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1/f differentiation- why am I not getting it

  1. Nov 20, 2007 #1
    What am I missing here


    Problem Statement:

    Let f be a differentiable at point x and assume f(x) is not equal to 0. Show that function 1/f is differentiable at x=0 and (1/f)' = -f'(x)/( f(x) ^2 )

    Solution
    1/f = f^-1.
    So taking its derivative I get the following: -f^-2. But I fail to see where the f'(x) term comes from. Plz advise.


    Thanks

    Asif
     
  2. jcsd
  3. Nov 20, 2007 #2

    Mute

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    You have to apply the chain rule when you take the derivative:

    If u is a function of x, then if you take the derivative of some function g(u) with respect to x,

    [tex]\frac{d}{dx}g(u) = \frac{dg}{du}\frac{du}{dx}[/tex]

    In this case, g(u) is your function 1/u, and u = f(x). Does this make sense?
     
  4. Nov 21, 2007 #3

    arildno

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    Dearly Missed

    Well, you could do it directly like this:
    [tex]\lim_{h\to{0}}\frac{\frac{1}{f(x+h)}-\frac{1}{f(x)}}{h}=\lim_{h\to{0}}\frac{f(x)-f(x+h)}{hf(x+h)f(x)}=-\lim_{h\to{0}}(\frac{f(x+h)-f(x)}{h})*\frac{1}{f(x+h)f(x)}[/tex]
    See if you manage to get something out of this.
     
  5. Nov 21, 2007 #4

    arildno

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    Because you differentiated with respect to f, rather than with respect to x.
     
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