倒数定则

${\displaystyle {\frac {1}{x^{2}+2x}}}$的导数为：

${\displaystyle {\frac {d}{dx}}\left({\frac {1}{x^{2}+2x}}\right)={\frac {-2x-2}{(x^{2}+2x)^{2}}}.}$

${\displaystyle {\frac {1}{\cos(x)}}}$的导数为：

${\displaystyle {\frac {d}{dx}}\left({\frac {1}{\cos(x)}}\right)={\frac {\sin(x)}{\cos ^{2}(x)}}={\frac {1}{\cos(x)}}{\frac {\sin(x)}{\cos(x)}}=\sec(x)\tan(x).}$

设 ${\displaystyle f(x)=1}$，则根据除法定则可得

${\displaystyle {\frac {d}{dx}}\left({\frac {1}{g(x)}}\right)={\frac {d}{dx}}\left({\frac {f(x)}{g(x)}}\right)}$	${\displaystyle ={\frac {f'(x)g(x)-f(x)g'(x)}{(g(x))^{2}}}}$
	${\displaystyle ={\frac {0\cdot g(x)-1\cdot g'(x)}{(g(x))^{2}}}}$
	${\displaystyle ={\frac {-g'(x)}{(g(x))^{2}}}.}$