Question

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

Answer 1

`f'(x) = 6cos(1/(3x-1)) * sin( 1 / (3x - 1)) * 1 /(3x-1)^2`

Explanation:

Let's define the chain of compositions:

`f(x) = cos^2(1/(3x-1)) = [color(orange)(cos(1/(3x-1)))]^2 = color(orange)(u)^2 = u^2`

where

`u = cos(1/(3x-1)) = cos(color(blue)(1/(3x-1))) = cos(color(blue)(v)) = cos(v)`

where

`v = 1 / (3x - 1) = 1 / color(red)(3x-1) = 1 / color(red)(w) = 1/w`

where

`w = 3x-1`

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

`[u^2]' = 2u = 2cos(1/(3x-1))`

`u' = [cos v]' = - sin v = - sin( 1 / (3x - 1))`

`v' = [1/w]' = [w^(-1)]' = -w^(-2) = - 1 /w^2 = - 1 /(3x-1)^2`

`w'= [3x - 1]' = 3`

The derivative of `f(x)` is defined as the product of those four derivatives:

`f'(x) = [u^2]' * u' * v' * w'`

`= 2cos(1/(3x-1)) * (- sin( 1 / (3x - 1))) * (- 1 /(3x-1)^2) * 3`

`= 6cos(1/(3x-1)) * sin( 1 / (3x - 1)) * 1 /(3x-1)^2`