Question

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

• A. $1$
• B. $\frac{3}{2}$
• C. $\frac{2}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C

$\frac{2}{3}$

Explanation For Correct Option

Finding derivative

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

Now putting $x=\tan \theta$ into the above equation,

$$\begin{gathered} \Rightarrow u={\cos }^{-1}\left[\frac{(1–{\tan }^2\theta )}{(1+{\tan }^2\theta )}\right] \\ ={\cos }^{-1}\left(\cos 2\theta \right) \\ =2\theta \\ =2{\tan }^{-1}x[\because x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x] \end{gathered}$$

Now, $\frac{du}{dx}=\frac{2}{(1+x^2)}....\left(i\right)$ after differentiating w.r.to $x$

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

Here putting $x=\tan \theta$

$$\begin{gathered} v=co{t}^{-1}\left[\frac{(1–3{\tan }^2\theta )}{(3\tan \theta –{\tan }^3\theta )}\right] \\ =co{t}^{-1}\left(cot3\theta \right) \\ =3\theta \\ =3{\tan }^{-1}x[\because x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x] \end{gathered}$$

Differentiating it w.r.t. $x$

$\Rightarrow \frac{dv}{dx}=\frac{3}{(1+x^2)}...\left(ii\right)$

Now dividing equation $\left(i\right)$by $\left(ii\right)$ we get,

$$\begin{gathered} \Rightarrow \frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{\frac{2}{(1+x^2)}}{\frac{3}{(1+x^2)}} \\ \Rightarrow \frac{du}{dv}=\frac{2}{3} \end{gathered}$$

Hence, the correct answer is Option (C).