Find the second-order derivatives of the function e^ 6x 3x - Vidyadip

Question

Find the second-order derivatives of the function $e^{6x}\cos 3x$

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Answer

Let $y = e^{6x}\cos 3x \therefore \frac{{dy}}{{dx}} = {e^{6x}}.\frac{d}{{dx}}\cos 3x + \cos 3x\frac{d}{{dx}}{e^{6x}}$
$= {e^{6x}}\left( { - \sin 3x} \right)\frac{d}{{dx}}\left( {3x} \right) + \cos 3x.{e^{6x}}\frac{d}{{dx}}\left( {6x} \right)$
$= -{e^{6x}}\sin 3x \times 3 + \cos 3x.{e^{6x}} \times 6$
$= e^{6x}(-3 \sin 3x + 6 \cos 3x)$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = {e^{6x}}\frac{d}{{dx}}\left( { - 3\sin 3x + 6\cos 3x} \right) $ $+ \left( { - 3\sin 3x + 6\cos 3x} \right)\frac{d}{{dx}}{e^{6x}}$
$= {e^{6x}}\left( { - 3\cos 3x \times 3 - 6\sin 3x \times 3} \right) $ $+ \left( { - 3\sin 3x + 6\cos 3x} \right){e^{6x}} \times 6$
$= e^{6x}(-9 \cos 3x - 18 \sin 3x - 18 \sin 3x + 36 \cos 3x)$
$= e^{6x}(27 \cos 3x - 36 \sin 3x)$
$= 9e^{6x}(3 \cos 3x - 4 \sin 3x)$

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