find y' and y'' for the equation below.

Show work please!!!

y= (x^3)/((x+1)^2)

https://www.wolframalpha.com/input/?i=derivative+of+%28x^3%29%2F%28%28x%2B1%29^2%29+

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

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

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

[ x^3 + 3 x^2] /(x+1)^3

x^2 (x+3) / (x+1)^3

then do it again :)

https://www.wolframalpha.com/input/?i=derivative+of+x^2%28x%2B3%29%2F%28x%2B1%29^3

6x / (x+1)^4

To find the first derivative y', we need to use the quotient rule.

The quotient rule states that for a function y(x) = u(x)/v(x), where u(x) and v(x) are differentiable functions, the derivative of y(x) is given by:

y'(x) = (v(x) * u'(x) - u(x) * v'(x)) / (v(x))^2

Now, let's find the first derivative y' for the given function y = (x^3)/((x+1)^2).

Step 1: Identify u(x) and v(x):
u(x) = x^3
v(x) = (x + 1)^2

Step 2: Find u'(x) and v'(x):
u'(x) = 3x^2
v'(x) = 2(x + 1)

Step 3: Apply the quotient rule formula:
y'(x) = ((x + 1)^2 * 3x^2 - x^3 * 2(x + 1)) / ((x + 1)^2)^2

Simplifying the expression further:
y'(x) = (3x^2(x + 1)^2 - 2x^3(x + 1)) / (x + 1)^4

Now, let's find the second derivative y'' by differentiating y' with respect to x.

Step 1: Differentiate y' with respect to x:
y''(x) = [(3x^2(x + 1)^2 - 2x^3(x + 1)) / (x + 1)^4]'

Step 2: Expand the expression:
y''(x) = [6x(x + 1)(x + 1) + 3x^2(2(x + 1)) - 12x^2(x + 1)^2] / (x + 1)^4

Simplifying the expression further:
y''(x) = [6x(x + 1)^2 + 6x^2(x + 1) - 12x^2(x + 1)^2] / (x + 1)^4

Now, we have found the first derivative y' and the second derivative y'' for the given equation y = (x^3)/((x+1)^2).