what is the derivative of (x^2+x)^2?

Answer 1

Even though I normally don't do it, I'm going to denote the derivative of a function as "prime" by adding an apostrophe at the end. That would mean

(sinx)' = cosx
(x^5)' = 5x^4

The way the chain rule works is that you would solve the derivative normally as if it were x, and then take the derivative of the inside. Therefore

(sin[junk])' = (cos[junk]) ([junk]')
([junk]^5)' = (5[junk]^4) ([junk]')

This is no different. If f(x) = (x^2 + x)^2, then
f'(x) = 2(x^2 + x) * (x^2 + x)'
f'(x) = 2(x^2 + x) (2x + 1)

Answer 2

Using the chain rule, you can get the derivative = 2(x^2+x)(2x+1).

Answer 3

First derivative is 2(x^2 + x)(2x + 1) by the chain rule

This can be simplified to 2x(x +1)(2x + 1)

Answer 4

The product is x^4 + 2x^3 + x^2,

and its derivative is therefore 4x^3 + 6x^2 + 2x.

Answer 5

Using the Chain Rule, which states that if if y=f(u) and u=g(x) then the derivative of f(u) would be
d/dx f(u)= f'(u) x g'(x)

so!

d/dx (x^2 + x)^2 = 2(x^2 + x) multiplied by (2x + 1)
=2x^2 + 2x (2x + 1)
= 4x^3 + 2x^2 + 4x^2 + 2x
= 4x^3 + 6x^2 + 2x
hope it helped =)

Answer 6

(x^2 + x)^2
(x^2 + x)(x^2 + x)
x^4 + x^3 + x^3 + x^2
x^4 + 2x^3 + x^2
4x^3 + 6x^2 + 2x

Answer 7

Go to http://www.math.com