You can differentiate this function by using the *sum* and *power rules*. Notice that you can rewrite this function as

#y = (x^2 + x)^2 = [x(x+1)]^2 = x^2 * (x+1)^2#

#y = x^2 * (x^2 + 2x + 1) = x^4 + 2x^2 + x^2#

Now, the sum rule tells you that for functions that take the form

#y = sum_(i=1)^(oo)f_i(x)#

you can find the derivative of #y# by adding the derivatives of those individual functions.

#color(blue)(d/dx(y) = f_1^'(x) + f_2^'(x) + ...#

In your case, you have

#y^' = d/dx(x^4 + 2x^2 + x^2)#

#y^' = d/dx(x^4) + d/dx(2x^2) + d/dx(x^2)#

#y^' = d/dx(x^4) * 2d/dx(x^3) * d/dx(x^2)#

To differentiate this fractions, use the power rule

#color(blue)(d/dx(x^a) = ax^(a-1))#

So, your derivative will come out to be

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

#y^' = color(green)(4x^3 + 6x^2 + 2x)#

**Alternatively**, you can use the chain rule to differentiate #y#.

#color(blue)(d/dx(y) = d/(du)(y) * d/dx(u))#

In your case, you have #y = u^2# and #u=x^2 + x#, so that you get

#dy/(dx) = d/(du)u^2 * d/dx(x^2 + x)#

#dy/dx = 2u * (2x + 1)#

#dy/dx = 2(x^2 + x) * (2x + 1)#

#dy/dx = (2x^2 + 2x) * (2x + 1)#

#dy/dx = 4x^3 + 2x^2 + 4x^2 + 2x = color(green)(4x^3 + 6x^2 + 2x)#