# How do you differentiate f(x)=(1-x)tan^2(2x) using the product rule?

Jan 11, 2016

Step by step working shown below.

#### Explanation:

$f \left(x\right) = \left(1 - x\right) {\tan}^{2} \left(2 x\right)$

This problem would require product rule and chain rule.

The product rule.

$\left(u v\right) ' = u v ' + v u '$

$f ' \left(x\right) = \left(1 - x\right) \frac{d}{\mathrm{dx}} \left({\tan}^{2} \left(2 x\right)\right) + {\tan}^{2} \left(2 x\right) \frac{d}{\mathrm{dx}} \left(1 - x\right)$

For derivating ${\tan}^{2} \left(2 x\right)$ we need to use the chain rule.

$f ' \left(x\right) = \left(1 - x\right) \left\{2 \tan \left(2 x\right) \frac{d}{\mathrm{dx}} \left(\tan \left(2 x\right)\right)\right\} + {\tan}^{2} \left(2 x\right) \left(- 1\right)$

$f ' \left(x\right) = \left(1 - x\right) \left\{2 \tan \left(2 x\right) {\sec}^{2} \left(2 x\right) \frac{d}{\mathrm{dx}} \left(2 x\right)\right\} - {\tan}^{2} \left(2 x\right)$

$f ' \left(x\right) = \left(1 - x\right) \left\{2 \tan \left(2 x\right) {\sec}^{2} \left(2 x\right) \left(2\right)\right\} - {\tan}^{2} \left(2 x\right)$

$f ' \left(x\right) = 4 \left(1 - x\right) \tan \left(2 x\right) {\sec}^{2} \left(2 x\right) - {\tan}^{2} \left(2 x\right)$