# How do you find the derivative of (sin2x)?

##### 1 Answer
Feb 26, 2015

We should use the chain rule:
$f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$.
In our situation $f \left(x\right) = \sin \left(x\right)$ and $g \left(x\right) = 2 x$. Then $f ' \left(x\right) = \cos \left(x\right)$ and $g ' \left(x\right) = 2$.

So $f \left(g \left(x\right)\right) = \sin \left(2 x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right) = \cos \left(2 x\right) \cdot 2 = 2 \cos \left(2 x\right)$