

이다.

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

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

이다.

마지막으로

$\begin{aligned} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=\frac{\partial }{\partial x} H(x,y) \\ &=\frac{\partial }{\partial x}\Big(9f_{11}(2x+3y,xy)+6xf_{12}(2x+3y,xy)+x^2f_{22}(2x+3y,xy)\Big)\\ &=9\frac{\partial }{\partial x}\Big(f_{11}(2x+3y,xy)\Big)\\& +6f_{12}(2x+3y,xy) +6x\frac{\partial }{\partial x}\Big(f_{12}(2x+3y,xy)\Big) \\& +2xf_{22}(2x+3y,xy) +x^2\frac{\partial }{\partial x}\Big(f_{22}(2x+3y,xy)\Big) \end{aligned}$

이다.

연쇄 법칙을 세 번 적용하면

$\begin{aligned} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=9\Big(2f_{111}+yf_{112}\Big)\\& +6f_{12}+6x\Big(2f_{121}+yf_{122}\Big)\\& +2xf_{22}+x^2\Big(2f_{221}+yf_{222}\Big)\\ &=6\,f_{12}+2x\,f_{22}+18\,f_{111}+(9y+12x)\,f_{112} +(6xy+2x^2)\,f_{122}+x^2y\,f_{222} \end{aligned}$

이다.

우변의 모든 함수의 인수는 $(2x+3y,xy)$ 이다.

원본 교재 풀이(영어)

By definition,

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

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

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

Here the subscript $1$ means take the partial derivative of $f$ with respect to the first argument while holding the second argument fixed, and the subscript $2$ means take the partial derivative of $f$ with respect to the second argument while holding the first argument fixed. Next call the middle derivative $H(x,y)=\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f(2x+3y,xy)\right)$ so that

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

By the chain rule (twice),

$\begin{aligned} \frac{\partial }{\partial y}\Big(f_1(2x+3y,xy)\Big) &=f_{11}(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_{12}(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_{11}(2x+3y,xy)+xf_{12}(2x+3y,xy) \\ \frac{\partial }{\partial y}\Big(f_2(2x+3y,xy)\Big) &=f_{21}(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_{22}(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_{21}(2x+3y,xy)+xf_{22}(2x+3y,xy) \end{aligned}$

so that

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

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

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

Finally, we get to

$\begin{aligned} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=\frac{\partial }{\partial x} H(x,y) \\ &=\frac{\partial }{\partial x}\Big(9f_{11}(2x+3y,xy)+6xf_{12}(2x+3y,xy)+x^2f_{22}(2x+3y,xy)\Big)\\ &=9\frac{\partial }{\partial x}\Big(f_{11}(2x+3y,xy)\Big)\\& +6f_{12}(2x+3y,xy) +6x\frac{\partial }{\partial x}\Big(f_{12}(2x+3y,xy)\Big) \\& +2xf_{22}(2x+3y,xy) +x^2\frac{\partial }{\partial x}\Big(f_{22}(2x+3y,xy)\Big) \end{aligned}$

By three applications of the chain rule

$\begin{aligned} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=9\Big(2f_{111}+yf_{112}\Big)\\& +6f_{12}+6x\Big(2f_{121}+yf_{122}\Big)\\& +2xf_{22}+x^2\Big(2f_{221}+yf_{222}\Big)\\ &=6\,f_{12}+2x\,f_{22}+18\,f_{111}+(9y+12x)\,f_{112} +(6xy+2x^2)\,f_{122}+x^2y\,f_{222} \end{aligned}$

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

한 번 더 — 비슷한 문제 풀기APEX · 12.5절 · 2번

# 다변수 연쇄법칙계산보통자동 분류

66.CLP-3 · 2.4절 · 8번대학 수준우리말 번역

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

영어 원문 보기

Find all second order derivatives of $g(s,t)=f(2s+3t,3s-2t)$ . 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

주어진 함수는

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

이다.

일계도함수는

$\begin{aligned} g_s(s,t)&=2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\\ g_t(s,t)&=3f_1(2s+3t,3s-2t)-2f_2(2s+3t,3s-2t) \end{aligned}$

이다.

