Question

If $y=si{n}^2\left(\log \right(2x+3\left)\right)$, find $\frac{dy}{dx}$.

Answer 1

Given,

$y={\sin }^2\left(log\right(2x+3\left)\right)$
$y=\left[\sin \right(log(2x+3)){]}^2$
$\therefore \frac{dy}{dx}=\frac{d}{dx}\left[\sin \right(log(2x+3)){]}^2$
$=2\sin \left(log\right(2x+3\left)\right)\frac{d}{dx}\left(\sin \right(log(2x+3)\left)\right)$

$=2\sin \left(log\right(2x+3\left)\right)\left(\cos \right(log(2x+3)\left)\right)\left(\frac{2}{2x+3}\right)$