Question

If $y={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)$,then $\frac{dy}{dx}$is equal to

• A. $\frac{x^2}{\sqrt{1-x^4}}$
• B. $\frac{x^2}{\sqrt{1+x^4}}$
• C. $\frac{x}{\sqrt{1+x^4}}$
• D. $\frac{x}{\sqrt{1-x^4}}$

Answer 1

The correct option is D

$\frac{x}{\sqrt{1-x^4}}$

Explanation for the correct option:

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

Put $x^2=\cos 2\theta \Rightarrow \theta =\frac{1}{2}{\cos }^{-1}{x}^2$

$$\begin{gathered} y={\tan }^{-1}\left(\frac{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{2{\cos }^2\theta }-\sqrt{2{\sin }^2\theta }}{\sqrt{2{\cos }^2\theta }+\sqrt{2{\sin }^2\theta }}\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{2}\cos \theta -\sqrt{2}\sin \theta }{\sqrt{2}\cos \theta +\sqrt{2}\sin \theta }\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{2}\left(\begin{array}{cc}\cos \theta -& \sin \theta \end{array}\right)}{\sqrt{2}\left(\begin{array}{cc}\cos \theta +& \sin \theta \end{array}\right)}\right) \\ ={\tan }^{-1}\left(\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }\right) \end{gathered}$$

Divide numerator and denominator by $\cos \theta$

$$\begin{gathered} y={\tan }^{-1}\left(\frac{1-\tan \theta }{1+\tan \theta }\right) \\ ={\tan }^{-1}\left(\frac{\tan {\displaystyle \frac{\pi }{4}}-\tan \theta }{\tan {\displaystyle \frac{\pi }{4}}+\tan \theta }\right) \\ ={\tan }^{-1}\tan \left(\frac{\pi }{4}-\theta \right) \\ =\frac{\pi }{4}-\theta \\ =\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}{x}^2 \end{gathered}$$

Now, differentiate with respect to $x$

$\begin{array}{rcl}\frac{dy}{dx}& =& 0-\frac{1}{2}\times \frac{-1}{\sqrt{1-x^4}}\times 2x\\ & \Rightarrow & \frac{dy}{dx}=\frac{x}{\sqrt{1-x^4}}\end{array}$

Hence, Option (D) is the correct answer.