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

1 Answer
Feb 8, 2016

Answer:

#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 #