# Help finding derivative

• Jun 29th 2010, 07:21 PM
ascendancy523
Help finding derivative
Find f''(x) if $f(x)=\frac{x^2}{\sqrt{x^2+4}}$

so which method would be easier to solve for? The quotient rule or should I bring $\sqrt{x^2+4}$ to the numerator as $(x^2+4)^-^\frac{1}{2}$ and use the product rule and chain rule?
• Jun 29th 2010, 07:45 PM
Prove It
Quote:

Originally Posted by ascendancy523
Find f''(x) if $f(x)=\frac{x^2}{\sqrt{x^2+4}}$

so which method would be easier to solve for? The quotient rule or should I bring $\sqrt{x^2+4}$ to the numerator as $(x^2+4)^-^\frac{1}{2}$ and use the product rule and chain rule?

Either way will involve using the chain rule. The product rule is always easier though, because you don't have to worry about the order.
• Jun 29th 2010, 07:58 PM
ascendancy523
Quote:

Originally Posted by Prove It
Either way will involve using the chain rule. The product rule is always easier though, because you don't have to worry about the order.

So then

$f'(x)=x^2(-\frac{1}{2})(x^2+4)^-^\frac{3}{2}(2x)+(x^2+4)^-^\frac{1}{2}(2x)$

$=(-x^3)(x^2+4)^-^\frac{3}{2}+(x^2+4)^-^\frac{1}{2}(2x)$

then I can't remember if it's acceptable to distribute, so
$=(-x^5-4x^3)^-^\frac{3}{2}+(2x^3+8x)^-^\frac{1}{2}$
• Jun 29th 2010, 08:06 PM
IKoRuPT
Quote:

Originally Posted by ascendancy523
So then

$f'(x)=x^2(-\frac{1}{2})(x^2+4)^-^\frac{3}{2}(2x)+(x^2+4)^-^\frac{1}{2}(2x)$

$=(-x^3)(x^2+4)^-^\frac{3}{2}+(x^2+4)^-^\frac{1}{2}(2x)$

then I can't remember if it's acceptable to distribute, so
$=(-x^5-4x^3)^-^\frac{3}{2}+(2x^3+8x)^-^\frac{1}{2}$

You can't distribute because what is in parenthesis is raised to a power. If you want to distribute, you'd need to raise the binomial to the exponent first (PEMDAS).

Edit: However, you can factor out $(x^2+4)^-^\frac{3}{2}$
• Jun 29th 2010, 08:46 PM
ascendancy523
Quote:

Originally Posted by IKoRuPT
You can't distribute because what is in parenthesis is raised to a power. If you want to distribute, you'd need to raise the binomial to the exponent first (PEMDAS).

Edit: However, you can factor out $(x^2+4)^-^\frac{3}{2}$

So then something like this...
$(x^2+4)^-^\frac{3}{2}[(-x^3)+(x^2+4)(2x)]$??
• Jun 29th 2010, 08:48 PM
Prove It
You can't take out $(x^2 + 4)^{-\frac{3}{2}}$, but you can take out $(x^2 + 4)^{-\frac{1}{2}}$.
• Jun 29th 2010, 09:00 PM
ascendancy523
Quote:

Originally Posted by Prove It
You can't take out $(x^2 + 4)^{-\frac{3}{2}}$, but you can take out $(x^2 + 4)^{-\frac{1}{2}}$.

ah, that's right..so maybe

$(x^2+4)^-^\frac{1}{2}[(-x^3)(x^2+4)^-^1+(2x)]$??