How do you differentiate f(x)=cos(3x)*(2/3tanx) using the product rule?

Dec 8, 2015

$f ' \left(x\right) = \frac{2 {\sec}^{2} x \cos 3 x - 6 \tan x \sin 3 x}{3}$

Explanation:

According to the product rule:

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

Find each derivative separately. The first will require the chain rule:

$\frac{d}{\mathrm{dx}} \left[\cos 3 x\right] = - \sin 3 x \frac{d}{\mathrm{dx}} \left[3 x\right] = - 3 \sin 3 x$

$\frac{d}{\mathrm{dx}} \left[\frac{2}{3} \tan x\right] = \frac{2}{3} \frac{d}{\mathrm{dx}} \left[\tan x\right] = \frac{2}{3} {\sec}^{2} x$

Plug these back in:

$f ' \left(x\right) = \frac{2}{3} \tan x \left(- 3 \sin 3 x\right) + \cos 3 x \left(\frac{2}{3} {\sec}^{2} x\right)$

$f ' \left(x\right) = \frac{2}{3} {\sec}^{2} x \cos 3 x - 2 \tan x \sin 3 x$

$f ' \left(x\right) = \frac{2 {\sec}^{2} x \cos 3 x - 6 \tan x \sin 3 x}{3}$