Question

Find the second order derivatives of the function
$e^{6x}cos3x$

Answer 1

The correct answer is $=9e^{6x}(3cos3x-4sin3x)$
Let $y=e^{6x}cos3x$
Then,
$\frac{dy}{dx}=\frac{d}{dx}(e^{6x}cos3x)=cos3x.\frac{d}{dx}(e^{6x})+e^{6x}.\frac{d}{dx}(cos3x)$
$=cos3x.e^{6x}.\frac{d}{dx}(6x)+e^{6x}.(-sin3x).\frac{d}{dx}(3x)$
$=6e^{6x}cos3x-3e^{6x}sin3x ......(1)$
$∴\frac{d^2y}{dx^2}=\frac{d}{dx}[6e^{6x}cos3x-3e^{6x}sin3x]=6.\frac{d}{dx}(e^{6x}cos3x)-3.\frac{d}{dx}(e^{6x}sin3x)$
$=6.[6e^{6x}cos3x-3e^{6x}sin3x]-3.[sin3x.\frac{d}{dx}(e^{6x})+e^{6x}.\frac{d}{dx}(sin3x)]$ [using(1)]
$=36e^{6x}cos3x-18e^{6x}sin3x-3[sin3x.e^{6x}.6+e^{6x}.cos3x.3]$
$=36e^{6x}cos3x-18e^{6x}sin3x-18e^{6x}sin3x-9e^{6x}cos3x$
$=27e^{6x}cos3x-36e^{6x}sin3x$
$=9e^{6x}(3cos3x-4sin3x)$