# Thread: Higher-order derivatives (simple problem)

1. ## Higher-order derivatives (simple problem)

I am having trouble finding the second derivative of this function. A step by step solution would be greatly appreciated.

$\displaystyle f(x) = 3(2-x^2)^3$

2. Originally Posted by Archduke01
I am having trouble finding the second derivative of this function. A step by step solution would be greatly appreciated.

$\displaystyle f(x) = 3(2-x^2)^3$
$\displaystyle y = f(x) = 3(2 - x^2)^3$.

Let $\displaystyle u = 2 - x^2$ so that $\displaystyle y = 3u^3$.

$\displaystyle \frac{du}{dx} = -2x$

$\displaystyle \frac{dy}{du} = 9u^2$

$\displaystyle = 9(2 - x^2)^2$.

Therefore $\displaystyle f'(x) = \frac{dy}{dx} = -18x(2 - x^2)^2$.

Now $\displaystyle f''(x) = \frac{d}{dx}\left(\frac{dy}{dx}\right)$

$\displaystyle = -18x\frac{d}{dx}[(2 - x^2)^2] + (2 - x^2)^2\frac{d}{dx}(-18x)$

$\displaystyle = -18x(-2x)(2)(2 - x^2) - 18(2 - x^2)^2$

$\displaystyle = 72x^2(2 - x^2) - 18(2 - x^2)^2$

$\displaystyle = 18(2-x^2)[4x^2 - (2 - x^2)]$

$\displaystyle = 18(2-x^2)(3x^2 - 2)$.

3. Originally Posted by Prove It
$\displaystyle y = f(x) = 3(2 - x^2)^3$.

Let $\displaystyle u = 2 - x^2$ so that $\displaystyle y = 3u^3$.

$\displaystyle \frac{du}{dx} = -2x$

$\displaystyle \frac{dy}{du} = 9u^2$

$\displaystyle = 9(2 - x^2)^3$.

Therefore $\displaystyle f'(x) = \frac{dy}{dx} = -18x(2 - x^2)^3$.

I'm grateful for your post, but I think you made a mistake. Shouldn't it be $\displaystyle f'(x) = \frac{dy}{dx} = -18x(2 - x^2)^2$?

BTW, the answer is $\displaystyle 18(2-x^2)(5x^2 - 2)$. I'm not sure where you went wrong with your method.

Here's what I did;
y' = 9(2-x^2)^2 (-2x)
y'' = 18(2-x^2)(-2x)(-2)

At this point I realized my method wasn't matching up to the given answer so I stopped. Could somebody tell me where I went wrong?

4. Originally Posted by Archduke01
I'm grateful for your post, but I think you made a mistake. Shouldn't it be $\displaystyle f'(x) = \frac{dy}{dx} = -18x(2 - x^2)^2$?

BTW, the answer is $\displaystyle 18(2-x^2)(5x^2 - 2)$. I'm not sure where you went wrong with your method.

Here's what I did;
y' = 9(2-x^2)^2 (-2x)
y'' = 18(2-x^2)(-2x)(-2)

At this point I realized my method wasn't matching up to the given answer so I stopped. Could somebody tell me where I went wrong?
Yes you are right. And since my first derivative was wrong, so is my second derivative. Will edit now.

5. Originally Posted by Prove It
$\displaystyle = -18x\frac{d}{dx}[(2 - x^2)^2] + (2 - x^2)^2\frac{d}{dx}(-18x)$

I'm with you up until before the plus sign. How did you get everything after it?

6. Originally Posted by Archduke01
I'm with you up until before the plus sign. How did you get everything after it?
It's a combination of the product chain rules.

7. Originally Posted by Prove It

$\displaystyle = -18x(-2x)(2)(2 - x^2) - 18(2 - x^2)^2$

$\displaystyle = 72x(2 - x^2) - 18(2 - x^2)^2$
Shouldn't that be a 72x^2?

EDIT: Nevermind, I understand the procedure now - thank you very much!

8. Originally Posted by Archduke01
Shouldn't that be a 72x^2?

EDIT: Nevermind, I understand the procedure now - thank you very much!
Yes it should be. Editing again.