Question

Differentiate given problems w.r.t.x.

$\frac{co{s}^{-}1\frac{x}{2}}{\sqrt{2x+7}}-2<x<2$.

Answer 1

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

Differentiating w.r.t . x, we get

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

$[\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}\left(u\right)-u\frac{d}{dx}\left(v\right)}{v^2}]$

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

=$\frac{-\frac{\sqrt{2x+7}}{\sqrt{4-x^2}}-\frac{co{s}^{-}1\left(\frac{x}{2}\right)}{\sqrt{2x+7}}}{2x+7}$ =$\frac{-\{2x+7+co{s}^{-}1\left(\frac{x}{2}\right)\sqrt{4-x^2}\}}{\sqrt{4-x^2(2x+7)3/2}}$

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