# How do you differentiate f(x)=csc3x * cos2x using the product rule?

Jan 2, 2016

$f ' \left(x\right) = - \csc 3 x \left(3 \cot 3 x \cos 2 x + 2 \sin 2 x\right)$

#### Explanation:

The product rule states that

$\frac{d}{\mathrm{dx}} \left[f \left(x\right) g \left(x\right)\right] = f ' \left(x\right) g \left(x\right) + g ' \left(x\right) f \left(x\right)$

Thus,

$f ' \left(x\right) = \cos 2 x \frac{d}{\mathrm{dx}} \left(\csc 3 x\right) + \csc 3 x \frac{d}{\mathrm{dx}} \left(\cos 2 x\right)$

Find each derivative, using the chain rule both times. Recall that

$\frac{d}{d} \left(\csc u\right) = - u ' \csc u \cot u$ and $\frac{d}{\mathrm{dx}} \left(\cos u\right) = - u ' \sin u$

Therefore,

$\frac{d}{\mathrm{dx}} \left(\csc 3 x\right) = - 3 \csc 3 x \cot 3 x$

and

$\frac{d}{\mathrm{dx}} \left(\cos 2 x\right) = - 2 \sin 2 x$

Plug these back in.

$f ' \left(x\right) = - 3 \csc 3 x \cot 3 x \cos 2 x - 2 \sin 2 x \csc 3 x$

$= - \csc 3 x \left(3 \cot 3 x \cos 2 x + 2 \sin 2 x\right)$