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Higher Order Derivatives

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Question

Find the second order derivatives of the following :
$e^{4 x} \cdot cos 5 x$

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Solution

Let $y = e^{4 x} \cdot cos 5 x$
Then $\frac{d y}{d x} = \frac{d}{d x} (e^{4 x} \cdot cos 5 x)$
$= e^{4 x} \cdot (cos 5 x) + cos 5 x \cdot \frac{d}{d x} (e^{4 x})$
$= e^{4 x} \cdot (- sin 5 x) \cdot \frac{d}{d x} (5 x) + cos 5 x \times e^{4 x} \cdot \frac{d}{d x} (4 x)$
$= - e^{4 x} \cdot sin 5 x \times 5 + e^{4 x} cos 5 x \times 4$
$= e^{4 x} (4 cos 5 x - 5 sin 5 x)$
and
$\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} [e^{4 x} (4 cos 5 x - 5 sin 5 x)]$
$= e^{4 x} \frac{d}{d x} (4 cos 5 x - 5 sin 5 x) + (4 cos 5 x - 5 sin 5 x) \frac{d}{d x} (4 x)$
$= e^{4 x} [[4 - sin 5 x) \cdot \frac{d}{d x} (5 x) - 5 cos 5 x \cdot \frac{d}{d x} (5 x)] + (4 cos 5 x - 5 sin 5 x) \times e^{4 x} \cdot \frac{d}{d x} (4 x)$
$= e^{4 x} [- 4 sin 5 x \times 5 - 5 cos 5 x \times 5] + (4 cos 5 x - 5 sin 5 x) e^{4 x} \times 4$
$= e^{4 x} (- 20 sin 5 x - 25 cos 5 x + 16 cos 5 x - sin 5 x)$
$= e^{4 x} (- 9 cos 5 x - 40 sin 5 x)$
$= - e^{4 x} (9 cos 5 x + 40 sin 5 x)$

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