Question

Differentiate

$\frac{x^2+2}{\sqrt{\cos x}}$

Answer 1

$\mathrm{Let}y=\frac{x^2+2}{\sqrt{\cos x}}$
Differentiate it with respect to x we get,
$$\begin{gathered} \frac{\mathit{d}y}{\mathit{d}x}=\frac{\sqrt{\cos x}{\displaystyle \frac{d}{dx}}\left(x^2+2\right)-\left(x^2+2\right){\displaystyle \frac{d}{dx}}\left(\sqrt{\cos x}\right)}{{\left(\sqrt{\cos x}\right)}^2}\left[\mathrm{Using}\mathrm{quotient}\mathrm{rule}\mathrm{and}\mathrm{chain}\mathrm{rule}\right] \\ =\frac{2x\sqrt{\cos x}-\left(x^2+2\right)\left(-{\displaystyle \frac{1}{2}}{\displaystyle \frac{\sin x}{\sqrt{\cos x}}}\right)}{\cos x} \\ =\frac{2x\sqrt{\cos x}+{\displaystyle \frac{\left(x^2+2\right)\sin x}{2\sqrt{\cos x}}}}{\cos x} \\ =\frac{4x\cos x+\left(x^2+2\right)\sin x}{2{\left(\cos x\right)}^{{\displaystyle \frac{3}{2}}}} \\ =\frac{2x}{\sqrt{\cos x}}+\frac{1}{2}\frac{\left(x^2+2\right)\sin x}{{\left(\cos x\right)}^{{\displaystyle \frac{3}{2}}}} \\ =\frac{1}{\sqrt{\cos x}}\left\{2x+\frac{1}{2}\frac{\left(x^2+2\right)\sin x}{\cos x}\right\} \\ =\frac{1}{\sqrt{\cos x}}\left\{2x+\frac{\left(x^2+2\right)\tan x}{2}\right\} \\ \mathrm{So},\frac{d}{dx}\left(\frac{x^2+2}{\sqrt{\cos x}}\right)=\frac{1}{\sqrt{\cos x}}\left\{2x+\frac{\left(x^2+2\right)\tan x}{2}\right\} \end{gathered}$$