Question

The derivative of $\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ with respect to $\cot ^{-1}\left(\frac{1-3 x^2}{3 x-x^3}\right)$, is

(A) 1

(B) $\frac{3}{2}$

(C) $\frac{2}{3}$

(D) $\frac{1}{2}$

Answer 1

Let

$y_1=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

$y_2=\cot ^{-1}\left(\frac{1-3 x^2}{3 x-x^3}\right)$

Put

$x=\tan \theta \Rightarrow \theta=\tan ^{-1} x$

$\Rightarrow y_1=2 \tan ^{-1} x$

and $y_2=3 \tan ^{-1} x$

On differentiating w.r.t.x,

$\begin{aligned}\frac{d y_1}{d x}= & \frac{2}{1+x^2} \text { and } \frac{d y_2}{d x}=\frac{3}{1+x^2} \\\therefore \quad \frac{d y_1}{d y_2} & =\frac{\left(\frac{d y_1}{d x}\right)}{\left(\frac{d y_2}{d x}\right)}=\frac{\left(\frac{2}{1+x^2}\right)}{\left(\frac{3}{1+x^2}\right)}=\frac{2}{3}\end{aligned}$

So, The correct option will be (C).