Question

If $y={\cot }^{-1}\left(\frac{\sqrt{1+x^2}+1}{x}\right),x>0,$ then $\frac{dy}{dx}$ is equal to

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

Answer 1

The correct option is B $\frac{1}{2}\cdot \frac{1}{1+x^2}$

Putting $x=\tan \theta ,\theta \in (0,\frac{\pi }{2})$ we have
$y={\cot }^{-1}\left(\frac{\sec \theta +1}{\tan \theta }\right)={\cot }^{-1}\left(\frac{1+\cos \theta }{\sin \theta }\right)={\cot }^{-1}\left(\cot \frac{\theta }{2}\right)$
$(\because \frac{\theta }{2}\in (0,\frac{\pi }{4}))$
$=\frac{\theta }{2}=\frac{1}{2}{\tan }^{-1}x$
$\Rightarrow$$\frac{dy}{dx}=\frac{1}{2}\cdot \frac{1}{1+x^2}$