[SOLVED] Is this double derivative correct?

• Feb 20th 2009, 02:37 PM
Some_One
[SOLVED] Is this double derivative correct?
Given:

$
f(x) = \frac{x^2}{x^2 + 4}
$

Using the quotient rule, I found that:

$
f'(x) = \frac{8x}{(x^2 + 4)^2}
$

$
f''(x) = \frac{-8(3x^4 + 8x^2 - 16)}{(x^2 + 4)^4}
$

Is this correct? I am a little suspicious of $f''(x)$. It seems to be a little too big, but I can't seem to find any mistake in my work.
• Feb 20th 2009, 02:58 PM
JoshHJ
Your 1st derivative is definitely correct. Now I'll check the second derivative:

$
f''(x) = \frac{8(x^2+4)^2 - (8x)(4x)(x^2+4)}{(x^2+4)^4}
$

Simplifying:
$
f''(x) = \frac{8*(x^2+4)^2 - 32x^2 * (x^2+4)}{(x^2+4)^4} =
\frac{8x^4 + 64x^2 + 128 - 32x^4 - 128x^2}{(x^2+4)^4}
$

More simplifying:
$
f''(x) = \frac{-8(3x^4 + 8x^2 - 16}{(x^2 + 4)^4}
$

So you are correct.