Question
Find the second order derivative of the following : $e^{2 x} \sin 3 x$

Answer

Let $y=e^{2 x} \sin 3 x$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(e^{2 x} \sin 3 x\right)=e^{2 x} \frac{d}{d x}(\sin 3 x)+\sin 3 x \frac{d}{d x}\left(e^{2 x}\right) \\
& \frac{d y}{d x}=e^{2 x}(\cos 3 x)(3)+\sin 3 x\left(e^{2 x}\right)(2) \\
& \frac{d y}{d x}=e^{2 x}(3 \cos 3 x+2 \sin 3 x)
\end{aligned}
$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left[e^{2 x}(3 \cos 3 x+2 \sin 3 x)\right] \\
& \frac{d^2 y}{d x^2}=e^{2 x} \frac{d}{d x}(3 \cos 3 x+2 \sin 3 x)+(3 \cos 3 x+2 \sin 3 x) \frac{d}{d x}\left(e^{2 x}\right) \\
& =e^{2 x}[3(-\sin 3 x)(3)+2(\cos 3 x)(3)]+(3 \cos 3 x+2 \sin 3 x) e^{2 x}(2) \\
& =e^{2 x}[-9 \sin 3 x+6 \cos 3 x+6 \cos 3 x+4 \sin 3 x] \\
& \frac{d^2 y}{d x^2}=e^{2 x}[12 \cos 3 x-5 \sin 3 x] \\
&
\end{aligned}
$

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