Giải thích các bước giải:
Ta có:
$y=\left(\dfrac{1-\tan^2x}{1+\tan^2x}\right)^{20}$
$\to y'=\left(\left(\dfrac{1-\tan^2x}{1+\tan^2x}\right)^{20}\right)'$
$\to y'=20\cdot \left(\dfrac{1-\tan^2x}{1+\tan^2x}\right)^{19}\cdot \left(\dfrac{1-\tan^2x}{1+\tan^2x}\right)'$
$\to y'=20\cdot \left(\dfrac{1-\tan^2x}{1+\tan^2x}\right)^{19}\cdot \left(\dfrac{\cos^2x-\cos^2x\cdot \tan^2x}{\cos^2x+\cos^2x\cdot\tan^2x}\right)'$
$\to y'=20\cdot \left(\dfrac{1-\tan^2x}{1+\tan^2x}\right)^{19}\cdot \left(\dfrac{\cos^2x-\sin^2x}{\cos^2x+\sin^2x}\right)'$
$\to y'=20\cdot \left(\dfrac{1-\tan^2x}{1+\tan^2x}\right)^{19}\cdot \left(\dfrac{\cos\left(2x\right)}{1}\right)'$
$\to y'=20\cdot \left(\dfrac{1-\tan^2x}{1+\tan^2x}\right)^{19}\cdot \left(\cos\left(2x\right)\right)'$
$\to y'=-40\cdot \left(\dfrac{1-\tan^2x}{1+\tan^2x}\right)^{19}\cdot \sin \left(2x\right)$
