Is the derivative of $\displaystyle y = 3(2x^2 + x)^2\Longrightarrow\; y' = 6(4x + 1)\Longrightarrow\; y' = 24x + 6$
Or did I screw up?
Sadly, I think you screwed up. When you brought the second power down, remember that you're just reducing the power by one, i.e., the parenthetical $\displaystyle 2x^2 + x$ is still being raised to the first power.
$\displaystyle y = 3(2x^2 + x)^2\Longrightarrow\; y' = 6(2x^2 + x)(4x + 1)\Longrightarrow\; y' = 6(8x^3 + 6x^2 + x)$
so
$\displaystyle y'=48x^3 + 36x^2 + 6x$.
Note your original equation is a fourth-degree polynomial (since it has an x-squared term squared), so your derivative should be a third degree polynomial.
yes let
$\displaystyle f(u)=3u^2 \mbox{ and } f'(u)=6u$
and
$\displaystyle u(x)=2x^2+x \mbox{ and } u'(x)=4x+1$
so
$\displaystyle y=f(u(x))=3(2x^2+x)^2$
so by defintion of the chain rule we get
$\displaystyle y'= f'(u(x)) \cdot u'(x)=6(2x^2+x) \cdot (4x+1) $
I hope this helps.
You can multiply the parts of the derivative in whatever order you want - multiplication is commutative and can be performed in any order. I chose to FOIL multiply the two binomials together (combining the outer and inner terms, as they were both x^2 terms) and then lastly distributing the 6 through.