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Find f'(x) of f(x) = 1/sqrt(x-3)

  1. Jun 15, 2003 #1
    I need to find f'(x) of f(x) = 1/sqrt(x-3) using the formal definition. I set the equation up as:
    f'(x) = lim ((1/ sqrt(x + h -3)) - (1/sqrt(x-3)))/h and I am not sure what the next step is...
     
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. Jun 15, 2003 #2
    To solve the limit, multiply the top and bottom of the fraction by the GCF [sqrt(x-3)*sqrt(x-3+h)]. Then multiply the top and bottom of the fraction by the conjugate of the new numerator
    [sqrt(x-3)+ sqrt(x-3+h)]. The radicals on the numerator will disappear, and the h's will cancel. Then you can substitute h=0 to find the limit. The answer is f'(x)= -1/[2*(x-3)^(3/2)].
     
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