Question

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

Answer 1

Use substitution to simplify the terms and then differentiate.
Put $x=\sec \theta \Rightarrow \theta ={\sec }^{-1}x$
$\because x>1\Rightarrow \theta \in (0,\frac{\pi }{2})$
$={\cot }^{-1}\sqrt{{\sec }^2\theta -1}+{\sec }^{-1}\sec \theta$
$={\cot }^{-1}\left(\tan \theta \right)+\theta$
$=\frac{\pi }{2}-\theta +\theta$
$=\frac{\pi }{2}$
$\Rightarrow$$\frac{dy}{dx}=0$