Help finding derivative

**ascendancy523** · June 29th 2010, 06:21 PM

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?

**Prove It** · June 29th 2010, 06:45 PM

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.

**ascendancy523** · June 29th 2010, 06:58 PM

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
$=(-x^5-4x^3)^-^\frac{3}{2}+(2x^3+8x)^-^\frac{1}{2}$

**IKoRuPT** · June 29th 2010, 07:06 PM

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

**ascendancy523** · June 29th 2010, 07:46 PM

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