Question

Differentiate the given function w.r.t. $x$:
$\frac{{\cos }^{-1}\frac{x}{2}}{\sqrt{2x+7}}$, $-2<x<2$

Answer 1

Let $y=\frac{{\cos }^{-1}\frac{x}{2}}{\sqrt{2x+7}}$
Thus using quotient rule, we obtain
$\frac{dy}{dx}=\frac{\sqrt{2x+7}\frac{d}{dx}\left({\cos }^{-1}\frac{x}{2}\right)-\left({\cos }^{-1}\frac{x}{2}\right)\frac{d}{dx}\left(\sqrt{2x+7}\right)}{(\sqrt{2x+7}{)}^2}$

$=\frac{\sqrt{2x+7}[\frac{-1}{\sqrt{1-{\left(\frac{x}{2}\right)}^2}}.\frac{d}{dx}\left(\frac{x}{2}\right)]-\left({\cos }^{-1}\frac{x}{2}\right).\frac{1}{2\sqrt{2x+7}}\frac{d}{dx}(2x+7)}{2x+7}$

$=\frac{\sqrt{2x+7}\frac{-1}{\sqrt{4-x^2}}-\left({\cos }^{-1}\frac{x}{2}\right)\frac{1}{2\sqrt{2x+7}}\times 2}{2x+7}$

$=-[\frac{1}{\sqrt{4-x^2}\sqrt{2x+7}}+\frac{{\cos }^{-1}\frac{x}{2}}{(2x+7{)}^{\frac{3}{2}}}]$