How solve Derivatives of Trigonometric Functions?

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1 Answer
Jun 18, 2018

$\frac{\mathrm{df}}{\mathrm{dx}} = 3 \sqrt{\csc \left(6 x\right)} {\sin}^{2} \left(4 x\right) \left[4 \cos \left(4 x\right) - \cot \left(6 x\right) \sin \left(4 x\right)\right]$

Explanation:

First use the product rule:

$\frac{\mathrm{df}}{\mathrm{dx}} = \sqrt{\csc \left(6 x\right)} \frac{d}{\mathrm{dx}} \left({\sin}^{3} \left(4 x\right)\right) + \frac{d}{\mathrm{dx}} \left(\sqrt{\csc \left(6 x\right)}\right) {\sin}^{3} \left(4 x\right)$

using now the chain rule:

$\frac{\mathrm{df}}{\mathrm{dx}} = 3 \sqrt{\csc \left(6 x\right)} {\sin}^{2} \left(4 x\right) \frac{d}{\mathrm{dx}} \left(\sin \left(4 x\right)\right) + \frac{d}{\mathrm{dx}} \left(\csc \left(6 x\right)\right) \frac{{\sin}^{3} \left(4 x\right)}{2 \sqrt{\csc \left(6 x\right)}}$

$\frac{\mathrm{df}}{\mathrm{dx}} = 12 \sqrt{\csc \left(6 x\right)} {\sin}^{2} \left(4 x\right) \cos \left(4 x\right) - 6 \cot \left(6 x\right) \csc \left(6 x\right) \frac{{\sin}^{3} \left(4 x\right)}{2 \sqrt{\csc \left(6 x\right)}}$

and as the function is defined only for $\csc x \ge 0$ we can simplify:

$\frac{\mathrm{df}}{\mathrm{dx}} = 12 \sqrt{\csc \left(6 x\right)} {\sin}^{2} \left(4 x\right) \cos \left(4 x\right) - 3 \cot \left(6 x\right) \sqrt{\csc \left(6 x\right)} {\sin}^{3} \left(4 x\right)$

$\frac{\mathrm{df}}{\mathrm{dx}} = 3 \sqrt{\csc \left(6 x\right)} {\sin}^{2} \left(4 x\right) \left[4 \cos \left(4 x\right) - \cot \left(6 x\right) \sin \left(4 x\right)\right]$