# What is the derivative of  (tan(8x))^2?

May 17, 2017

The derivative is $= 16 \tan \left(8 x\right) {\sec}^{2} \left(8 x\right)$

#### Explanation:

We need

$\left({u}^{n}\right) ' = n {u}^{n - 1} \cdot u '$

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

We calculate this derivative by the chain rule

Let $y = {\left(\tan \left(8 x\right)\right)}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \tan \left(8 x\right) \cdot {\sec}^{2} \left(8 x\right) \cdot 8$

$= 16 \tan \left(8 x\right) {\sec}^{2} \left(8 x\right)$