Question

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

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

Answer 1

The correct option is B $\frac{-1}{2(1+x^2)}$

$\begin{array}{c}y-{\tan }^{-1}\left(\frac{2}{\sqrt{1-x^2-1}}\right)\\ x=\tan \theta \\ {\tan }^{-1}\left(\frac{\tan \theta }{\sec \theta -1}\right)={\tan }^{-1}\left(\frac{\sin \theta }{1-\cos \theta }\right)\\ ={\tan }^{-1}\left(\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2\sin \frac{\theta }{2}}\right)\\ ={\tan }^{-1}\left(\tan \left(\frac{\pi }{2}-\frac{\theta }{2}\right)\right)\\ =\frac{\pi }{2}-\frac{\theta }{2}\\ t=\frac{\pi }{2}-\frac{1}{2}{\tan }^{-1}x\\ \frac{dy}{dx}=-\frac{1}{2\left(1+x^2\right)}\\ OptionBiscorrectanswer.\end{array}$