Question

Differentiate

$e^x\log \sin 2x$

Answer 1

$\mathrm{Let}y=e^x\log \sin 2x$
Differentiate it with respect to x we get,

$$\begin{gathered} \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{d}{dx}\left[e^x\log \sin 2x\right] \\ =e^x\frac{d}{dx}\left(\log \sin 2x\right)+\left(\log \sin 2x\right)\frac{d}{dx}\left(e^x\right)\left[\mathrm{Using}\mathrm{product}\mathrm{rule}\mathrm{and}\mathrm{chain}\mathrm{rule}\right] \\ =e^x\frac{1}{\sin 2x}\frac{d}{dx}\left(\sin 2x\right)+\log \sin 2x\left(e^x\right) \\ =\frac{e^x}{\sin 2x}\cos 2x\frac{d}{dx}\left(2x\right)+e^x\log \sin 2x \\ =\frac{2\cos 2x{e}^x}{\sin 2x}+e^x\log \sin 2x \\ =2e^x\cot 2x+e^x\log \sin 2x \\ \mathrm{So},\frac{d}{dx}\left[e^x\log \sin 2x\right]=2e^x\cot 2x+e^x\log \sin 2x \end{gathered}$$