Question

If $y=In(\sqrt{x}+{\sin }^{2}x),$ then $\frac{dy}{dx}$ is equal to :

Answer 1

$y=\ln (\sqrt{x}+{\sin }^{2}x)$

Differenciate with respect to $x\Rightarrow$

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

$=\left(\frac{1}{\sqrt{x}+{\sin }^{2}x}\right)\left(\frac{1}{2\sqrt{x}}2\sin x\cos x\right)$

$=\left(\frac{1}{\sqrt{x}+{\sin }^{2}x}\right)\left(\frac{1+2\sqrt{x}\sin 2x}{2\sqrt{x}}\right)$