证明复合函数求导法则
定理
\[\begin{align}
复合函数求导法则(亦称链式法则):\\
设\mu=\psi(x)在点x处可导,\quad y=f(\mu)在对应点\mu=\psi(x)处可导,\\
那么复合函数y=f[\psi(x)]在点x处可导,\quad 则有: \quad y'(x)=f'(\mu) \cdot \psi'(x)\\
或: \quad \frac{dy}{dx}=\frac{dy}{d\mu} \cdot \frac{d\mu}{dx}
\end{align}
\]
推导过程
\[\begin{align}
即推导: \quad 若设 \mu=\psi(x),y=f(\mu), \ y'=?
\\ \\
y=f(\mu)=f[\psi(x)] 为复合函数, \ y'=f'[\psi(x)]
\\ \\
\Delta \mu=\psi(x+\Delta x)-\psi(x)\\
则 \psi(x+\Delta x)= \psi(x)+\Delta \mu=\mu+\Delta \mu
\\ \\
f'[\psi(x)]=\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}
=\lim_{\Delta x \to 0} \frac{f[\psi(x+ \Delta x)] -f[\psi(x)] } { \Delta x }
\\ \\
=\lim_{\Delta x \to 0}\frac{ f(\mu+\Delta \mu)-f(\mu)}{\Delta x}
\\ \\
分子分母同乘以一个式子:\\
\lim_{\Delta x \to 0}\frac{ f(\mu+\Delta \mu)-f(\mu) \cdot [\psi(x+\Delta x)-\psi(x)] }
{\Delta x \cdot [\psi(x+\Delta x)-\psi(x)]}
\\ \\
\lim_{\Delta x \to 0}\frac{ f(\mu+\Delta \mu)-f(\mu) }{[\psi(x+\Delta x)-\psi(x)]} \cdot
\frac{ \psi(x+\Delta x)-\psi(x) }
{\Delta x}
\\ \\
\lim_{\Delta x \to 0} \frac{[\psi(x+\Delta x)-\psi(x)]}{\Delta x}=\psi'(x)
\\ \\
\lim_{\Delta x \to 0} \frac{ f(\mu+\Delta \mu)-f(\mu) }{[\psi(x+\Delta x)-\psi(x)]}
=\frac{ f(\mu+\Delta \mu)-f(\mu) }{ \mu+\Delta \mu-\mu }
\\ \\
=\lim_{\Delta x \to 0} \frac{ f(\mu+\Delta \mu)-f(\mu) }{\Delta \mu}=f'(\mu)
\\ \\
\therefore y'=f'(\mu)\psi'(x)
\end{align}
\]
