Rena Giron

2021-12-04

Find the second order derivatives of the function f(x)
${e}^{x}\mathrm{sin}x\mathrm{cos}2x$

Sarythe

Step 1
Given
${e}^{x}\mathrm{sin}x\mathrm{cos}2x$
Step 2 Finding the second order derivatives
Let $f\left(x\right)={e}^{x}\mathrm{sin}x\mathrm{cos}2x$
$=\frac{1}{2}{e}^{x}\left(2\mathrm{cos}2x\mathrm{sin}x\right)$
$=\frac{1}{2}{e}^{x}\left(\mathrm{sin}3x-\mathrm{sin}x\right)$
Differentiating w.r.t x
${f}^{\prime }\left(x\right)=\frac{{e}^{x}}{2}\left[3\mathrm{cos}3x-\mathrm{cos}x\right]+\frac{1}{2}\left(\mathrm{sin}3x-\mathrm{sin}x\right){e}^{x}$
$=\frac{{e}^{x}}{2}\left(3\mathrm{cos}3x-\mathrm{cos}x+\mathrm{sin}3x-\mathrm{sin}x\right)$
Again Differentiating w.r.t x
$f{}^{″}\left(x\right)=\frac{{e}^{x}}{2}\left(-9\mathrm{sin}3x+\mathrm{sin}x+3\mathrm{cos}3x-\mathrm{cos}x\right)$
$+\frac{1}{2}\left(3\mathrm{cos}3x-\mathrm{cos}x+\mathrm{sin}3x-\mathrm{sin}x\right){e}^{x}$
$=\frac{{e}^{x}}{2}\left[6\mathrm{cos}3x-8\mathrm{sin}3x-2\mathrm{cos}x\right]$
$={e}^{x}\left(3\mathrm{cos}3x-4\mathrm{sin}3x-\mathrm{cos}x\right)$

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