다변수 연쇄법칙

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

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

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 $(x,y,z)$.