# How do you find the derivative of [3 cos 2x + sin^2 x]?

Nov 27, 2015

The derivative is $- 4 \sin \left(x\right) \cos \left(x\right)$.

#### Explanation:

You need to use the chain rule for differentiation:

if $f \left(x\right) = \left(u \circ v\right) \left(x\right) = u \left(v \left(x\right)\right)$, then $f ' \left(x\right) = u ' \left(v \left(x\right)\right) \cdot v ' \left(x\right)$.

With this rule in mind, you can compute the derivative as follows:

$f ' \left(x\right) = 3 \cdot \left(- \sin \left(2 x\right)\right) \cdot 2 + 2 \cdot \sin \left(x\right) \cdot \cos \left(x\right)$

$\textcolor{w h i t e}{\times \times} = - 6 \sin \left(2 x\right) + 2 \sin \left(x\right) \cos \left(x\right)$

... use the transformation $\sin \left(2 x\right) = 2 \sin \left(x\right) \cos \left(x\right)$ ...

$\textcolor{w h i t e}{\times \times} = - 12 \sin \left(x\right) \cos \left(x\right) + 2 \sin \left(x\right) \cos \left(x\right)$

$\textcolor{w h i t e}{\times \times} = - 10 \sin \left(x\right) \cos \left(x\right)$