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

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We have the function f(x) = (2x+4/3x-1)^3.

The derivative f'(x) can be found using the chain rule.

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

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

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

f'(x) = -3*(2x+4/3x-1)^2*(14/(3x - 1)^2)

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

**The required derivative is [-168*(x +2)^2]/(3x - 1)^4**

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**

Okay so your function is [(2x+4)/(3x-1)]^3.

Use the product rule with the chain rule where u = (2x+4)^3 and v = 1/(3x-1)^3.

So then we have (2x+4)^3 d/dx(1/(3x-1)^3)+1/(3x-1)^3 d/dx((2x+4)^3)

Then we use the chain rule again where u = 2x+4 and n = 3.

(2x+4)^3 d/dx(1/(3x-1)^3) + (3(2x+4)^2)/(3x-1)^3 d/dx(2x+4)

The derivative of a sum is a sum of the derivatives.

(2x+4)^3 d/dx(1/(3x-1)^3) + (3(2x+4)^2(d/dx(4)+d/dx(2x))/(3x-1)^3

The derivative of the constant 4 is 0.

(2x+4)^3 d/dx(1/(3x-1)^3) + (3(2x+4)^2)/(3x-1)^3 d/dx(2x)

The derivative of a constant times a function is the constant times the derivative of the function.

(2x+4)^3 d/dx(1/(3x-1)^3) + (6(2x+4)^2)/(3x-1)^3

Use the chain rule where u = 3x-1 and n = -3

(6(2x+4)^2)/(3x-1)^3 - (3(2x+4)^3)/(3x-1)^4 d/dx(3x-1)

d/dx of 3x-1 is 3.

(6(2x+4)^2)/(3x-1)^3 - (9(2x+4)^3)/(3x-1)^4

Now we simplify assuming the variables is in an appropriatte range.

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