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Another basic derivative problem

  1. Dec 29, 2011 #1
    1. The problem statement, all variables and given/known data

    Find f'(x) of f(x)= 1 over x-1
    using the definition of a derivative


    2. Relevant equations
    definition of a derivative is
    f'(x)= lim as h→ 0 of f(x+h) - f(X) ALL OVER h


    3. The attempt at a solution
    I have no idea how to do this using the definition of a derivative...
     
  2. jcsd
  3. Dec 29, 2011 #2

    SammyS

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    If [itex]\displaystyle f(x)=\frac{1}{x-1}\,,[/itex] then what is f(x+h) ?
     
  4. Dec 29, 2011 #3
    This is where I draw a blank.....

    is it simpily (1/x-1 +h) - f(x) over h meaning it cancels out to be h/h or 0?
     
  5. Dec 29, 2011 #4

    Mark44

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    No.
    All Sammy asked you was what is f(x + h)?

    In the right side of the formula equation, replace x by x + h. That's how function notation works.
     
  6. Dec 29, 2011 #5

    SammyS

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    Don't forget that f(x) is 1/(x-1), and No, h does not cancel the way you are supposing it does.

    You will have:
    [itex]\displaystyle\frac{f(x+h)-f(x)}{h}=\frac{\displaystyle\frac{1}{x+h-1}-\frac{1}{x-1}}{h}[/itex]​

    You need to add the two fractions in the numerator -- use a common denominator.

    After some simplification, you will get the h in the overall denominator to cancel.

    Now we see why algebra skills are important for Calculus.
     
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