Finding dervivative of 4x^5(3x^2+3)^3

1. Oct 25, 2014

1. The problem statement, all variables and given/known data
Find the derivative of $4x^5(3x^2+3)^3$ by applying the chain rule twice.

2. Relevant equations

3. The attempt at a solution
I have had a go, but I have not used the chain rule twice so it probably will not be accepted even if it is mathmatically correct. But that is what I wanted to check, if the maths and thinking behind what I done is correct, as when I put it into a derivative calculator the answer is different.

What I did was break it down in to two parts, using the chain rule to find the derivative of the second term (the one in paranatheses) and then applied the product rule.

I said...

If $g=(3x^2+3)^3$ and $h=3x^2+3$ and that $c=h^3$

Then, $g'=h' \cdot c' = 6x \cdot 3(3x^2+3)^2 = 18x(3x^2+3)^2$

Then if $m=4x^5$ then $y'=m' \cdot g + g' \cdot m$
$y'=20x^4(3x^2+3)^3 + 18x(3x^2+3)^2 \\ y'=20x[x^3(3x^2+3)^3+\frac{9}{10}(3x^2+3)^2 ]$

Then I was not sure if doing this makes it look neater or even if its possible, as I am not too sure but if the above is correct then could I do...
$20x(3x^2+3)^2[x(3x^2+3)+\frac{9}{10}]$

I would appreciate it if someone could check my maths and my method, and also to give advice on how to solve it by using the chain rule twice.

Thanks :)

2. Oct 25, 2014

vela

Staff Emeritus
You didn't multiply g' by m in the second term.

3. Oct 25, 2014

Oh yeah, duh!

So then it is...

$y'=20x^4(3x^2+3)^3 + 18x(3x^2+3)^2 4x^5 \\ y'=20x^4(3x^2+3)^3 + 72x^6(3x^2+3)^2 \\ y'=x^4[20(3x^2+3)^3 + 72x^2(3x^2+3)^2 ] \\$

4. Oct 25, 2014

vela

Staff Emeritus
I don't see how you can apply the chain rule twice in this problem.

One suggestion I have would be to pull the constants out front in the beginning to make the numbers easier to deal with.
$$y = 4x^5(3x^2+3)^3 = (4\times3^3)[x^5(x^2+1)^3]$$ Then you just differentiate $x^5(x^2+1)^3$ and multiply the result by 108 in the end.

5. Oct 25, 2014