# How do you differentiate 3sin^5(2x)  using the chain rule?

Dec 30, 2015

$30 {\sin}^{4} \left(2 x\right) \cos \left(2 x\right)$

#### Explanation:

The first issue is that the sine function is to the fifth power.

According to the chain rule, $\frac{d}{\mathrm{dx}} \left({u}^{5}\right) = 5 {u}^{4} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

So,

$\frac{d}{\mathrm{dx}} \left(3 {\sin}^{5} \left(2 x\right)\right) = 5 \left(3 {\sin}^{4} \left(2 x\right)\right) \frac{d}{\mathrm{dx}} \left(\sin \left(2 x\right)\right)$

$\implies 15 {\sin}^{4} \left(2 x\right) \cdot \frac{d}{\mathrm{dx}} \left(\sin \left(2 x\right)\right)$

To find $\frac{d}{\mathrm{dx}} \left(\sin \left(2 x\right)\right)$, use the chain rule again:

$\frac{d}{\mathrm{dx}} \left(\sin \left(u\right)\right) = \cos \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Thus,

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

Plug this back in:

$15 {\sin}^{4} \left(2 x\right) \cdot \frac{d}{\mathrm{dx}} \left(\sin \left(2 x\right)\right)$

$\implies 15 {\sin}^{4} \left(2 x\right) \cdot 2 \cos \left(2 x\right)$

$\implies 30 {\sin}^{4} \left(2 x\right) \cos \left(2 x\right)$