Question

Differentiate  \[e^x \log \sin 2x\] ?

Answer 1

\[\text{Let } y = e^x \log \sin2x\]

Differentiate it with respect to x we get,

\[\frac{d y}{d x} = \frac{d}{dx}\left[ e^x \log \sin2x \right]\]

\[ = e^x \frac{d}{dx}\left( \log \sin2x \right) + \left( \log \sin2x \right)\frac{d}{dx}\left( e^x \right) \left[ \text{ Using product rule and chain rule} \right]\]

\[ = e^x \frac{1}{\sin2x}\frac{d}{dx}\left( \sin2x \right) + \log \sin2x\left( e^x \right)\]

\[ = \frac{e^x}{\sin2x}\cos2x \frac{d}{dx}\left( 2x \right) + e^x \log \sin2x\]

\[ = \frac{2\cos2x e^x}{\sin2x} + e^x \log \sin2x\]

\[ = 2 e^x \cot2x + e^x \log \sin2x\]

\[So, \frac{d}{dx}\left[ e^x \log \sin2x \right] = 2 e^x \cot2x + e^x \log \sin2x\]