# 高等数学——复杂函数的求导方法

2019/04/10 10:10

<br>

<br>

## 函数四则运算求导法则

<br>

\begin{aligned} \left[u(x) \pm v(x)\right]'&= u'(x) \pm v'(x) \ \left[u(x)v(x)\right]' &= u'(x)v(x) + u(x)v'(x) \ \left[\frac{u(x)}{v(x)}\right] &= \frac{u'(x)v(x)-u(x)v'(x)}{v^2(x)} (v(x) \neq 0) \end{aligned}

\begin{aligned} \left[u(x) \pm v(x) \right]' &= \lim_{\Delta x \to 0} \frac{\left[u(x+\Delta x) \pm v(x + \Delta x) \right] - \left[u(x) \pm v(x) \right] }{\Delta x} \ &= \lim_{\Delta x \to 0}\frac{u(x+\Delta x)}{\Delta x} \pm \lim_{\Delta x \to 0} \frac{v(x+\Delta x)}{\Delta x} \ &= u'(x) \pm v'(x) \end{aligned}

\begin{aligned} \left[u(x)v(x)\right]' &= \lim_{\Delta x \to 0} \frac{u(x+\Delta x) v(x + \Delta x) - u(x) v(x)}{\Delta x} \ &= \lim_{\Delta x \to 0} \frac{u(x+\Delta x) v(x + \Delta x) - u(x)v(x+ \Delta x) + u(x)v(x+\Delta x) - u(x) v(x)}{\Delta x} \ &= \lim_{\Delta x \to 0} \frac{(u(x+\Delta x) - u(x))v(x+\Delta x) + u(x)(v(x+\Delta x) - v(x))}{\Delta x} \ &= \lim_{\Delta x \to 0}v(x+\Delta x) \frac{u(x+\Delta x) - u(x)}{\Delta x} + \lim_{\Delta x \to 0}u(x)\frac{v(x+\Delta x) - v(x)}{\Delta x}\ &=v(x+\Delta x)u'(x) + u(x)v'(x) \ &=u(x)v'(x) + u'(x)v(x) \end{aligned}

\displaystyle \begin{aligned} \left[\frac{u(x)}{v(x)}\right] &= \lim_{\Delta x \to 0}\frac{\frac{u(x+\Delta x)}{v(x+\Delta x)} - \frac{u(x)}{v(x)}}{\Delta x} \ &= \lim_{\Delta x \to 0}\frac{v(x)u(x+\Delta x)-v(x+\Delta x)u(x)}{v(x+\Delta x)v(x)\Delta x} \ &=\lim_{\Delta x \to 0} \ &= \lim_{\Delta x \to 0}\frac{v(x)u(x+\Delta x)-v(x)u(x)+v(x)u(x)-v(x+\Delta x)u(x)}{v(x+\Delta x)v(x)\Delta x} \ &=\lim_{\Delta x \to 0} \frac{\frac{u(x+\Delta x)-u(x)}{\Delta x}v(x)-\frac{v(x+\Delta x)-v(x)}{\Delta x}u(x)}{v(x+\Delta x)v(x)}\ &=\frac{u'(x)v(x)-u(x)v'(x)}{v^2(x)} \end{aligned}

<br>

## 反函数求导法则

<br>

$$\left[f^{-1}(x)\right]'=\frac{1}{f'(y)}$$

\begin{aligned} \Delta y=f^{-1}(x+\Delta x)-f^{-1}x \neq 0 \ \frac{\Delta y}{\Delta x} = \frac{1}{\frac{\Delta x}{\Delta y}}=\frac{1}{f'(y)} \end{aligned}

$$(\arcsin x)'=\frac{1}{(\sin y)'}=\frac{1}{\cos y}$$

$$(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}$$

<br>

## 复合函数求导

<br>

$$\frac{dy}{dx}=f'(u)\cdot g'(x)=\frac{dy}{du}\cdot \frac{du}{dx}$$

$$\displaystyle\lim_{\Delta u \to 0}\frac{\Delta y}{\Delta u} = f'(u)$$

$$\frac{\Delta y}{\Delta u} = f'(u) + a$$

$$\Delta y = f'(u)\Delta u + a\cdot \Delta u$$

$$\displaystyle\frac{\Delta y}{\Delta x}=f'(u)\frac{\Delta u}{\Delta x} + a\cdot\frac{\Delta u}{\Delta x}$$

$$\displaystyle \lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x \to 0}[f'(u)\frac{\Delta u}{\Delta x}+a\frac{\Delta u}{\Delta x}]$$

$$\displaystyle \lim_{\Delta x \to 0}\frac{\Delta u}{\Delta x}=g'(x)$$

$$\displaystyle \lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}=f'(u)\cdot \frac{\Delta u}{\Delta x}=f'(u)\cdot g'(x)$$

$y=\ln \sin 3x$，求$\frac{dy}{dx}$

\begin{aligned} \frac{dy}{dx}&=\frac{dy}{dg}\cdot \frac{dg}{du}\cdot\frac{du}{dx}\ &=\frac{1}{g}\cdot \cos u\cdot 3\ &=3\frac{\cos 3x}{\sin 3x} \ &=3 \cot 3x \end{aligned}

$$f(\theta) = \frac{1}{m}(\theta X-Y)^2$$

\begin{aligned} f'(\theta) &= \frac{1}{m}\cdot 2 \cdot (\theta X - Y)\cdot X \ &=\frac{2}{m}X^T(\theta X - Y) \end{aligned}

0
0 收藏

0 评论
0 收藏
0