What is the derivative of #(x^2+x)^2#?

Question

What is the derivative of  #(x^2+x)^2#?

Isabella Ackerman · Accepted Answer

#y^' = 4x^3 + 6x^2 + 2x#

Using the sum and power rules, you can differentiate this function. You'll see that this function can be rewritten 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#

Currently, you can see from the sum rule that for functions that have 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) + ...#

As for you, you've

#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 distinguish between these fractions, apply the power rule.

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

Consequently, your derivative will 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)#

Ezra Adams · Answer

undefined To find the derivative of the function (x^2+x)^2, you can apply the chain rule. First, differentiate the outer function (u^2) with respect to u, and then multiply by the derivative of the inner function (x^2+x) with respect to x.

Using the chain rule, the derivative is:

2*(x^2+x)*(2x+1)