# Find dy/dx as a function of x for y= x^3*(x+1)/(3x-2)^4Find dy/dx as a function of x for y= x^3*(x+1)/(3x-2)^4

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### 2 Answers

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).`

To find dy/dx, we'll differentiate y with respect to x:

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

We'll remove the brackets form numerator:

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

dy/dx = [d/dx(x^4 + x^3)]*(3x - 2)^4 - [d/dx(3x - 2)^4]*(x^4 + x^3)/(3x - 2)^8

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

We'll factorize the numerator by (3x-2)^3

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

We'll simplify, we'll combine like terms and we'll get:

dy/dx = (- 12x^4 - 8x^3 + 3x^2)/(3x - 2)^5

We'll factorize by x^2:

dy/dx = x^2*(-12x^2 - 8x + 3)/(3x - 2)^5