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I just can't get this

  1. Jul 19, 2006 #1
    ok, the reciprocal rule: g(x)' / g(x)^2

    In this book I'm reading:

    problem: -1/x^2

    This seems simple, it should be -2/x^3 right?

    But the answer says: 2x^(-3)

    Am I missing something here?
     
  2. jcsd
  3. Jul 19, 2006 #2

    Office_Shredder

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    Ok.... so we have -1/x^2. Now, the reciprocal rule is that [1/g(x)]' = -g'(x)/g2(x).

    So (-1/x^2)' = -(-2x/x^4) = 2/x^3
     
  4. Jul 19, 2006 #3

    NateTG

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    You're off by a sign.

    A quick derivation of the 'reciprocal rule':

    [tex]f(x)=\frac{1}{x}[/tex]
    [tex]\frac{d}{dx} f(x)= - \frac{1}{x^2}[/tex]
    (This follows from the power rule: [itex]\frac{1}{x}=x^{-1}[/itex])

    Now, let's say [itex]g(x)[/itex] is a differentiable non-zero function with non-zero derivative on some domain, then on that domain we have:
    [tex]\frac{d}{dx} f(g(x)) = g'(x) f'(g(x))[/tex]
    by the chain rule. Applying what we know about [itex]f(x)[/itex] gives:
    [tex]\frac{d}{dx} \frac{1}{g(x)}=g'(x) \times - \frac{1}{{g(x)}^2}=-\frac{g'(x)}{{g(x)}^2}[/tex]
    (With apolgies for sloppy notation.)
     
    Last edited: Jul 19, 2006
  5. Jul 19, 2006 #4
    oh ok, thanks guys, my reciprocal rule was wrong, I missed the negative.
     
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