이계도함수는

$\begin{aligned} g_{ss}(s,t) &=\frac{\partial }{\partial s}\Big(2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\Big)\\ &=2\Big(2f_{11}+3f_{12}\Big) +3\Big(2f_{21}+3f_{22}\Big)\\ &=4f_{11}+6f_{12}+6f_{21}+9f_{22}\\ &=4f_{11}(2s+3t,3s-2t)+12f_{12}(2s+3t,3s-2t)+9f_{22}(2s+3t,3s-2t)\\ g_{st}(s,t)&=\frac{\partial }{\partial t}\Big(2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\Big)\\ &=2\Big(3f_{11}-2f_{12}\Big) +3\Big(3f_{21}-2f_{22}\Big)\\ &=6f_{11}(2s+3t,3s-2t)+5f_{12}(2s+3t,3s-2t)-6f_{22}(2s+3t,3s-2t)\\ g_{tt}(s,t)&=\frac{\partial }{\partial t}\Big(3f_1(2s+3t,3s-2t)-2f_2(2s+3t,3s-2t)\Big)\\ &=3\Big(3f_{11}-2f_{12}\Big) -2\Big(3f_{21}-2f_{22}\Big)\\ &=9f_{11}(2s+3t,3s-2t)-12f_{12}(2s+3t,3s-2t)+4f_{22}(2s+3t,3s-2t) \end{aligned}$

이다.

여기서 $f_1$ 은 $f$ 를 첫 번째 변수에 대해 편미분한 편도함수를 뜻하고, $f_{12}$ 는 $f$ 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

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

The first order derivatives are

$\begin{aligned} g_s(s,t)&=2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\\ g_t(s,t)&=3f_1(2s+3t,3s-2t)-2f_2(2s+3t,3s-2t) \end{aligned}$

The second order derivatives are

$\begin{aligned} g_{ss}(s,t) &=\frac{\partial }{\partial s}\Big(2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\Big)\\ &=2\Big(2f_{11}+3f_{12}\Big) +3\Big(2f_{21}+3f_{22}\Big)\\ &=4f_{11}+6f_{12}+6f_{21}+9f_{22}\\ &=4f_{11}(2s+3t,3s-2t)+12f_{12}(2s+3t,3s-2t)+9f_{22}(2s+3t,3s-2t)\\ g_{st}(s,t)&=\frac{\partial }{\partial t}\Big(2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\Big)\\ &=2\Big(3f_{11}-2f_{12}\Big) +3\Big(3f_{21}-2f_{22}\Big)\\ &=6f_{11}(2s+3t,3s-2t)+5f_{12}(2s+3t,3s-2t)-6f_{22}(2s+3t,3s-2t)\\ g_{tt}(s,t)&=\frac{\partial }{\partial t}\Big(3f_1(2s+3t,3s-2t)-2f_2(2s+3t,3s-2t)\Big)\\ &=3\Big(3f_{11}-2f_{12}\Big) -2\Big(3f_{21}-2f_{22}\Big)\\ &=9f_{11}(2s+3t,3s-2t)-12f_{12}(2s+3t,3s-2t)+4f_{22}(2s+3t,3s-2t) \end{aligned}$

Here $f_1$ denotes the partial derivative of $f$ with respect to its first argument, $f_{12}$ is the result of first taking one partial derivative of $f$ 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

67.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

68.CLP-3 · 2.4절 · 10번대학 수준우리말 번역

$z = f(x,y)$ 이고 $x = 2s + t$ , $y = s - t$ 라 하자. 다음 등식이 성립하도록 하는 상수 $a$ , $b$ , $c$ 의 값을 구하세요.

$a\frac{\partial^2 z}{\partial x^2} +b\frac{\partial^2 z}{\partial x\,\partial y} +c\frac{\partial^2 z}{\partial y^2} =\frac{\partial^2 z}{\partial s^2} +\frac{\partial^2 z}{\partial t^2}$

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

영어 원문 보기

Let $z = f(x,y)$ where $x = 2s + t$ and $y = s - t$ . Find the values of the constants $a$ , $b$ and $c$ such that

$a\frac{\partial^2 z}{\partial x^2} +b\frac{\partial^2 z}{\partial x\,\partial y} +c\frac{\partial^2 z}{\partial y^2} =\frac{\partial^2 z}{\partial s^2} +\frac{\partial^2 z}{\partial t^2}$

