1. ## 2nd Derivative

How do you solve the 2nd derivative of this problem?
$\displaystyle f'(x)=-48x/(x^2 +12)^2$
I know you have to use the quotient rule and chain rule. I just don't know how to get to the answer correctly. I keep screwing up, so if you could just walk me through this it would be great.

2. Originally Posted by sgonzalez90
How do you solve the 2nd derivative of this problem?
$\displaystyle f'(x)=-48x/(x^2 +12)^2$
I know you have to use the quotient rule and chain rule. I just don't know how to get to the answer correctly. I keep screwing up, so if you could just walk me through this it would be great.
Let g(x) = f'(x)

Let $\displaystyle u = -48x$ and so $\displaystyle u' = -48$

Let $\displaystyle v = (x^2+12)^2$ and using the chain rule we get $\displaystyle v' = 2(x^2+12) \times 2x = 4x(x^2+12) = 4x^3+48x$

The quotient rule says: $\displaystyle g'(x) = f''(x) = \frac{u'v - uv'}{v^2}$

Can you take it from there?

3. Originally Posted by sgonzalez90
How do you solve the 2nd derivative of this problem?
$\displaystyle f'(x)=-48x/(x^2 +12)^2$
I know you have to use the quotient rule and chain rule. I just don't know how to get to the answer correctly. I keep screwing up, so if you could just walk me through this it would be great.

i prefer to write it like this

$\displaystyle f'(x)=-48x(x^2+12)^{-2}$

$\displaystyle f''(x)=-48(x^2+12)^{-2}+(-2)(2x)(x^2+12)^{-3}(-48x)$

or use

$\displaystyle f'(x)=\frac{-48x}{(x^2+12)^2}$

$\displaystyle f''(x)=\frac{-48(x^2+12)^{2}-(-48x)(2)(2x)(x^2+12)}{(x^2+12)^{4}}$

you simplify

4. Originally Posted by e^(i*pi)
Let g(x) = f'(x)

Let $\displaystyle u = -48x$ and so $\displaystyle u' = -48$

Let $\displaystyle v = (x^2+12)^2$ and using the chain rule we get $\displaystyle v' = 2(x^2+12) \times 2x = 4x(x^2+12) = 4x^3+48x$

The quotient rule says: $\displaystyle g'(x) = f''(x) = \frac{u'v - uv'}{v^2}$

Can you take it from there?

The answer is: $\displaystyle 144x^2-576/(x^2 +12)^3$

5. Originally Posted by Amer
i prefer to write it like this

$\displaystyle f'(x)=-48x(x^2+12)^{-2}$

$\displaystyle f''(x)=-48(x^2+12)^{-2}+(-2)(2x)(x^2+12)^{-3}(-48x)$

or use

$\displaystyle f'(x)=\frac{-48x}{(x^2+12)^2}$

$\displaystyle f''(x)=\frac{-48(x^2+12)^{2}-(-48x)(2)(2x)(x^2+12)}{(x^2+12)^{4}}$
Okay, like i said. I already know that this is what I'm supposed to get. But How do i simplify this to get the answer I just posted?

6. $\displaystyle f''(x)=\frac{-48(x^2+12)^{2}-(-48x)(2)(2x)(x^2+12)}{(x^2+12)^{4}}$

Factor out $\displaystyle (x^2 + 12)$:

$\displaystyle f''(x)=\frac{-48(x^2+12)-(-48x)(2)(2x)}{(x^2+12)^{3}}$

Expand:

$\displaystyle f''(x)=\frac{-48x^2 - 576 + 192x^2}{(x^2+12)^{3}}$

You get:

$\displaystyle f''(x)=\frac{144x^2 - 576}{(x^2+12)^{3}}$