# Find second derivitive

1. May 30, 2012

### ForeverMo

Find second derivitive!!

1. The problem statement, all variables and given/known data
f(x)=4(x^2-2)^3

2. Relevant equations
Chain rule??

3. The attempt at a solution
f'=12(x^2-2)^2(2x)
=24x(x^2-2)^2

f''=2(24x)(x^2-2)(2x)
96x^2(x^2-2)
After this, I got lost...

2. May 30, 2012

### Norfonz

Re: Find second derivitive!!

Looks good to me.

3. May 30, 2012

### ForeverMo

Re: Find second derivitive!!

Really? I even went further & distributed and got: 96x^4-192x^2 ... & still got it wrong!

4. May 30, 2012

### Dick

Re: Find second derivitive!!

You'll need to use the product rule when you go from f' to f''. f' is a product of the two functions (24x) and (x^2-2)^2.

5. May 30, 2012

### ForeverMo

Re: Find second derivitive!!

Ok.. so would I set it up like this>> (24)[2(x^2-2)]+(x^2-2)^2(24) ??

6. May 30, 2012

### Dick

Re: Find second derivitive!!

No. That's not right. Review how the product rule works and try it again.

7. May 30, 2012

### HallsofIvy

Staff Emeritus
Re: Find second derivitive!!

No, that is also wrong. Please show exactly how you are trying to do this.

8. May 30, 2012

### ForeverMo

Re: Find second derivitive!!

Product rule: d/dx[fs]=fs'+sf'
24x×2(x^2-2)+(x^2-2)^2×24
Is that the right way?

9. May 30, 2012

### Staff: Mentor

Re: Find second derivitive!!

No, that isn't right either. You also have to use the chain rule when you differentiate (x2 - 2)2.

It's NOT a good idea to use x for multiplication, especially when x is the variable. You've made it slightly easier by bolding some of the variables. In calculus, we generally don't use x for multiplication.