Giải thích các bước giải:
Ta có:
$y=f(x)=\tan x-\dfrac13\tan^3x+\dfrac15\tan^5x$
Đặt $\tan x=u$
$\to f(u)=u-\dfrac13u^3+\dfrac15u^5$
$\to y=f(u)$
$\to y'=f'(u)\cdot u'$
$\to y'=(1-u^2+u^4)\cdot \dfrac{1}{\cos^2x}$
$\to y'=(1-\tan^2x+\tan^4x)\cdot \dfrac{1}{\cos^2x}$
$\to y'=(1-\tan^2x+\tan^4x)\cdot (1+\dfrac{\sin^2x}{\cos^2x})$
$\to y'=(1-\tan^2x+(\tan^2x)^2)\cdot (1+\tan^2x)$
$\o y'=1+\tan^6x$