You may assume that $z = f(x,y)$ 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 $z$ 가 한 등식 안에서 두 가지 다른 뜻으로 쓰였기 때문이다. 좌변에서는 $x$ 와 $y$ 의 함수이고, 우변에서는 $s$ 와 $t$ 의 함수이다. 안타깝게도 이러한 표기의 남용은 매우 흔하다. $s$ 와 $t$ 의 함수를 $F(s,t)$ 로 바꾸어 표기의 충돌을 없애자. 즉,

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

이 새 표기에서는 다음이 성립하도록 $a$ , $b$ , $c$ 를 구하면 된다.

$\begin{aligned} a\frac{\partial^2 f}{\partial x^2} +b\frac{\partial^2 f}{\partial x\partial y} +c\frac{\partial^2 f}{\partial y^2} &=\frac{\partial^2 F}{\partial s^2} + \frac{\partial^2 F}{\partial t^2} \end{aligned}$

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

연쇄 법칙에 의해

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

이고

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

이다. 인수를 생략하면

$\begin{aligned} \frac{\partial^2 F}{\partial s^2} + \frac{\partial^2 F}{\partial t^2} &=5\frac{\partial^2 f}{\partial x^2} +2\frac{\partial^2 f}{\partial x\partial y} +2\frac{\partial^2 f}{\partial y^2} \end{aligned}$

이다. 마지막으로 문제의 (좋지 않은) 표기로 되돌리면

$\begin{aligned} \frac{\partial^2 z}{\partial s^2} + \frac{\partial^2 z}{\partial t^2} &=5\frac{\partial^2 z}{\partial x^2} +2\frac{\partial^2 z}{\partial x\partial y} +2\frac{\partial^2 z}{\partial y^2} \end{aligned}$

이므로 $a=5$ 이고 $b=c=2$ 이다.

원본 교재 풀이(영어)

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

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

In this new notation, we are being asked to find $a$ , $b$ and $c$ so that

$\begin{aligned} a\frac{\partial^2 f}{\partial x^2} +b\frac{\partial^2 f}{\partial x\partial y} +c\frac{\partial^2 f}{\partial y^2} &=\frac{\partial^2 F}{\partial s^2} + \frac{\partial^2 F}{\partial t^2} \end{aligned}$

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

By the chain rule,

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

and

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

Suppressing the arguments

$\begin{aligned} \frac{\partial^2 F}{\partial s^2} + \frac{\partial^2 F}{\partial t^2} &=5\frac{\partial^2 f}{\partial x^2} +2\frac{\partial^2 f}{\partial x\partial y} +2\frac{\partial^2 f}{\partial y^2} \end{aligned}$

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

$\begin{aligned} \frac{\partial^2 z}{\partial s^2} + \frac{\partial^2 z}{\partial t^2} &=5\frac{\partial^2 z}{\partial x^2} +2\frac{\partial^2 z}{\partial x\partial y} +2\frac{\partial^2 z}{\partial y^2} \end{aligned}$

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

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 26번

# 다변수 연쇄법칙계산보통자동 분류

69.CLP-3 · 2.4절 · 11번대학 수준우리말 번역

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

$\begin{aligned} \frac{\partial^2}{\partial x\, \partial y} F(x^2 - y^2 , 2xy) \end{aligned}$

를 함수 $F$ 의 편도함수를 사용하여 나타내세요.

영어 원문 보기

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

$\begin{aligned} \frac{\partial^2}{\partial x\, \partial y} F(x^2 - y^2 , 2xy) \end{aligned}$

in terms of partial derivatives of the function $F$ .

