What is the derivative of ?

$y = \sin x {\cos}^{2} 2 x$ ?

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Explanation:

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Feb 9, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{3} x - 2 \sin x \sin 4 x$

Explanation:

We use here product rule. Here $y = f \left(x\right) g \left(x\right)$ and according to product rule $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dx}} g \left(x\right) + f \left(x\right) \frac{\mathrm{dg}}{\mathrm{dx}}$

Here we have $f \left(x\right) = \sin x$ and $g \left(x\right) = {\cos}^{2} 2 x$

and therefore $\frac{\mathrm{df}}{\mathrm{dx}} = \cos x$ and $\frac{\mathrm{dg}}{\mathrm{dx}} = 2 \cos 2 x \times \left(- \sin 2 x\right) \times 2 = - 4 \sin 2 x \cos 2 x = - 2 \sin 4 x$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \cos x \times {\cos}^{2} x + \sin x \times \left(- 2 \sin 4 x\right)$

= ${\cos}^{3} x - 2 \sin x \sin 4 x$

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