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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability3 Marks
Question
Find the second-order derivatives of the function e6x cos 3x
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Answer

Let y = e6xcos3x

$\therefore \frac{{dy}}{{dx}} = {e^{6x}}.\frac{d}{{dx}}\cos 3x + \cos 3x\frac{d}{{dx}}{e^{6x}}$

$= {e^{6x}}\left( { - \sin 3x} \right)\frac{d}{{dx}}\left( {3x} \right) + \cos 3x.{e^{6x}}\frac{d}{{dx}}\left( {6x} \right)$

$= -{e^{6x}}\sin 3x \times 3 + \cos 3x.{e^{6x}} \times 6$

= e6x(-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) $ $+ \left( { - 3\sin 3x + 6\cos 3x} \right){e^{6x}} \times 6$

= e6x(-9 cos 3x - 18 sin 3x - 18 sin 3x + 36 cos 3x)

= e6x(27cos 3x - 36sin3x)

= 9e6x(3 cos 3x - 4 sin 3x)

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