# How do you differentiate f(x)=cos^2(1/(3x-1)) using the chain rule?

Feb 8, 2016

$f ' \left(x\right) = 6 \cos \left(\frac{1}{3 x - 1}\right) \cdot \sin \left(\frac{1}{3 x - 1}\right) \cdot \frac{1}{3 x - 1} ^ 2$

#### Explanation:

Let's define the chain of compositions:

$f \left(x\right) = {\cos}^{2} \left(\frac{1}{3 x - 1}\right) = {\left[\textcolor{\mathmr{and} a n \ge}{\cos \left(\frac{1}{3 x - 1}\right)}\right]}^{2} = {\textcolor{\mathmr{and} a n \ge}{u}}^{2} = {u}^{2}$

where

$u = \cos \left(\frac{1}{3 x - 1}\right) = \cos \left(\textcolor{b l u e}{\frac{1}{3 x - 1}}\right) = \cos \left(\textcolor{b l u e}{v}\right) = \cos \left(v\right)$

where

$v = \frac{1}{3 x - 1} = \frac{1}{\textcolor{red}{3 x - 1}} = \frac{1}{\textcolor{red}{w}} = \frac{1}{w}$

where

$w = 3 x - 1$

Now, you need to compute the derivatives of those four functions:

$\left[{u}^{2}\right] ' = 2 u = 2 \cos \left(\frac{1}{3 x - 1}\right)$

$u ' = \left[\cos v\right] ' = - \sin v = - \sin \left(\frac{1}{3 x - 1}\right)$

$v ' = \left[\frac{1}{w}\right] ' = \left[{w}^{- 1}\right] ' = - {w}^{- 2} = - \frac{1}{w} ^ 2 = - \frac{1}{3 x - 1} ^ 2$

$w ' = \left[3 x - 1\right] ' = 3$

The derivative of $f \left(x\right)$ is defined as the product of those four derivatives:

$f ' \left(x\right) = \left[{u}^{2}\right] ' \cdot u ' \cdot v ' \cdot w '$

$= 2 \cos \left(\frac{1}{3 x - 1}\right) \cdot \left(- \sin \left(\frac{1}{3 x - 1}\right)\right) \cdot \left(- \frac{1}{3 x - 1} ^ 2\right) \cdot 3$

$= 6 \cos \left(\frac{1}{3 x - 1}\right) \cdot \sin \left(\frac{1}{3 x - 1}\right) \cdot \frac{1}{3 x - 1} ^ 2$