函数的求导法则

函数的和差积商的求导法则

如果函数$u=u(x)$和$v=v(x)$都在点$x$处导，那么它们的和差积商（分母不为0）都在点$x$处可导，

$$(u(x) \pm v(x))'=u'(x) \pm v'(x) \\ (u(x)v(x))'=u'(x)v(x)+u(x)v'(x) \\ [\frac{u(x)}{v(x)}]' = \frac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}$$ $$(tan x)' = sec^2 x \\ (sec x)' = sec x \cdot tan x \\ (csc x)' = - csc x \cdot cot x \\ (cot x)' = - csc^2 x \\$$

反函数的求导法则

如果函数$x =f(y)$在区间I内单调可导且$f'(y) \neq 0$，则它的反函数$y=f^{-1}(x)$在区间$f(I)$内也可导，且

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

简单来说就是反函数的导数等于直接函数导数的倒数

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

复合函数的求导法则

如果函数$u=u(x)$在点$x$处可导，而函数$y=f[u(x)]$在点$u(x)$处也可导，那么复合函数$y=f[u(x)]$就在点$x$处可导，其导数为

$$[f[u(x)]]' = f'(u(x)) \cdot u'(x) \text{或者} \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

基本初等函数的求导公式

(C)' = 0 \\ (x^n)' = nx^{n-1} \\ (sin x)' = cos x \\ (cos x)' = -sin x \\ (tan x)' = sec^2 x \\ (cot x)' = -csc^2 x \\ (sec x)' = sec x \cdot tan x \\ (csc x)' = - csc x \cdot cot x \\ (a^x)' = a^x \cdot ln(a) \\ (e^x)' = e^x \\ (ln x)' = \frac{1}{x} (log_a x)' = \frac{1}{x \cdot ln(a)} \\ (arcsin x)' = \frac{1}{\sqrt{1-x^2}} \\ (arccos x)' = -\frac{1}{\sqrt{1-x^2}} \\ (arctan x)' = \frac{1}{1+x^2} \\ (arccot x)' = -\frac{1}{1+x^2} $$

sh x = \frac{e^x - e^{-x}}{2} \ ch x = \frac{e^x + e^{-x}}{2} \ th x = \frac{e^x - e^{-x}}{e^x + e^{-x}} \ (sh x)' = ch x \ (ch x)' = sh x \ (th x)' = -th x + \frac{1}{2} $$

$$arsh x = ln(x + \sqrt{1+x^2}) \\ arch x = ln(x + \sqrt{x^2 - 1}) \\ (arsh x)' = \frac{1}{\sqrt{1+x^2}} \\ (arch x)' = \frac{1}{\sqrt{x^2 - 1}}$$