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Wierd Q to take a derivative of

  1. Apr 23, 2005 #1
    wierd Q to take a derivative of....

    y=(x)/((x+2)(x+3)(x+4)).
    how to do u take the derivative??? HELP!!!!!! :confused:
     
  2. jcsd
  3. Apr 23, 2005 #2

    dextercioby

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    So is it

    [tex] y(x)=\frac{x}{(x+2)(x+3)(x+4)} [/tex]

    Do u know to apply the

    1.Quotient's derivative rule.
    2.Leibniz product rule...?

    Daniel.
     
  4. Apr 23, 2005 #3

    arildno

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    AARGH!!!!!!!!!!!!!!!!!!!!!!!!!

    DO NOT DOUBLE, TRIPLE, OR QUADRUPLE POST IN THE FUTURE!!!!!!!!!!!!!!!!!!!!!
    :grumpy:
     
  5. Apr 23, 2005 #4
    The easiest way would be to rewrite it like this first:

    y=(x) * (x+2)(^-1) * (x+3)(^-1) * (x+4)(^-1)
     
  6. Apr 23, 2005 #5
    haha, i guess i'd get mroe help dat way, but, oops hehe
     
  7. Apr 23, 2005 #6
    the easiest way is actually probably to just multiply out the denominator:

    [tex]\frac{x}{(x+2)(x+3)(x+4)} = \frac{x}{x^3 + 9x^2 + 26x + 24}[/tex]

    and then just use the product rule and the fact that

    [tex]\frac{d}{dx}\frac{1}{f(x)} = -\frac{f^\prime (x)}{\left[f(x)\right]^2}[/tex]
     
  8. Apr 24, 2005 #7

    arildno

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    At least, now you've learnt your lesson, right? :wink:
     
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