The derivative of ${\mathrm{e}}^{3\mathrm{x}} \sin 4\mathrm{x}$ with respect to $\mathrm{x},$ is

Suppose that $y=e^{3x}\sin 4x$

Differentiate both sides 

$\begin{array}{rcl}\frac{dy}{dx}& =& e^{3x}\left(4\cos 4x\right)+3e^{3x}\left(\sin 4x\right)\\ & =& e^{3x}\left(3\sin 4x+4\cos 4x\right)\\ & =& e^{3x}\left(\frac{3}{5}\sin 4x+\frac{4}{5}\cos 4x\right) \left(devide with \sqrt{3^2+4^2}=5\right)\\ & =& 5e^{3x}\left(\cos 4x\sin A+\sin 4x\cos A\right)\\ & =& 5e^{3x}\left(\sin \left(A+4x\right)\right)\end{array}$

Here $\sin A=\frac{4}{5},\cos A=\frac{3}{5}\Rightarrow \tan A=\frac{4}{3}$

Therefore, $\frac{dy}{dx}=5e^{3x}\left(\sin \left(4x+{\tan }^{-1}\left(\frac{4}{3}\right)\right)\right)$