Second Derivative of 4(x^2-2)^3: Apply Chain Rule for f(x)

[ForeverMo]

Find second derivitive!

Homework Statement

f(x)=4(x^2-2)^3

Homework Equations

Chain rule??

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...

 

[Norfonz]

Looks good to me.

 

[ForeverMo]

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

 

[Dick]

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.

 

[ForeverMo]

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.

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

 

[Dick]

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

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

 

[HallsofIvy]

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

 

[ForeverMo]

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

 

[Mark44]

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

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.