다변수 연쇄법칙

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 $(x,y,z)$.