# How do you differentiate f(x)= sin2xtan4x using the product rule?

Dec 29, 2015

$f ' \left(x\right) = 2 \cos 2 x \tan 4 x + 4 \sin 2 x {\sec}^{2} 4 x$

#### Explanation:

The product rule states that:

$f ' \left(x\right) = \tan 4 x \frac{d}{\mathrm{dx}} \left(\sin 2 x\right) + \sin 2 x \frac{d}{\mathrm{dx}} \left(\tan 4 x\right)$

Finding either derivative requires the chain rule.

According to the chain rule,

$\frac{d}{\mathrm{dx}} \left(\sin u\right) = u ' \cos u$ and $\frac{d}{\mathrm{dx}} \left(\tan u\right) = u ' {\sec}^{2} u$.

Thus,

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

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

Plug these back in.

$f ' \left(x\right) = 2 \cos 2 x \tan 4 x + 4 \sin 2 x {\sec}^{2} 4 x$