# 复合函数的导数（链式法则）

\begin{align*}\frac{dy}{dx}&=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta u}\cdot \frac{\Delta u}{\Delta x}\\&=\lim_{\Delta u\to 0}\frac{\Delta y}{\Delta u}\cdot\lim_{\Delta x\to 0} \frac{\Delta u}{\Delta x}\\&=\frac{dy}{du}\cdot\frac{du}{dx}=f'(u)\cdot g'(x)\\&=f'(g(x))\cdot g'(x)\end{align*}

\begin{align*}\frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}=(\ln u)'(\sin x)’\\&=\frac{1}{u}\cdot \cos x=\frac{\cos x}{\sin x}\\&=\tan x\end{align*}

$y’=\frac{dy}{dx}=(e^u)'(x^3)’=e^u\cdot 3x^2=3x^2e^{x^3}$

\begin{align*}f'(x)=\left((x^2+4x-7)^{\frac{1}{2}}\right)’&=\frac{1}{2}(x^2+4x-7)^{-\frac{1}{2}}\cdot (x^2+4x-7)’\\ &=\frac{1}{2}(x^2+4x-7)^{-\frac{1}{2}}\cdot(2x+4)\end{align*}

\begin{align*}f'(x)&=(\sin(e^{x^2}))’=\cos(e^{x^2})(e^{x^2})’\\ &=\cos(e^{x^2})e^{x^2}(x^2)’=\cos(e^{x^2})e^{x^2}2x\end{align*}

$f'(x)=\frac{1}{2}\left(x+\sqrt{x+\sqrt{x}}\right)^{-\frac{1}{2}}\cdot \left(x+\sqrt{x+\sqrt{x}}\right)’$ 这时候就要注意了，括号里面是两部分的和，

\begin{align*}f'(x)&=\frac{1}{2}\left(x+\sqrt{x+\sqrt{x}}\right)^{-\frac{1}{2}}\cdot \left(x+\sqrt{x+\sqrt{x}}\right)’\\ &=\frac{1}{2}\left(x+\sqrt{x+\sqrt{x}}\right)^{-\frac{1}{2}}\cdot \left(1+\left(\sqrt{x+\sqrt{x}}\right)’\right)\\&=\frac{1}{2}\left(x+\sqrt{x+\sqrt{x}}\right)^{-\frac{1}{2}}\cdot \left(1+\left(x+\sqrt{x}\right)^{-\frac{1}{2}}\cdot\left(x+\sqrt{x}\right)’\right)\\&=\frac{1}{2}\left(x+\sqrt{x+\sqrt{x}}\right)^{-\frac{1}{2}}\cdot \left(1+\left(x+\sqrt{x}\right)^{-\frac{1}{2}}\cdot\left(1+\frac{1}{2}x^{-\frac{1}{2}}\right)\right)\end{align*}