[SOLVED] Is this double derivative correct?

**Some_One** · February 20th 2009, 02:37 PM

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.

**JoshHJ** · February 20th 2009, 02:58 PM

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.

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