# How do you differentiate f(x)=sin2x * cotx using the product rule?

Jan 24, 2016

$2 \cos 2 x \cdot \cot x - {\csc}^{2} x \cdot \sin 2 x$

#### Explanation:

• The product rule:
$\frac{d}{\mathrm{dx}} \left[f \left(x\right)\right] = \left(\text{derivative of the first term" * "the second term")+("derivative of the second term"*"the first term}\right)$

• $\frac{d}{\mathrm{dx}} \left[\cot x\right] = - {\csc}^{2} x$

• $\frac{d}{\mathrm{dx}} \left[f \left(x\right)\right] = \left(\frac{d}{\mathrm{dx}} \left[\sin \left(2 x\right)\right] \cdot \cot x\right) + \left(\frac{d}{\mathrm{dx}} \left[\cot x\right] \cdot \sin 2 x\right)$
$= \left(2 \cos 2 x \cdot \cot x\right) + \left(- {\csc}^{2} x \cdot \sin 2 x\right)$
$= 2 \cos 2 x \cdot \cot x - {\csc}^{2} x \cdot \sin 2 x$

• You could just stop there, or you could simplify the answer further by uniting all the angles as $x$ using double angle formulas.