Question

The derivative of $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ with respect to $\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, is

(A) -1

(B) 1

(C) 2

(D) 4

Answer 1

SOLUTION — Let

$p=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)=2 \tan ^{-1} x$

and $\operatorname{ttq}=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2 \tan ^{-1} x$

$\begin{array}{ll}\Rightarrow \quad & \frac{d p}{d x}=\frac{2}{1+x^2}, \frac{d q}{d x}=\frac{2}{1+x^2} \\\therefore \quad & \frac{d p}{d q}=\frac{d p / d x}{d q / d x}=\frac{2 /\left(1+x^2\right)}{2 /\left(1+x^2\right)}=1\end{array}$

So, The correct option will be (B).