정답·풀이 보기

$u(x,y) = x^2 - y^2$ , $v(x,y) = 2xy$ 라 하자. 그러면 $F(x^2 - y^2 , 2xy) =F\big(u(x,y),v(x,y)\big)$ 이다. 연쇄 법칙에 의해

$\begin{aligned} \frac{\partial }{\partial y}F(x^2 - y^2 , 2xy) &=\frac{\partial }{\partial y}F(u(x,y) , v(xy)) \\ &=F_u(u(x,y) , v(xy))\frac{\partial u}{\partial y}(x,y) + F_v(u(x,y) , v(xy))\frac{\partial v}{\partial y}(x,y) \\ &= F_u(x^2 - y^2 , 2xy)\ (-2y) + F_v(x^2 - y^2 , 2xy)\, (2x) \\ \frac{\partial^2}{\partial x\, \partial y} F(x^2 - y^2 , 2xy) &=\frac{\partial }{\partial x}\left\{-2y F_u(x^2 - y^2 , 2xy) + 2x F_v(x^2 - y^2 , 2xy) \right\} \\ &= -2y\frac{\partial }{\partial x}\left[ F_u(x^2 - y^2 , 2xy)\right] +2 F_v(x^2 - y^2 , 2xy) \\& + 2x \frac{\partial }{\partial x}\left[F_v(x^2 - y^2 , 2xy)\right] \\ &= -4xy\,F_{uu}(x^2 - y^2 , 2xy) -4y^2 F_{uv}(x^2 - y^2 , 2xy) +2 F_v(x^2 - y^2 , 2xy) \\& + 4x^2 F_{vu}(x^2 - y^2 , 2xy) +4xy\,F_{vv}(x^2 - y^2 , 2xy) \\ &= 2 \, F_v(x^2 - y^2 , 2xy) -4xy\, F_{uu}(x^2 - y^2 , 2xy) \\& +4(x^2-y^2)\, F_{uv}(x^2 - y^2 , 2xy) \\& +4xy\, F_{vv}(x^2 - y^2 , 2xy) \end{aligned}$

원본 교재 풀이(영어)

Let $u(x,y) = x^2 - y^2$ , and $v(x,y) = 2xy$ . Then $F(x^2 - y^2 , 2xy) =F\big(u(x,y),v(x,y)\big)$ . By the chain rule

$\begin{aligned} \frac{\partial }{\partial y}F(x^2 - y^2 , 2xy) &=\frac{\partial }{\partial y}F(u(x,y) , v(xy)) \\ &=F_u(u(x,y) , v(xy))\frac{\partial u}{\partial y}(x,y) + F_v(u(x,y) , v(xy))\frac{\partial v}{\partial y}(x,y) \\ &= F_u(x^2 - y^2 , 2xy)\ (-2y) + F_v(x^2 - y^2 , 2xy)\, (2x) \\ \frac{\partial^2}{\partial x\, \partial y} F(x^2 - y^2 , 2xy) &=\frac{\partial }{\partial x}\left\{-2y F_u(x^2 - y^2 , 2xy) + 2x F_v(x^2 - y^2 , 2xy) \right\} \\ &= -2y\frac{\partial }{\partial x}\left[ F_u(x^2 - y^2 , 2xy)\right] +2 F_v(x^2 - y^2 , 2xy) \\& + 2x \frac{\partial }{\partial x}\left[F_v(x^2 - y^2 , 2xy)\right] \\ &= -4xy\,F_{uu}(x^2 - y^2 , 2xy) -4y^2 F_{uv}(x^2 - y^2 , 2xy) +2 F_v(x^2 - y^2 , 2xy) \\& + 4x^2 F_{vu}(x^2 - y^2 , 2xy) +4xy\,F_{vv}(x^2 - y^2 , 2xy) \\ &= 2 \, F_v(x^2 - y^2 , 2xy) -4xy\, F_{uu}(x^2 - y^2 , 2xy) \\& +4(x^2-y^2)\, F_{uv}(x^2 - y^2 , 2xy) \\& +4xy\, F_{vv}(x^2 - y^2 , 2xy) \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 10번

# 다변수 연쇄법칙계산보통자동 분류

70.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

71.CLP-3 · 2.4절 · 13번대학 수준우리말 번역

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

영어 원문 보기

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

답 쓰고 채점

정답·풀이 보기

연쇄 법칙에 의해

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

특히

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

이다.

원본 교재 풀이(영어)

By the chain rule,

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

In particular

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

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 12번

# 다변수 연쇄법칙계산보통자동 분류

72.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

73.CLP-3 · 2.4절 · 15번대학 수준우리말 번역

$z = f (x, y)$ 가 연속인 이계 편도함수를 갖고, $x = r \cos t$ , $y = r \sin t$ 라 하자. 다음 편도함수를 $r$ , $t$ , 그리고 $f$ 의 편도함수로 나타내세요.

