Question
Find the second $-$ order derivative of the function $e^x \sin 5x$

Answer

Let $ y = e^xsin 5x \therefore \frac{{dy}}{{dx}} = {e^x}\frac{d}{{dx}}\sin 5x + \sin 5x\frac{d}{{dx}}{e^x}$
$= {e^x}\cos 5x\frac{d}{{dx}}5x + \sin 5x.{e^x} $
$= {e^x}\cos 5x \times 5 + {e^x}\sin 5x$
$= e^x(5 \cos 5x + \sin 5x)$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = {e^x}\frac{d}{{dx}}\left( {5\cos 5x + \sin 5x} \right) + \left( {5\cos 5x + \sin 5x} \right)\frac{d}{{dx}}{e^x}$
$= {e^x}\left[ {5\left( { - \sin 5x} \right) \times 5 + \left( {\cos 5x} \right) \times 5} \right] + (5 \cos 5x + \sin 5x)e^x$
$= e^x(-25 \sin 5x + 5 \cos 5x + 5 \cos 5x + \sin 5x)$
$= e^x(10 \cos 5x - 24 \sin 5x)$
$= 2e^x(5 \cos 5x - 12 \sin 5x)$

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