Question

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

Answer 1

$\begin{array}{c}y={\cos }^{-1}2x\sqrt{1-x^2}\left[x=\sin \theta \right]\\ ={\cos }^{-1}2\sin \theta \sqrt{1-{\sin }^1\theta }\\ ={\cos }^{-1}2\sin \theta .\cos \theta \\ ={\cos }^{-1}\sin 2\theta \\ ={\cos }^{-1}\cos \left(\frac{\pi }{2}-2\theta \right)\\ =\frac{\pi }{2}-2\theta \\ =\frac{\pi }{2}-2{\sin }^{-1}x\\ \frac{dy}{dx}=0-\frac{2}{\sqrt{1-x^2}}\\ =\frac{-2}{\sqrt{1-x^2}}\end{array}$

Hence, this is the answer.