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

영어 원문 보기

Suppose $z = f (x, y)$ has continuous second order partial derivatives, and $x = r \cos t$ , $y = r \sin t$ . Express the following partial derivatives in terms $r$ , $t$ , and partial derivatives of $f$ .

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

정답·풀이 보기

정의에 의해 $z(r,t) = f(r\cos t\,,\,r\sin t)$ 이다.

(a) 연쇄 법칙에 의해

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

(b) 선형성, 곱의 미분법, 연쇄 법칙에 의해

$\begin{aligned} \frac{\partial^2 z}{\partial t^2}(r,t) &= -\frac{\partial }{\partial t}\left[r\sin t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\right] +\frac{\partial }{\partial t}\left[r\cos t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\right] \\ &= -r\cos t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) -r\sin t\ \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\right] \\ & -r\sin t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) +r\cos t\ \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\right] \\ &= -r\cos t \ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) \\ & +r^2\sin^2 t\ \frac{\partial^2 f}{\partial x^2}(r\cos t\,,\,r\sin t) -r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial y\partial x}(r\cos t\,,\,r\sin t) \\ &\phantom{=}-r\sin t \ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) \\ & -r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial x\partial y}(r\cos t\,,\,r\sin t) +r^2\cos^2 t\ \frac{\partial^2 f}{\partial y^2}(r\cos t\,,\,r\sin t) \\ &= -r\cos t \ \frac{\partial f}{\partial x} -r\sin t \ \frac{\partial f}{\partial y} \\ & +r^2\sin^2 t\ \frac{\partial^2 f}{\partial x^2} -2r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial x\partial y} +r^2\cos^2 t\ \frac{\partial^2 f}{\partial y^2} \end{aligned}$

여기서 $f$ 의 모든 편도함수는 $(r\cos t\,,\,r\sin t)$ 에서의 값이다.

원본 교재 풀이(영어)

By definition $z(r,t) = f(r\cos t\,,\,r\sin t)$ .

(a) By the chain rule

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

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

$\begin{aligned} \frac{\partial^2 z}{\partial t^2}(r,t) &= -\frac{\partial }{\partial t}\left[r\sin t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\right] +\frac{\partial }{\partial t}\left[r\cos t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\right] \\ &= -r\cos t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) -r\sin t\ \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\right] \\ & -r\sin t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) +r\cos t\ \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\right] \\ &= -r\cos t \ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) \\ & +r^2\sin^2 t\ \frac{\partial^2 f}{\partial x^2}(r\cos t\,,\,r\sin t) -r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial y\partial x}(r\cos t\,,\,r\sin t) \\ &\phantom{=}-r\sin t \ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) \\ & -r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial x\partial y}(r\cos t\,,\,r\sin t) +r^2\cos^2 t\ \frac{\partial^2 f}{\partial y^2}(r\cos t\,,\,r\sin t) \\ &= -r\cos t \ \frac{\partial f}{\partial x} -r\sin t \ \frac{\partial f}{\partial y} \\ & +r^2\sin^2 t\ \frac{\partial^2 f}{\partial x^2} -2r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial x\partial y} +r^2\cos^2 t\ \frac{\partial^2 f}{\partial y^2} \end{aligned}$

with all of the partial derivatives of $f$ evaluated at $(r\cos t\,,\,r\sin t)$ .

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 25번

# 다변수 연쇄법칙활용보통자동 분류

74.CLP-3 · 2.4절 · 16번대학 수준우리말 번역

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

$f_x (2, 1) = 5, \qquad f_y(2, 1) =-2, \qquad f_{xx}(2, 1) = 2,\qquad f_{xy}(2, 1) = 1, \qquad f_{yy}(2, 1) = -4$

이다.

