If $u=sin^{-1}(\frac{2x}{1+x^2})$ and $v= tan^{-1}x , x\in (-1, 1)$ then , $\frac{du}{dv}$ is equal to :

2

The correct answer is Option (4) → 2

$u=\sin^{-1}(\frac{2x}{1+x^2})$

let $x=\tan θ⇒θ=\tan^{-1}x$

$u=\sin^{-1}(\frac{2\tan θ}{1+\tan^2 θ})⇒u=\sin^{-1}(\frac{2\tan θ}{\sec^2θ})$

so $u=\sin^{-1}(2\sin θ\cos θ)=\sin^{-1}(\sin 2θ)$

$u=2θ=2\tan^{-1}x$   (1)

$v=\tan^{-1}x⇒\frac{du}{dv}=2$