Ta có
$y = \sin(\tan^{13}x)$
Vậy
$y' = [\sin (\tan^{13} x)]'$
$ = \cos(\tan^{13}x) . (\tan^{13}x)'$
$= \cos(\tan^{13}x) . 13\tan^{12}x (\tan x)'$
$= \cos(\tan^{13}x) . 13\tan^{12}x . \dfrac{1}{\cos^2x}$
Vậy
$y' = \dfrac{\cos(\tan^{13}x) . 13\tan^{12}x}{\cos^2x}$