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

영어 원문 보기

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

$f_x (2, 1) = 5, \qquad f_y(2, 1) =-2, \qquad f_{xx}(2, 1) = 2,\qquad f_{xy}(2, 1) = 1, \qquad f_{yy}(2, 1) = -4$

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

답 쓰고 채점

정답·풀이 보기

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

$\begin{aligned} \frac{\mathrm{d}w}{\mathrm{d}t}(t)&= \frac{\mathrm{d}}{\mathrm{d}t} f\big(x(t),y(t)\big) \\ &= f_x(x(t)\,,\,y(t))\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) +f_y(x(t)\,,\,y(t))\,\frac{\mathrm{d}y}{\mathrm{d}t}(t) \\ &= 4t\,f_x(x(t)\,,\,y(t)) +3t^2\,f_y(x(t)\,,\,y(t)) \end{aligned}$

이다. 선형성, 곱의 미분법, 연쇄 법칙에 의해

$\begin{aligned} \frac{\mathrm{d^2}}{\mathrm{d}{t}^2} f\big(x(t),y(t)\big) &= \frac{\mathrm{d}}{\mathrm{d}t}\left[4t\,f_x\big(x(t),y(t)\big) \right] +\frac{\mathrm{d}}{\mathrm{d}t}\left[3t^2\,f_y\big(x(t),y(t)\big) \right] \\ &= 4\,f_x\big(x(t),y(t)\big) + 4t \frac{\mathrm{d}}{\mathrm{d}t}\left[f_x\big(x(t),y(t)\big) \right] \\& +6t\,f_y\big(x(t),y(t)\big) + 3t^2\frac{\mathrm{d}}{\mathrm{d}t}\left[f_y\big(x(t),y(t)\big) \right] \\ &= 4\,f_x(2t^2\,,\,t^3) + 4t\Big[f_{xx}\big(x(t),y(t)\big)\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) + f_{xy}\big(x(t),y(t)\big)\,\frac{\mathrm{d}y}{\mathrm{d}t}(t)\Big] \\ &\phantom{=} + 6t\,f_y(2t^2\,,\,t^3) + 3t^2\Big[f_{yx}\big(x(t),y(t)\big)\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) + f_{yy}\big(x(t),y(t)\big)\,\frac{\mathrm{d}y}{\mathrm{d}t}(t)\Big] \\ &= 4\,f_x(2t^2\,,\,t^3) + 16t^2\,f_{xx}(2t^2\,,\,t^3) + 12t^3\,f_{xy}(2t^2\,,\,t^3) \\ &\phantom{=} + 6t\,f_y(2t^2\,,\,t^3) + 12t^3\,f_{yx}(2t^2\,,\,t^3) + 9t^4\,f_{yy}(2t^2\,,\,t^3) \end{aligned}$

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

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

이다.

원본 교재 풀이(영어)

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

$\begin{aligned} \frac{\mathrm{d}w}{\mathrm{d}t}(t)&= \frac{\mathrm{d}}{\mathrm{d}t} f\big(x(t),y(t)\big) \\ &= f_x(x(t)\,,\,y(t))\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) +f_y(x(t)\,,\,y(t))\,\frac{\mathrm{d}y}{\mathrm{d}t}(t) \\ &= 4t\,f_x(x(t)\,,\,y(t)) +3t^2\,f_y(x(t)\,,\,y(t)) \end{aligned}$

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

$\begin{aligned} \frac{\mathrm{d^2}}{\mathrm{d}{t}^2} f\big(x(t),y(t)\big) &= \frac{\mathrm{d}}{\mathrm{d}t}\left[4t\,f_x\big(x(t),y(t)\big) \right] +\frac{\mathrm{d}}{\mathrm{d}t}\left[3t^2\,f_y\big(x(t),y(t)\big) \right] \\ &= 4\,f_x\big(x(t),y(t)\big) + 4t \frac{\mathrm{d}}{\mathrm{d}t}\left[f_x\big(x(t),y(t)\big) \right] \\& +6t\,f_y\big(x(t),y(t)\big) + 3t^2\frac{\mathrm{d}}{\mathrm{d}t}\left[f_y\big(x(t),y(t)\big) \right] \\ &= 4\,f_x(2t^2\,,\,t^3) + 4t\Big[f_{xx}\big(x(t),y(t)\big)\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) + f_{xy}\big(x(t),y(t)\big)\,\frac{\mathrm{d}y}{\mathrm{d}t}(t)\Big] \\ &\phantom{=} + 6t\,f_y(2t^2\,,\,t^3) + 3t^2\Big[f_{yx}\big(x(t),y(t)\big)\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) + f_{yy}\big(x(t),y(t)\big)\,\frac{\mathrm{d}y}{\mathrm{d}t}(t)\Big] \\ &= 4\,f_x(2t^2\,,\,t^3) + 16t^2\,f_{xx}(2t^2\,,\,t^3) + 12t^3\,f_{xy}(2t^2\,,\,t^3) \\ &\phantom{=} + 6t\,f_y(2t^2\,,\,t^3) + 12t^3\,f_{yx}(2t^2\,,\,t^3) + 9t^4\,f_{yy}(2t^2\,,\,t^3) \end{aligned}$

