导数列表

線性法则

${\displaystyle {{\mbox{d}}(Mf) \over {\mbox{d}}x}=M{{\mbox{d}}f \over {\mbox{d}}x};\qquad [Mf(x)]'=Mf'(x)}$

${\displaystyle {{{\mbox{d}}(f\pm g)} \over {{\mbox{d}}x}}={{\mbox{d}}f \over {\mbox{d}}x}\pm {{\mbox{d}}g \over {\mbox{d}}x}\ }$

乘法定则

${\displaystyle {{\mbox{d}}fg \over {\mbox{d}}x}={{\mbox{d}}f \over {\mbox{d}}x}g+f{\frac {{\mbox{d}}g}{{\mbox{d}}x}}}$

除法定则

${\displaystyle {\frac {{\mbox{d}}{\dfrac {f}{g}}}{{\mbox{d}}x}}={\frac {{\dfrac {{\mbox{d}}f}{{\mbox{d}}x}}g-f{\dfrac {{\mbox{d}}g}{{\mbox{d}}x}}}{g^{2}}}\qquad (g\neq 0)}$

倒数定则

${\displaystyle {\frac {{\mbox{d}}{\dfrac {1}{g}}}{{\mbox{d}}x}}={\frac {-{\dfrac {{\mbox{d}}g}{{\mbox{d}}x}}}{g^{2}}}\qquad (g\neq 0)}$

复合函数求导法则（連鎖定則）

${\displaystyle (f\circ g)'(x)=f'(g(x))g'(x).}$

${\displaystyle {\frac {{\mbox{d}}f[g(x)]}{{\mbox{d}}x}}={\frac {{\mbox{d}}f(g)}{{\mbox{d}}g}}{\frac {{\mbox{d}}g}{{\mbox{d}}x}}=f'[g(x)]g'(x)}$

反函数的导数

由于 ${\displaystyle g(f(x))=x}$，故 ${\displaystyle g(f(x))'=1}$，根據复合函数求导法则，則 ${\displaystyle g(f(x))'={\frac {{\mbox{d}}g[f(x)]}{{\mbox{d}}x}}={\frac {{\mbox{d}}g(f)}{{\mbox{d}}f}}{\frac {{\mbox{d}}f}{{\mbox{d}}x}}=1}$

所以 ${\displaystyle {\frac {{\mbox{d}}f}{{\mbox{d}}x}}={\frac {1}{\dfrac {{\mbox{d}}g(f)}{{\mbox{d}}f}}}=[{\frac {{\mbox{d}}g(f)}{{\mbox{d}}f}}]^{-1}=[g'(f)]^{-1}}$

同理 ${\displaystyle {\frac {{\mbox{d}}g}{{\mbox{d}}x}}={\frac {1}{\dfrac {{\mbox{d}}f(g)}{{\mbox{d}}g}}}=[{\frac {{\mbox{d}}f(g)}{{\mbox{d}}g}}]^{-1}=[f'(g)]^{-1}}$

广义幂法则

${\displaystyle (f^{g})'=\left(e^{g\ln f}\right)'=f^{g}\left(g'\ln f+{\frac {g}{f}}f'\right)}$

(n为任意实常数)

${\displaystyle {{\mbox{d}}M \over {\mbox{d}}x}=0}$

${\displaystyle {{\mbox{d}}x^{n} \over {\mbox{d}}x}=nx^{n-1}\qquad x\neq 0}$

${\displaystyle {{\mbox{d}}|x| \over {\mbox{d}}x}={x \over |x|}=\operatorname {sgn} x\qquad x\neq 0}$

${\displaystyle {d \over dx}e^{f(x)}=f'(x)e^{f(x)}}$.

${\displaystyle {\begin{aligned}{\frac {{\mbox{d}}\ e^{x}}{{\mbox{d}}x}}&=\lim _{\Delta x\to 0}{\frac {e^{x}-e^{x-\Delta x}}{\Delta x}}\\&=e^{x}\lim _{\Delta x\to 0}{\frac {1-e^{-\Delta x}}{\Delta x}}\\&=e^{x}\end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {{\mbox{d}}\ \alpha ^{x}}{{\mbox{d}}x}}&={\frac {{\mbox{d}}\ e^{x\!\ln \!\alpha }}{{\mbox{d}}x}}\\&={\frac {{\mbox{d}}e^{x\!\ln \!\alpha }}{{\mbox{d}}\ x\!\ln \!\alpha }}\cdot {\frac {{\mbox{d}}\ x\!\ln \!\alpha }{{\mbox{d}}x}}\\&=e^{x\!\ln \!\alpha }\!\ln \!\alpha \\&=\alpha ^{x}\!\ln \!\alpha \end{aligned}}}$

${\displaystyle {\begin{aligned}{\frac {{\mbox{d}}\ln x}{{\mbox{d}}x}}&=\lim _{h\to 0}{\frac {\ln(x+h)-\ln x}{h}}\\&=\lim _{h\to 0}({\frac {1}{h}}\ln({\frac {x+h}{x}}))\\&=\lim _{h\to 0}({\frac {x}{xh}}\ln(1+{\frac {h}{x}}))\\&={\frac {1}{x}}\ln(\lim _{h\to 0}(1+{\frac {h}{x}})^{\frac {x}{h}})\\&={\frac {1}{x}}\ln e\\&={\frac {1}{x}}\end{aligned}}}$

${\displaystyle {\frac {{\mbox{d}}\log _{\alpha }|x|}{{\mbox{d}}x}}={1 \over \ln \alpha }{\frac {{\mbox{d}}\ln |x|}{{\mbox{d}}x}}={1 \over x\ln \alpha }}$

