Question

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

Answer 1

Let $y=\frac{{\cos }^{-1}\frac{x}{2}}{\sqrt{2x+7}}$

Differentiating both sides w.r.t. $x$, we get

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

By quotient rule : ${\left(\frac{u}{v}\right)}^{\prime }=\frac{u^{\prime }v-v^{\prime }u}{v^2}$

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


$\frac{dy}{dx}=\frac{\frac{-\sqrt{2x+7}}{2\sqrt{\frac{4-x^2}{4}}}-\frac{{\cos }^{-1}\frac{x}{2}}{2\sqrt{2x+7}}.(2+0)}{(\sqrt{2x+7}{)}^2}$

$\frac{dy}{dx}=\frac{\frac{-\sqrt{2x+7}}{\sqrt{4-x^2}}-\frac{{\cos }^{-1}\frac{x}{2}}{\sqrt{2x+7}}}{(2x+7)}$

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

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

$\frac{dy}{dx}=\frac{-1}{\sqrt{2x+7}\sqrt{4-x^2}}-\frac{{\cos }^{-1}\frac{x}{2}}{(2x+7)\sqrt{2x+7}}$

$\frac{dy}{dx}=\frac{-1}{\sqrt{2x+7}\sqrt{4-x^2}}-\frac{{\cos }^{-1}\frac{x}{2}}{(2x+7{)}^{\frac{3}{2}}}$.