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

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

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 24번

# 다변수 연쇄법칙활용보통자동 분류

75.CLP-3 · 2.4절 · 17번대학 수준우리말 번역

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

영어 원문 보기

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

답 쓰고 채점

정답·풀이 보기

연쇄 법칙에 의해

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

이고

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

이고

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

이다.

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

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

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

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

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

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

따라서

$\begin{aligned} \frac{\partial^2G}{\partial\gamma ^2}(\gamma ,s,t) &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \\ &\phantom{=}+ \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \\ &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +2\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{aligned}$

이고

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

이다.

그러므로 변수를 생략하고 나타내면,

$\begin{aligned} \frac{\partial^2 G}{\partial \gamma^2} + \frac{\partial^2 G}{\partial s^2} -\frac{\partial G}{\partial t} &=2\frac{\partial^2 F}{\partial x^2} + 2\frac{\partial^2 F}{\partial y^2} -A \frac{\partial F}{\partial z} =2\frac{\partial F}{\partial z}-A \frac{\partial F}{\partial z} =0 \end{aligned}$

이다. 이는 $A=2$ 이면 성립한다.

원본 교재 풀이(영어)

By the chain rule

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

and

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

and

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

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

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

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

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

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

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

Consequently

$\begin{aligned} \frac{\partial^2G}{\partial\gamma ^2}(\gamma ,s,t) &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \\ &\phantom{=}+ \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \\ &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +2\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{aligned}$

and

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

So, suppressing the arguments,

$\begin{aligned} \frac{\partial^2 G}{\partial \gamma^2} + \frac{\partial^2 G}{\partial s^2} -\frac{\partial G}{\partial t} &=2\frac{\partial^2 F}{\partial x^2} + 2\frac{\partial^2 F}{\partial y^2} -A \frac{\partial F}{\partial z} =2\frac{\partial F}{\partial z}-A \frac{\partial F}{\partial z} =0 \end{aligned}$

if $A=2$ .

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 16번

# 다변수 연쇄법칙계산보통자동 분류

76.CLP-3 · 2.4절 · 18번대학 수준우리말 번역

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

영어 원문 보기

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

답 쓰고 채점

정답·풀이 보기

연쇄 법칙에 의해

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

특히

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

원본 교재 풀이(영어)

By the chain rule

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

In particular

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

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.5절 · 23번

# 다변수 연쇄법칙계산보통자동 분류

77.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

78.CLP-3 · 2.4절 · 20번대학 수준우리말 번역

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

(a) $w_{ss}$ 를 $u_{xx}$ , $u_{xy}$ , $u_{yy}$ 로 나타내세요 ( $u_{xy} = u_{yx}$ 라고 가정해도 됩니다). (b) $u_{xx} + u_{yy} = 0$ 이라 하자. 상수 $A$ 가 얼마일 때 $w_{ss} = Aw_{tt}$ 가 되는지 구하세요.

영어 원문 보기

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

(a) Find $w_{ss}$ in terms of $u_{xx}$ , $u_{xy}$ , and $u_{yy}$ (you can assume that $u_{xy} = u_{yx}$ ). (b) Suppose $u_{xx} + u_{yy} = 0$ . For what constant $A$ will $w_{ss} = Aw_{tt}$ ?

