# How do you differentiate f(x)=(sin2x^2)/4 using the chain rule?

Feb 9, 2016

$f ' \left(x\right) = x \cos \left(2 {x}^{2}\right)$

#### Explanation:

According to the chain rule, since the derivative of $\sin \left(x\right)$ is $\cos \left(x\right)$, we will differentiate the outside $\sin$ function to give us $\cos$, and then multiply that by the derivative of the inside function, which is $2 {x}^{2}$.

Formally, this can be written as

$\frac{d}{\mathrm{dx}} \left[\sin \left(g \left(x\right)\right)\right] = \cos \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

Here, notice first that $\sin \left(2 {x}^{2}\right)$ is being multiplied by $1 \text{/} 4$, which we can bring out of the expression:

$f ' \left(x\right) = \frac{1}{4} \cos \left(2 {x}^{2}\right) \cdot \frac{d}{\mathrm{dx}} \left[2 {x}^{2}\right]$

Since the derivative of $2 {x}^{2}$ is $4 x$, this gives us

$f ' \left(x\right) = \frac{1}{4} \cos \left(2 {x}^{2}\right) \cdot 4 x$

The $4$s will cancel, leaving us with

$f ' \left(x\right) = x \cos \left(2 {x}^{2}\right)$