Question

The differential coefficient of ${\tan }^{-1}\sqrt{\frac{1-x^2}{1+x^2}}$ with respect to ${\cos }^{-1}\left(x^2\right)$ is

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $1$
• D. $0$

Answer 1

The correct option is A

$\frac{1}{2}$

Explanation for the correct option :

Step-1 : Simplification:

Put $x^2=\cos \theta$. Then we obtain

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

and

$$\begin{gathered} v={\cos }^{-1}\left(x^2\right) \\ ={\cos }^{-1}\left(\cos \theta \right) \\ =\theta \end{gathered}$$

Step-2 : Differentiating $u$ and $v$ with respect to $x$

Differentiating $u$ with respect to $x$, we get

$$\begin{gathered} \frac{du}{dx}=\frac{d}{dx}\left(\frac{\theta }{2}\right) \\ =\frac{d}{d\theta }\left(\frac{\theta }{2}\right)\frac{d\theta }{dx} \\ =\frac{1}{2}\frac{d\theta }{dx} \end{gathered}$$

Again, differentiating $v$ with respect to $x$, we get

$$\begin{gathered} \frac{dv}{dx}=\frac{d}{dx}\left(\theta \right) \\ =\frac{d}{d\theta }\left(\theta \right)\frac{d\theta }{dx} \\ =\frac{d\theta }{dx} \end{gathered}$$

Step-3 : Finding the differential coefficient of $u$ with respect to $v$

The differential coefficient of $u$ with respect to $v$ is given by

$$\begin{gathered} \frac{du}{dv}=\frac{{\displaystyle \frac{1}{2}\frac{d\theta }{dx}}}{{\displaystyle \frac{d\theta }{dx}}} \\ =\frac{1}{2} \end{gathered}$$

Hence, option (A) is the correct answer.