${\displaystyle {\frac {{\mbox{d}}\ x^{x}}{{\mbox{d}}x}}=x^{x}(1+\ln x)}$

${\displaystyle {\begin{aligned}(\sin x)'&=\lim _{h\to 0}{\frac {\sin(x+h)-\sin x}{h}}\\&=\lim _{h\to 0}{\frac {\sin x\cos h+\cos x\sin h-\sin x}{h}}\\&=\lim _{h\to 0}(\sin x{\frac {\cos h-1}{h}}+\cos x{\frac {\sin h}{h}})\\&=\cos x\end{aligned}}}$

${\displaystyle {\begin{aligned}(\cos x)'&=\lim _{h\to 0}{\frac {\cos(x+h)-\cos x}{h}}\\&=\lim _{h\to 0}{\frac {\cos x\cos h-\sin x\sin h-\cos x}{h}}\\&=\lim _{h\to 0}(\cos x{\frac {\cos h-1}{h}}-\sin x{\frac {\sin h}{h}})\\&=-\sin x\end{aligned}}}$

${\displaystyle {\begin{aligned}(\tan x)'&=({\frac {\sin x}{\cos x}})'\\&={\frac {(\sin x)'\cos x-\sin x(\cos x)'}{\cos ^{2}x}}\\&={\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}\\&={\frac {1}{\cos ^{2}x}}=\sec ^{2}x\end{aligned}}}$

${\displaystyle {\begin{aligned}(\cot x)'&=({\frac {\cos x}{\sin x}})'\\&={\frac {(\cos x)'\sin x-\cos x(\sin x)'}{\sin ^{2}x}}\\&={\frac {-\sin ^{2}x-\cos ^{2}x}{\sin ^{2}x}}\\&=-{\frac {1}{\sin ^{2}x}}=-\csc ^{2}x\end{aligned}}}$

${\displaystyle {\begin{aligned}(\sec x)'&=({\frac {1}{\cos x}})'\\&={\frac {\sin x}{\cos ^{2}x}}\\&=\sec x\tan x\end{aligned}}}$

${\displaystyle {\begin{aligned}(\csc x)'&=({\frac {1}{\sin x}})'\\&={\frac {-\cos x}{\sin ^{2}x}}\\&=-\csc x\cot x\end{aligned}}}$

${\displaystyle {\begin{aligned}(\arcsin x)'&={\frac {1}{\cos(\arcsin x)}}\Leftrightarrow \sin(\arcsin x)=x\Leftrightarrow \cos(\arcsin x)(\arcsin x)'=1\\&={\frac {1}{\sqrt {1-\sin ^{2}(\arcsin x)}}}\\&={\frac {1}{\sqrt {1-x^{2}}}}\ \ (\left|x\right|<1)\end{aligned}}}$

${\displaystyle {\begin{aligned}(\arccos x)'&={\frac {1}{-\sin(\arccos x)}}\Leftrightarrow \cos(\arccos x)=x\Leftrightarrow -\sin(\arccos x)(\arccos x)'=1\\&=-{\frac {1}{\sqrt {1-\cos ^{2}(\arccos x)}}}\\&=-{\frac {1}{\sqrt {1-x^{2}}}}\ \ (\left|x\right|<1)\end{aligned}}}$

${\displaystyle {\begin{aligned}(\arctan x)'&={\frac {1}{\sec ^{2}(\arctan x)}}\Leftrightarrow \tan(\arctan x)=x\Leftrightarrow \sec ^{2}(\arctan x)(\arctan x)'=1\\&={\frac {1}{1+\tan ^{2}(\arctan x)}}\\&={\frac {1}{1+x^{2}}}\end{aligned}}}$

${\displaystyle {\begin{aligned}(\operatorname {arccot} x)'&={\frac {1}{-\csc ^{2}(\operatorname {arccot} x)}}\Leftrightarrow \cot(\operatorname {arccot} x)=x\Leftrightarrow -\csc ^{2}(\operatorname {arccot} x)(\operatorname {arccot} x)'=1\\&=-{\frac {1}{1+\cot ^{2}(\operatorname {arccot} x)}}\\&=-{\frac {1}{1+x^{2}}}\end{aligned}}}$

${\displaystyle {\begin{aligned}(\operatorname {arcsec} x)'&={\frac {1}{\sec(\operatorname {arcsec} x)\tan(\operatorname {arcsec} x)}}\Leftrightarrow \sec(\operatorname {arcsec} x)=x\Leftrightarrow \sec(\operatorname {arcsec} x)\tan(\operatorname {arcsec} x)(\operatorname {arcsec} x)'=1\\&={\frac {1}{|x|{\sqrt {\sec ^{2}(\operatorname {arcsec} x)-1}}}}\\&={\frac {1}{|x|{\sqrt {x^{2}-1}}}}\ \ (\left|x\right|>1)\end{aligned}}}$

${\displaystyle {\begin{aligned}(\operatorname {arccsc} x)'&={\frac {1}{-\csc(\operatorname {arccsc} x)\cot(\operatorname {arccsc} x)}}\Leftrightarrow \csc(\operatorname {arccsc} x)=x\Leftrightarrow -\csc(\operatorname {arccsc} x)\cot(\operatorname {arccsc} x)(\operatorname {arccsc} x)'=1\\&=-{\frac {1}{|x|{\sqrt {\csc ^{2}(\operatorname {arcsec} x)-1}}}}\\&=-{\frac {1}{|x|{\sqrt {x^{2}-1}}}}\ \ (\left|x\right|>1)\end{aligned}}}$