Question

If $y={\sin }^{-1}\frac{2x}{1+x^2}+{\sec }^{-1}\frac{1+x^2}{1-x^2},$ then show that $\frac{dy}{dx}=\frac{4}{1+x^2}.$

Answer 1

$y={\sin }^{-1}\frac{2x}{1+x^2}+{\sec }^{-1}\frac{1+x^2}{1-x^2},$
$\Rightarrow y={\sin }^{-1}\frac{2x}{1+x^2}+{\cos }^{-1}\frac{1-x^2}{1+x^2},$
Put $x=\tan \theta$
$y={\sin }^{-1}\sin 2\theta +{\cos }^{-1}\cos 2\theta$

$\Rightarrow y=2\theta +2\theta$

$=4\theta$

$\Rightarrow y=4{\tan }^{-1}x$
$\therefore \frac{dy}{dx}=\frac{4}{1+x^2}$