【数学分析笔记】第4章第4节 复合函数求导法则及其应用（1）

4. 微分

4.4 复合函数求导法则及其应用

4.4.1 复合函数求导法则

【定理4.4.1】$u=g(x)$在$x=x_0$可导，$g(x_0)=u(0)$，$y=f(u)$在$u=u_0$可导，则$y=f(g(x))$在$x=x_0$可导且$[f(g(x)){]}_{x=x_0}^{\prime }=f^{\prime }(u_0)g^{\prime }(x_0)$
【一个有缺陷的证明！！！】$g(x_0+\Delta x)-g(x_0)=\Delta u$，$\Delta y=f(u_0+\Delta u)-f(u_0)$，$[f(g(x)){]}_{x=x_0}^{\prime }=\underset{\Delta x\to 0}{\lim }\frac{\Delta y}{\Delta x}$
当$\Delta \to 0,\Delta u\to 0$，则：
$[f(g(x)){]}_{x=x_0}^{\prime }=\underset{\Delta x\to 0}{\lim }\frac{\Delta y}{\Delta x}=\underset{\Delta x\to 0}{\lim }(\frac{\Delta y}{\Delta u}\cdot \frac{\Delta u}{\Delta x})=\underset{\Delta u\to 0}{\lim }\frac{\Delta y}{\Delta u}\cdot \underset{\Delta x\to 0}{\lim }\frac{\Delta u}{\Delta x}=f^{\prime }(u_0)g^{\prime }(x_0)$
【问题所在】$\Delta u\to 0,\Delta u\ne 0$，但是$\underset{\Delta u\to 0}{\lim }\frac{\Delta y}{\Delta u}$中，$\Delta u$可能为0，比如有下面的反例：

$u(x)=\{\begin{array}{c}{x}^2\sin \frac{1}{x}\\ 0\end{array}$，如果求0点的复合函数导数，则
$\Delta u=u(0+\Delta x)-u(0)=(\Delta x{)}^2\sin \frac{1}{\Delta x}$
则$u^{\prime }(0)=\underset{\Delta x\to 0}{\lim }\frac{(\Delta x{)}^2\sin \frac{1}{\Delta x}}{\Delta x}=0$，根据海涅定理，取$\Delta x_n=\frac{1}{n\pi }\ne 0$，但是此时$\Delta u=\frac{1}{n^2{\pi }^2}\sin (n\pi )=0$，这就出现了$\Delta u=0$的情况，如果这个函数和别的函数$f(u)$复合，则在计算$\underset{\Delta u\to 0}{\lim }\frac{\Delta y}{\Delta u}$时，会出现分母确确实实是真0而不是趋近于0的情况，所以不能这样证明。
【更改】由$f(u)$可导，则$f(u)$可微，$\Delta y=f(u_0+\Delta u)-f(u_0)=f^{\prime }(u_0)\Delta u+o(\Delta u)$
令$\alpha =\frac{o(\Delta u)}{\Delta u},\underset{\Delta u\to 0,\Delta u\ne 0}{\lim }\alpha =0$
则$\Delta y=f(u_0+\Delta u)-f(u_0)=f^{\prime }(u_0)\Delta u+\alpha \Delta u(\Delta u\to 0)$
当$\Delta u=0,\alpha =0$，对上式$\Delta y=f^{\prime }(u_0)\Delta u+\alpha \Delta u$也成立
$\frac{\Delta y}{\Delta x}=f^{\prime }(u_0)\frac{\Delta u}{\Delta x}+\alpha \frac{\Delta u}{\Delta x}$
令$\Delta x\to 0\Rightarrow \Delta u\to 0(\Delta u=0或\Delta u\ne 0)\Rightarrow \underset{\Delta u\to 0}{\lim }\alpha =0$
$\underset{\Delta x\to 0}{\lim }\frac{\Delta y}{\Delta x}=\underset{\Delta x\to 0}{\lim }{f}^{\prime }(u_0)\frac{\Delta u}{\Delta x}+\underset{\Delta x\to 0}{\lim }\alpha \frac{\Delta u}{\Delta x}=f^{\prime }(u_0)g^{\prime }(x_0)+0\times g^{\prime }(x_0)=f^{\prime }(u_0)g^{\prime }(x_0)$
【复合函数链式求导法则】
$y=f(u),u=g(x),y=f(g(x))=f\circ g(x)$
$[f(g(x)){]}^{\prime }=\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$
【例】$y=f(u),u=g(x),x=h(t),\frac{dy}{dt}=\frac{dy}{du}\cdot \frac{du}{dx}\cdot \frac{dx}{dt}=f^{\prime }(u_0)g^{\prime }(x_0)h^{\prime }(t_0)$

【例4.4.1】$y=x^{\alpha }=e^{\alpha \ln x},y=e^u,u=\alpha \ln x,\frac{dy}{dx}=(e^u{)}^{\prime }(\alpha \ln x{)}^{\prime }=e^u\cdot \frac{\alpha }{x}=x^{\alpha }\cdot \frac{\alpha }{x}=\alpha x^{\alpha -1}$


【例4.4.2】$y=e^{\cos x}$，求$y^{\prime }$
【解】$\frac{dy}{dx}=(e^u{)}^{\prime }(\cos x{)}^{\prime }=-e^{\cos x}\sin x$

【例】求$(\sqrt{1+x^2}{)}^{\prime }$
【解】$(\sqrt{1+x^2}{)}^{\prime }=\frac{1}{2\sqrt{1+x^2}}\cdot 2x=\frac{x}{\sqrt{1+x^2}}$