求导公式
(xn)′=nxn−1(x^n)'=nx^{n-1}(xn)′=nxn−1
(ex)′=ex(e^x)'=e^x(ex)′=ex
(lnx)′=1x(\ln x)'=\dfrac 1x(lnx)′=x1
(ax)′=axlna(a^x)'=a^x\ln a(ax)′=axlna
(logax)′=1xlna(log_ax)'=\dfrac{1}{x \ln a}(logax)′=xlna1
(sinx)′=cosx(\sin x)'=\cos x(sinx)′=cosx
(cosx)′=sinx(\cos x)'=\sin x(cosx)′=sinx
(tanx)′=sec2x(\tan x)'=\sec^2 x(tanx)′=sec2x
(cotx)′=−csc2x(\cot x)'=-\csc^2 x(cotx)′=−csc2x
(secx)′=secxtanx(\sec x)'=\sec x\tan x(secx)′=secxtanx
(cscx)′=−cscxcotx(\csc x)'=-\csc x\cot x(cscx)′=−cscxcotx
(arcsinx)′=11−x2(\arcsin x)'=\dfrac{1}{\sqrt{1-x^2}}(arcsinx)′=1−x21
(arccosx)′=−11−x2(\arccos x)'=-\dfrac{1}{\sqrt{1-x^2}}(arccosx)′=−1−x21
(arctanx)′=11+x2(\arctan x)'=\dfrac{1}{1+x^2}(arctanx)′=1+x21
(arccot x)′=−11+x2(\text{arccot} \ x)'=-\dfrac{1}{1+x^2}(arccot x)′=−1+x21
有关三角函数
sinxcscx=1\sin x\csc x=1sinxcscx=1
cosxsecx=1\cos x\sec x=1cosxsecx=1
tanxcotx=1\tan x\cot x=1tanxcotx=1
求导法则
[f(x)±g(x)]′=f′(x)±g′(x)[f(x)\pm g(x)]'=f'(x)\pm g'(x)[f(x)±g(x)]′=f′(x)±g′(x)
[f(x)⋅g(x)]′=f′(x)g(x)+f(x)g′(x)[f(x)\cdot g(x)]'=f'(x)g(x)+f(x)g'(x)[f(x)⋅g(x)]′=f′(x)g(x)+f(x)g′(x)
[f(x)g(x)]′=f′(x)g(x)−f(x)g′(x)g2(x)[\dfrac{f(x)}{g(x)}]'=\dfrac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}[g(x)f(x)]′=g2(x)f′(x)g(x)−f(x)g′(x)