"Find the second order derivatives of the function..." solved and explained - Plainmath

Rena Giron

Answered question

2021-12-04

Find the second order derivatives of the function f(x)

e^{x} \sin x \cos 2 x

Answer & Explanation

Sarythe

Beginner2021-12-05Added 11 answers

Step 1
Given

e^{x} \sin x \cos 2 x

Step 2 Finding the second order derivatives
Let

f (x) = e^{x} \sin x \cos 2 x

= \frac{1}{2} e^{x} (2 \cos 2 x \sin x)

= \frac{1}{2} e^{x} (\sin 3 x - \sin x)

Differentiating w.r.t x

f^{'} (x) = \frac{e^{x}}{2} [3 \cos 3 x - \cos x] + \frac{1}{2} (\sin 3 x - \sin x) e^{x}

= \frac{e^{x}}{2} (3 \cos 3 x - \cos x + \sin 3 x - \sin x)

Again Differentiating w.r.t x

f^{″} (x) = \frac{e^{x}}{2} (- 9 \sin 3 x + \sin x + 3 \cos 3 x - \cos x)

+ \frac{1}{2} (3 \cos 3 x - \cos x + \sin 3 x - \sin x) e^{x}

= \frac{e^{x}}{2} [6 \cos 3 x - 8 \sin 3 x - 2 \cos x]

= e^{x} (3 \cos 3 x - 4 \sin 3 x - \cos x)

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