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Higher Order Derivatives — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSHigher Order Derivatives5 Marks
Question
Find the second order derivatives of the following functions:
$\text{e}^{6\text{x}} \cos \text{x}$
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Answer

Let $\text{y}=\text{e}^{6\text{x}}\cos3\text{x}$
Then,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{e}^{6\text{x}}.\frac{\text{d}}{\text{dx}}(6\text{x})+\text{e}^{6\text{x}}.\frac{\text{d}}{\text{dx}}(3\text{x})$
$=\cos3\text{x}.\text{e}^{6\text{x}}.\frac{\text{d}}{\text{dx}}(6\text{x})+\text{e}^{6\text{x}}.(-\sin3\text{x}).\frac{\text{d}}{\text{dx}}(\cos3\text{x})$
$=6\text{e}^{6\text{x}}\cos3\text{x}-3\text{e}^{6\text{x}}\sin3\text{x}...(1)$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}(6\text{e}^{6\text{x}}\cos3\text{x}-3\text{e}^{6\text{x}}\sin3\text{x})$
$=6.\frac{\text{d}}{\text{dx}}(\text{e}^{6\text{x}}\cos3\text{x})-3.\frac{\text{d}}{\text{dx}}(\text{e}^{6\text{x}}\sin3\text{x})$
$=6.[6\text{e}^{6\text{x}}\cos3\text{x}-3\text{e}^{6\text{x}}\sin3\text{x}]$
$-3.\Big[\sin3\text{x}.\frac{\text{d}}{\text{dx}}(\text{e}^{6\text{x}})+\text{e}^{6\text{x}}.\frac{\text{d}}{\text{dx}}(\text{e}^{6\text{x}}\sin3\text{x})$
$=36\text{e}^{6\text{x}}\cos3\text{x}-18\text{e}^{6\text{x}}\sin3\text{x}-3[\sin3\text{x}.\text{e}^{6\text{x}}.\cos3\text{x}.3$
$=36\text{e}^{6\text{x}}\cos3\text{x}-18\text{e}^{6\text{x}}\sin3\text{x}-18\text{e}^{6\text{x}}\sin3\text{x}-9\text{e}^{6\text{x}}\cos3\text{x}$
$=27\text{e}^{6\text{x}}\cos3\text{x}-36\text{e}^{6\text{x}}\sin3\text{x}$
$=9\text{e}^{6\text{x}}(3\cos3\text{x}-4\sin3\text{x})$

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