1. ## Chain Rule

Is the derivative of $y = 3(2x^2 + x)^2\Longrightarrow\; y' = 6(4x + 1)\Longrightarrow\; y' = 24x + 6$
Or did I screw up?

2. Originally Posted by R3ap3r
Is the derivative of $y = 3(2x^2 + x)^2\Longrightarrow\; y' = 6(4x + 1)\Longrightarrow\; y' = 24x + 6$
Or did I screw up?
Not quite

$y=3(2x^2+x)^2$

using the chain rule we get

$
\frac{dy}{dx}=\underbrace{6(2x^2+x)}_{PowerRule} \cdot \underbrace{4x+1}_{derivative Inside}
$

3. 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 $2x^2 + x$ is still being raised to the first power.

$y = 3(2x^2 + x)^2\Longrightarrow\; y' = 6(2x^2 + x)(4x + 1)\Longrightarrow\; y' = 6(8x^3 + 6x^2 + x)$

so

$y'=48x^3 + 36x^2 + 6x$.

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.

4. Originally Posted by TheEmptySet
Not quite

$y=3(2x^2+x)^2$

using the chain rule we get

$\frac{dy}{dx}=\underbrace{6(2x^2+x)}_{PowerRule} \cdot \underbrace{4x+1}_{derivative Inside}
$
So your saying multiply the $6(2x^2+x)$ out then multiply that by $(4x+1)$ then theres the derivative?

5. Originally Posted by R3ap3r
So your saying multiply the $6(2x^2+x)$ out then multiply that by $(4x+1)$ then theres the derivative?
When using the chain rule you follow this general formula:

The derivative of $f(g(x))$ is equal to $f'(g(x))\cdot g'(x)$.

6. Originally Posted by topher0805
When using the chain rule you follow this general formula:

The derivative of $f(g(x))$ is equal to $f'(g(x))\cdot g'(x)$.
Alright so the poster above me is correct then.

7. Originally Posted by R3ap3r
So your saying multiply the $6(2x^2+x)$ out then multiply that by $(4x+1)$ then theres the derivative?
yes let

$f(u)=3u^2 \mbox{ and } f'(u)=6u$

and

$u(x)=2x^2+x \mbox{ and } u'(x)=4x+1$

so

$y=f(u(x))=3(2x^2+x)^2$

so by defintion of the chain rule we get

$y'= f'(u(x)) \cdot u'(x)=6(2x^2+x) \cdot (4x+1)$

I hope this helps.

8. 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.