use the chain rule to find the derivative of f(x)=(2x+4/3x-1)^3

First, we'll have to re-write the given fraction in a right way,
using the brackets:

f(x)
=[(2x+4)/(3x-1)]^3

We'll use the chain rule considering that the
function f(x) is the result of composing 2 functions:

f(x) =
u(v(x))

u(v) = v^3 => u'(v) =
3v^2

v(x) = (2x+4)/(3x-1) => v'(x) = [(2x+4)'*(3x-1) -
(2x+4)*(3x-1)']/(3x-1)^2

v'(x) = [2(3x-1) -
3(2x+4)]/(3x-1)^2

v'(x) = (6x - 2 - 6x -
12)/(3x-1)^2

v'(x) =
-14/(3x-1)^2

u(v(x)) =
3*[(2x+4)/(3x-1)]^2*[-14/(3x-1)^2]

f'(x) =
-42(2x+4)^2/(3x-1)^4

The derivative of the given function
is:

f'(x) =
-168(x+2)^2/(3x-1)^4