Question

If $y={\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}$ then $\frac{dy}{dx}=$

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

Answer 1

Answer: (C)

$\frac{1}{2(1+x^2)}$

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

$\frac{dx}{d\theta }={\sec }^2\theta$

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

$={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)={\tan }^{-1}\frac{2{\sin }^2\theta /2}{2\sin \theta /2\cos \theta /2}$

$={\tan }^{-1}\tan \frac{\theta }{2}=\frac{\theta }{2}$

$\frac{dy}{d\theta }=\frac{1}{2}$So $\frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}=\frac{1}{2{\sec }^2\theta }=\frac{1}{2(1+x^2)}$

Option (C)