Section02_求导工具

Notes

1. 基本初等函数： $$\begin{cases}x^{a}\\a^{x}& (a>0\ \&\ a\neq1 )\\\log_{a}x&(a>0\ \& a\neq 1)\\\sin x,\cos x, \tan x,\cot x,\sec x, \csc x\\\arcsin x,\arccos x, \arctan x,\text{arccot } x\end{cases}$$
2. 初等函数：由$\begin{cases}\text{常数}\\\text{基本初等函数}\end{cases}$经过$\begin{cases}\text{四则运算}\\\text{复合}\end{cases}$而成的函数

基本公式

（一）基本公式

$n$ 阶导数

• $(\sin x)^{(n)} = \sin(x+\frac{n\pi}{2})$
• $(\cos x)^{(n)} = \cos(x+\frac{n\pi}{2})$
• $\displaystyle{\bigg(\frac{1}{ax+b}\bigg)^{n}=\frac{(-1)^{n}n!a^{n}}{(ax+b)^{n+1}}}$

（二）求导法则

四则运算

1. $(u\pm v)' = u'\pm v'$
2. $(uv)'=u'v+v'u\quad (uvw)'=u'vw+uv'w+uvw'$
3. $(\frac{u}{v})'=\frac{u'v-uv'}{v^{2}}$

复合法则

• $y=f(u)$ 可导，$u=\phi(x)$ 可导，且 $\phi'(x)\neq 0$ 则 $y=f(\phi(x))$ 可导，且 $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=f'(u)\phi'(x)=f'(\phi(x))\phi'(x)$
• Proof $$\begin{split} & f'(u) = \lim_{\Delta u\to 0}\frac{\Delta y}{\Delta u},\ \phi'(x)=\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x}\neq 0\Rightarrow\Delta u = O(\Delta x)\\ \Rightarrow & \frac{dy}{dx} = \lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x\to0}\frac{\Delta y}{\Delta u}\frac{\Delta u}{\Delta x} = \lim_{\Delta u\to 0}\frac{\Delta y}{\Delta u}\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x} \\ & = f'(u)\phi'(x) = f'(\phi(x))\phi'(x) \end{split}$$

（三）反函数求导

• 定理 $y=f(x)$ 可导且 $f'(x)\neq 0$ （$f(x)$ 严格单调）$\Rightarrow x=\phi(y)$可导，且：$\phi'(y)=\frac{1}{f'(x)}$
  • Proof $$\begin{split}&y=f(x), x=\phi(y)\\ \Rightarrow & y=f(\phi(y))\\ \Rightarrow& 1=f'(\phi(y))\phi'(y)\\ \Rightarrow &\phi'(y) = \frac{1}{f'(x)}\end{split}$$
• 例 求 $y=\ln(x+\sqrt{1+x^{2}})$ 的反函数
• 解 $$\begin{array}{ll} \because & y = \ln(x+\sqrt{1+x^{2}}) \\ \therefore & e^{y} = x+\sqrt{1+x^{2}}\\ \because & (-x+\sqrt{1+x^{2}})(x+\sqrt{1+x^{2}})=1 \\ \therefore & -x+\sqrt{1+x^{2}}=\frac{1}{x+\sqrt{1+x^{2}}} = e^{-y} \\ \therefore & 2x = e^{y}-e^{-y} \\ & x= \frac{1}{2}(e^{y}-e^{-y}) \end{array}$$