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

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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**

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**

To find the first derivative of y = x^3*(x+1)/(3x-2)^4.

We know (u(x)/v(x))' = {u'(x) v(x) - u(x)v(x)}/(v(x))^2

u(x) = x^3(x+1)= x^4+x^3.

u'(x) = (x^4+x^3) ' = 4x^3+3x^2

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

Therefore y' = {x^3(x+1)/(3x-2)^4)}' = {( 4x^3+3x^2)(3x-2)^4-(x^4+x^3)( 12(3x-2)^3.)}/(3x-2)^8

{x^3(x+1)/(3x-2)^4)}'={x^2(3x-2)^3[4x+3+12x^2+12x]}/(3x-2)^8

{x^3(x+1)/(3x-2)^4)}' = { x^2(12x^2+16x+3}/(3x-2)^5.

{x^3(x+1)/(3x-2)^4)}' = x^2{12x^2+16x+3}/(3x-2)^5.