derivative of $y=(x^2+x^3)^4$

Question

I can't figure out where I am going wrong.

$$y=(x^2+x^3)^4$$ chain rule it first $$4(x^2+x^3)^3* \frac d{dx}(x^2+x^3)$$

which should become: $$4(x^2+x^3)^3(2x+3x^2)$$ factoring out should give me: $$4*x^2*x(1+x)^3(2+3x)$$

which to me says the answer is: $4x^3(1+x)^3(2+3x)$

but the book says: $4x^7(1+x)^3(2+3x)$

where did I go wrong?

This is your mistake:$$(x^2+x^3)^3=(x^2(1+x))^3=\color{red}{x^6}(1+x)^3$$ — DonAntonio, May 19 '14 at 19:05
@DonAntonio It should be only $x$ at the end instead of $x^3$ — The very fluffy Panda, May 19 '14 at 19:06

score 1 · Answer 1 · answered May 19 '14 at 19:06

1

Your mistake is in the factoring out. When you factor $x^2$ out from the $(x^2+x^3)^3$ term, it becomes $(x^2)^3(1+x)^3=x^6(1+x)^3$ instead of $x^2(1+x)^3$

answered May 19 '14 at 19:06

Aidan F. Pierce

1,465

derivative of $y=(x^2+x^3)^4$

1 Answers1