Question

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

Answer 1

The correct answer is $=-\bigg[\frac{1}{\sqrt{4-x^2}\sqrt{2x+7}}+\frac{cos^{-1}\frac{x}{2}}{(2x+7)^{\frac{3}{2}}}\bigg]$
Let $y=\frac{cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}$
By quotient rule,we obtain
$\frac{dy}{dx}=\frac{\sqrt{2x+7}\frac{d}{dx}(cos^{-1}\frac{x}{2})-(cos^{-1}\frac{x}{2})\frac{d}{dx}(\sqrt{2x+7})}{\sqrt{(2x+7)^2}}$
$=\frac{\sqrt{2x+7}\bigg[\frac{-1}{\sqrt{1-(\frac{x}{2})^2}}.\frac{d}{dx}(\frac{x}{2})\bigg]-(cos^{-1}\frac{x}{2})\frac{1}{2\sqrt{2x+7}}.\frac{d}{dx}(2x+7)}{2x+7}$
$=\frac{\sqrt{2x+7}\frac{-1}{\sqrt{4-x^2}}-(cos^{-1}\frac{x}{2})\frac{2}{2\sqrt{2x+7}}}{2x+7}$
$=\frac{-\sqrt{2x+7}}{\sqrt{4-x^2}(2x+7)}-\frac{cos^{-1}\frac{x}{2}}{(\sqrt{2x+7})(2x+7)}$
$=-\bigg[\frac{1}{\sqrt{4-x^2}\sqrt{2x+7}}+\frac{cos^{-1}\frac{x}{2}}{(2x+7)^{\frac{3}{2}}}\bigg]$