Aug 15, 2017

$f ' \left(x\right) = \cos \left(x\right) \cos \left(2 x\right) - 2 \sin \left(x\right) \sin \left(2 x\right)$

Explanation:

I believe you mean $f \left(x\right) = \sin \left(x\right) \cdot \cos \left(2 x\right)$

We can differentiate this function using the chain rule and the product rule.

Let $u = 2 x R i g h t a r r o w u ' = 2$ and $v = \cos \left(u\right) R i g h t a r r o w v ' = - \sin \left(u\right)$:

$R i g h t a r r o w f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\sin \left(x\right)\right) \cdot \left(\cos \left(2 x\right)\right) + \frac{d}{\mathrm{dx}} \left(\cos \left(2 x\right)\right) \cdot \left(\sin \left(x\right)\right)$

$R i g h t a r r o w f ' \left(x\right) = \cos \left(x\right) \cdot \cos \left(2 x\right) + u ' \cdot v ' \cdot \sin \left(x\right)$

$R i g h t a r r o w f ' \left(x\right) = \cos \left(x\right) \cos \left(2 x\right) + 2 \cdot - \sin \left(u\right) \cdot \sin \left(x\right)$

$R i g h t a r r o w f ' \left(x\right) = \cos \left(x\right) \cos \left(2 x\right) - 2 \sin \left(u\right) \sin \left(x\right)$

Let's replace $u$ with $2 x$:

$R i g h t a r r o w f ' \left(x\right) = \cos \left(x\right) \cos \left(2 x\right) - 2 \sin \left(2 x\right) \sin \left(x\right)$

$\therefore f ' \left(x\right) = \cos \left(x\right) \cos \left(2 x\right) - 2 \sin \left(x\right) \sin \left(2 x\right)$