# How do you differentiate f(x)=sin(cos(tanx)) using the chain rule?

Dec 20, 2015

$f ' \left(x\right) = - {\sec}^{2} x \sin \left(\tan x\right) \cos \left(\cos \left(\tan x\right)\right)$

#### Explanation:

The first major issue is the sin function. According to the chain rule,

$\frac{d}{\mathrm{dx}} \left[\sin u\right] = u ' \cos u$, so

$f ' \left(x\right) = \cos \left(\cos \left(\tan x\right)\right) \cdot \frac{d}{\mathrm{dx}} \left[\cos \left(\tan x\right)\right]$

Now to differentiate $\cos \left(\tan x\right)$, know that

$\frac{d}{\mathrm{dx}} \left[\cos u\right] = - u ' \sin u$

Thus,

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

Also recall that $\frac{d}{\mathrm{dx}} \left[\tan x\right] = {\sec}^{2} x$, so

$\frac{d}{\mathrm{dx}} \left[\cos \left(\tan x\right)\right] = - {\sec}^{2} x \sin \left(\tan x\right)$

Plug this back into the $f ' \left(x\right)$ equation.

$f ' \left(x\right) = - {\sec}^{2} x \sin \left(\tan x\right) \cos \left(\cos \left(\tan x\right)\right)$