정답·풀이 보기

(a) 연쇄 법칙에 의해

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

이고

$\begin{aligned} w_{ss}(s,t) &= 2\frac{\partial }{\partial s}\big[u_x(2s + 3t, 3s - 2t)\big] + 3\frac{\partial }{\partial s}\big[u_y(2s + 3t, 3s - 2t)\big] \\ &= \big[4u_{xx}(2s + 3t, 3s - 2t) +6u_{xy}(2s + 3t, 3s - 2t) \big] \\& + \big[6u_{yx}(2s + 3t, 3s - 2t) +9u_{yy}(2s + 3t, 3s - 2t) \big] \\ &=4\,u_{xx}(2s + 3t, 3s - 2t) +12\,u_{xy}(2s + 3t, 3s - 2t) +9\,u_{yy}(2s + 3t, 3s - 2t) \end{aligned}$

이다.

(b) 다시 연쇄 법칙에 의해

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

이고

$\begin{aligned} w_{tt}(s,t) &= 3\frac{\partial }{\partial t}\big[u_x(2s + 3t, 3s - 2t)\big] - 2\frac{\partial }{\partial t}\big[u_y(2s + 3t, 3s - 2t)\big] \\ &= \big[9u_{xx}(2s + 3t, 3s - 2t) -6u_{xy}(2s + 3t, 3s - 2t) \big] \\& - \big[6u_{yx}(2s + 3t, 3s - 2t) -4u_{yy}(2s + 3t, 3s - 2t) \big] \\ &=9\,u_{xx}(2s + 3t, 3s - 2t) -12\,u_{xy}(2s + 3t, 3s - 2t) +4\,u_{yy}(2s + 3t, 3s - 2t) \end{aligned}$

이다.

따라서 임의의 상수 $A$ 에 대해

$\begin{aligned} w_{ss} - Aw_{tt} &= (4-9A) u_{xx} +(12+12A) u_{xy} + (9-4A) u_{yy} \end{aligned}$

이다.

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

원본 교재 풀이(영어)

(a) By the chain rule

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

and

$\begin{aligned} w_{ss}(s,t) &= 2\frac{\partial }{\partial s}\big[u_x(2s + 3t, 3s - 2t)\big] + 3\frac{\partial }{\partial s}\big[u_y(2s + 3t, 3s - 2t)\big] \\ &= \big[4u_{xx}(2s + 3t, 3s - 2t) +6u_{xy}(2s + 3t, 3s - 2t) \big] \\& + \big[6u_{yx}(2s + 3t, 3s - 2t) +9u_{yy}(2s + 3t, 3s - 2t) \big] \\ &=4\,u_{xx}(2s + 3t, 3s - 2t) +12\,u_{xy}(2s + 3t, 3s - 2t) +9\,u_{yy}(2s + 3t, 3s - 2t) \end{aligned}$

(b) Again by the chain rule

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

and

$\begin{aligned} w_{tt}(s,t) &= 3\frac{\partial }{\partial t}\big[u_x(2s + 3t, 3s - 2t)\big] - 2\frac{\partial }{\partial t}\big[u_y(2s + 3t, 3s - 2t)\big] \\ &= \big[9u_{xx}(2s + 3t, 3s - 2t) -6u_{xy}(2s + 3t, 3s - 2t) \big] \\& - \big[6u_{yx}(2s + 3t, 3s - 2t) -4u_{yy}(2s + 3t, 3s - 2t) \big] \\ &=9\,u_{xx}(2s + 3t, 3s - 2t) -12\,u_{xy}(2s + 3t, 3s - 2t) +4\,u_{yy}(2s + 3t, 3s - 2t) \end{aligned}$

Consquently, for any constant $A$ ,

$\begin{aligned} w_{ss} - Aw_{tt} &= (4-9A) u_{xx} +(12+12A) u_{xy} + (9-4A) u_{yy} \end{aligned}$

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

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.5절 · 15번

# 다변수 연쇄법칙계산보통자동 분류

79.CLP-3 · 2.4절 · 22번대학 수준우리말 번역

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

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

다음을 알고 있다고 하자.

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

또한

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

이고

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

이라고 하자.

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

${ \boldsymbol{\nabla} } z(1,2)$

영어 원문 보기

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

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

Suppose that we know

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

Suppose also that

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

and

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

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

${ \boldsymbol{\nabla} } z(1,2)$

정답·풀이 보기

연쇄 법칙에 의해

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

특히

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

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

원본 교재 풀이(영어)

By the chain rule

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

In particular

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

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

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 23번

# 다변수 연쇄법칙계산보통자동 분류

80.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