Question

If $y={\tan }^{-1}\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}$, 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 $\frac{x}{\sqrt{1-x^4}}$

$y={\tan }^{-1}\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}$
Put $x^2=\cos 2\theta$

$\therefore y={\tan }^{-1}\frac{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}$

$={\tan }^{-1}\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }$

$={\tan }^{-1}\tan (\frac{\pi }{4}-\theta )$

$\Rightarrow y=\frac{\pi }{4}-\theta =\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}{x}^2$

On differentiating both sides, we get
$\frac{dy}{dx}=0-\frac{1}{2}(-\frac{\left(2x\right)}{\sqrt{1-x^4}})=\frac{x}{\sqrt{1-x^4}}$