Question

$y{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+{\sec }^{-1}\left(\frac{1+x^2}{1+x^2}\right)$, show that $\frac{dy}{dx}=\frac{4}{1+x^2}$.

Answer 1

Let $y={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+{\sec }^{-1}\left(\frac{1+x^2}{1+x^2}\right)$

Let $x=\tan \theta ;\theta ={\tan }^{-1}x$

Now,

$y={\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)+\frac{\pi }{2}$

$y={\sin }^{-1}\left(\sin 2\theta \right)+\frac{\pi }{2}$

$y=2\theta +\frac{\pi }{2}$

$y=2{\tan }^{-1}x+\frac{\pi }{2}$

$\frac{dy}{dx}=2\times \frac{1}{1+x^2}+0$

$\frac{dy}{dx}=\frac{2}{1+x^2}$