Question

Differentiate the following with respect to $x$:
${\cos }^{-1}\left[\sqrt{\frac{1+x}{2}}\right],-1<x<1$

Answer 1

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

Hence, this is the answer.