You need to find derivative of the given function, hence, you need to use the product rule, quotient rule and chain rule such that:

`(dy)/(dx) = ((x^3*(x+1))'(3x-2)^4 - (x^3*(x+1))((3x-2)^4)')/((3x-2)^8)`

`(dy)/(dx) = ((3x^2(x + 1) + x^3)(3x-2)^4 - (x^3*(x+1))(12(3x-2)^3))/((3x-2)^8)`

Factoring out `(3x-2)^3` yields:

`(dy)/(dx) = (3x-2)^3((4x^3 + 3x^2)(3x - 2) -...

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You need to find derivative of the given function, hence, you need to use the product rule, quotient rule and chain rule such that:

`(dy)/(dx) = ((x^3*(x+1))'(3x-2)^4 - (x^3*(x+1))((3x-2)^4)')/((3x-2)^8)`

`(dy)/(dx) = ((3x^2(x + 1) + x^3)(3x-2)^4 - (x^3*(x+1))(12(3x-2)^3))/((3x-2)^8)`

Factoring out `(3x-2)^3` yields:

`(dy)/(dx) = (3x-2)^3((4x^3 + 3x^2)(3x - 2) - 12(x^4 + x^3))/((3x-2)^8)`

Reducing duplicate factors yields:

`(dy)/(dx) = x^2((4x + 3)(3x - 2) - 12(x^2 + x))/((3x-2)^5)`

`(dy)/(dx) = x^2(12x^2 + x - 6 - 12x^2 - 12x)/((3x-2)^5)`

`(dy)/(dx) = (x^2(-11x - 6))/((3x-2)^5)`

**Hence, evaluating the derivative of the given function yields **`(dy)/(dx) = (x^2(-11x - 6))/((3x-2)^5).`

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

Let us simplify the function first:

y = (x^3(x_1)/ (3x-2)64

= (x^4 + x^3)/ (3x-2)^4

Now we will assume that:

y = u/v such that:

u = x^4 + x^3 ==> u' = 4x^3 + 3x^2

v = (3x-2)^4 ==? v' = 4(3x-2)^3 (3) = 12(3x-2)^3

Then we know that:

y' = (u'v - uv')/ v^2

= ( 4x^3 + 3x^2)(3x-2)^4 - (x^4+x^3)(12(3x-2)^3 / (3x-2)^8

= (x^2(x + 3)(3x-2)^4 - 12(x^3(x^2 + 1)(3x-2)^3 / (3x-2)^8

Let us factor x^2(3x-2)^3:

==> y' = (x^2(3x-2)^3 [ (x+3)(3x-2) - 12x(x^2+1) ]/ (3x-2)^8

= x^2[ (3x^2 + 7x - 6) - 12x^3 - 12x) / (3x-2)^5

**= x^2[ -12x^3 + 3x^2 - 5x - 6)/ (3x-2)^5**