# Chain rule for $y = (x^2 + x^3)^4$

I am trying to find the derivative of $y = (x^2 + x^3)^4$

and it seems pretty simple I get

$4(x^2+x^3)^3 (2x+3x^2)$ This seems to be the proper answer to me but the book gets

$4x^7 (x+1)^3 (3x+2)$ and I have no idea how that happened, what process the author went through or why. My answer seems to be a better and more accurate answer since that is what the chain rule will give you.

-

To obtain the book's answer from yours: factor $x^2$ from the $(x^2+x^3)$ term in your expression and apply the rule $(ab)^n=a^nb^n$, factor $x$ from the $(2x+3x^2)$ term, and finally combine the $x^6$ and $x$ terms. \eqalign{ 4(\color{maroon}{x^2+x^3})^3 (\color{darkgreen}{2x+3x^2}) &= 4 \bigl(\color{maroon}{ x^2(1+x)}\bigr)^3\cdot\color{darkgreen}{ x(2+3x)}\cr &=4\cdot (x^2)^3(1+x)^3\cdot x(2+3x)\cr &=4x^6(1+x)^3\cdot x(2+3x)\cr &=4x^7(x+1)^3(3x+2). }
@Jordan No, it wont. Not if you use the exponent rule $(ab)^n=a^nb^n$ –  Galois Group May 5 '12 at 17:10
@Jordan Use $(a\cdot b)^n=a^nb^n$. Here $\bigl(\color{maroon}{x^2}\color{darkgreen}{(1+x)}\bigr)^3 = \color{maroon}{(x^2)^3}\cdot\color{darkgreen}{(1+x)}^3$. –  David Mitra May 5 '12 at 17:10
You're right. So is the book. Because a little algebra reveals that $4(x^2+x^3)^3(2x+3x^2)=4(x^2(x+1))^3(x(2+3x))=4x^6(x+1)^3x(3x+2)=4x^7(x+1)^3(3x+2)$