# What is the derivative of cos(sin(x))?

Jul 7, 2016

$\frac{d}{\mathrm{dx}} \cos \left(\sin \left(x\right)\right) = - \cos \left(x\right) \sin \left(\sin \left(x\right)\right)$

#### Explanation:

The chain rule states that given two functions $f$ and $g$, we have $\left(f \circ g\right) ' \left(x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$.

In this case, letting $f \left(x\right) = \cos \left(x\right)$ and $g \left(x\right) = \sin \left(x\right)$, we have

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

Thus

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

$= f ' \left(g \left(x\right)\right) g ' \left(x\right)$

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