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Problem on finding the second derivative.

  1. Feb 26, 2005 #1


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    I have a problem on finding the second derivative for this function:

    [tex] \frac {x}{x^2-4} [/tex]

    For the first derivative, I got:

    [tex] \frac {-x^2-4}{(x^2-4)^2} [/tex]

    Now here is where I am stuck! So far for the second derivative, I got this (Please check!):

    [tex] \frac {-2x(x^2-4)^2-4x(-x^2-4)(x^2-4)}{(x^2-4)^4} [/tex]

    I need this simplified! I know, it's an easy question. I may have lost my mind! :rofl:
    Also please show me the steps. Thanks!

    BTW. I am new to this forum and just learned the latex system. It is very complicated. Took me a while just to write out the problems above. I guess I gotta get used to it.
  2. jcsd
  3. Feb 26, 2005 #2


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    Gold Member

    First derivative is:

    [tex]\frac{1}{x^2 -4} - \frac{2x^2}{(x^2-4)^2}[/tex]

    Second derivative is:
    [tex]\frac{-6x}{(x^2-4)^2} + \frac{8x^3}{(x^2-4)^3}[/tex]
  4. Feb 28, 2005 #3


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    it would be easier if you rewrite the original equation like this:
    x(x^2 -4)^-1. So you don't need to deal with fractions
  5. Feb 28, 2005 #4


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    Not necessarily. It is actually more complicated using the chain rule for this particular function. But either way is fine. I solved the problem already. Thanks for the advice though.

    Here's a formula for finding quotient derivatives: (I'm sure you know it already)

    Function: [tex]\frac {f(x)}{g(x)}[/tex]

    Formula: [tex]\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}[/tex]
    Last edited: Feb 28, 2005
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