# How do you differentiate  f(x) =2cosx+sin2x ?

Mar 15, 2018

$2 \cos \left(2 x\right) - 2 \sin \left(x\right)$

#### Explanation:

the derivative of $\cos \left(x\right)$ is defined as $- \sin \left(x\right)$
therefore for the first term, the derivative of a constant multiplied by $\cos \left(x\right)$ gives that same constant multiplied by $- \sin \left(x\right)$

therefore derivative of the first term is
$- 2 \sin \left(x\right)$

the derivative of the second term can be found by using The Chain Rule

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

therefore,
let $f \left(x\right) = \sin \left(x\right)$
and $g \left(x\right) = 2 x$

therefore,
$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = \frac{d}{\mathrm{dx}} \sin \left(2 x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right) = \cos \left(2 x\right) \cdot 2$

$= 2 \cos \left(2 x\right)$

therefore, the entire derivative is,
$- 2 \sin \left(x\right) + 2 \cos \left(2 x\right)$
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