# Question #910e9

Feb 23, 2018

$8 {\sec}^{2} \left(x\right) {\tan}^{3} \left(x\right) + 8 {\sec}^{4} \left(x\right) \tan \left(x\right)$

#### Explanation:

Here, we'll have to use the Chain Rule, Product Rule, and rules for differentiating trigonometric functions. Let's factor out the constant so the differentiation process is much cleaner:

$4 \frac{d}{\mathrm{dx}} \left({\sec}^{2} \left(x\right) {\tan}^{2} \left(x\right)\right)$

Using the product rule, we get:

$4 \left(\frac{d}{\mathrm{dx}} \left({\sec}^{2} \left(x\right)\right) {\tan}^{2} \left(x\right) + {\sec}^{2} \left(x\right) \frac{d}{\mathrm{dx}} {\tan}^{2} \left(x\right)\right)$

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

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

Plugging in the above derivatives, our solution becomes:

$4 \left(2 \sec \left(x\right) \sec \left(x\right) \tan \left(x\right) {\tan}^{2} \left(x\right) + {\sec}^{2} \left(x\right) 2 \tan \left(x\right) {\sec}^{2} \left(x\right)\right)$

Simplifying, we get:

$4 \left(2 {\sec}^{2} \left(x\right) {\tan}^{3} \left(x\right) + 2 {\sec}^{4} \left(x\right) \tan \left(x\right)\right)$

$8 {\sec}^{2} \left(x\right) {\tan}^{3} \left(x\right) + 8 {\sec}^{4} \left(x\right) \tan \left(x\right)$