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Find the derivative

  1. Mar 14, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]f(x) = x - \sqrt{x}[/tex]


    2. Relevant equations
    [tex]f'(x) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}[/tex]


    3. The attempt at a solution
    I'm not exactly sure how to solve this problem. I plugged the problem into the equation and got stuck... I have a class right now, so I'll have to post my work later.
     
    Last edited: Mar 14, 2007
  2. jcsd
  3. Mar 14, 2007 #2
    It's not easy to find the derivative of sqrt(x) in that way. (You can't use a Taylor expansion- because that makes use of the answer.)

    I think you have to do a binomial expansion of sqrt(x)
    http://www.rism.com/Trig/binomial.htm

    There may be a very easy solution- but I don't see it.
     
  4. Mar 14, 2007 #3
    Duh, wait a minute....

    if g(x)=x^1/2, then g(x)^2=x. Now take derivatives of both sides- which can be proved to give:

    2g(x)dg(x)/dx=1

    dg(x)/dx=1./2g(x)=1/2 (x^-1/2)

    So a method which is a little easier involves proving
    d[g(x)^2]/dx=2g(x)dg(x)/dx, for any fn. g(x)
     
  5. Mar 16, 2007 #4
    I just tried it and got a wrong answer. Your goal is the get everything in the numerator so that it has an h in it. That way u can cancel out the h on the bottom.
     
    Last edited: Mar 16, 2007
  6. Mar 16, 2007 #5
    Using the limit definition is so tedious :(
     
  7. Mar 16, 2007 #6
    Got it...working on the tex
     
    Last edited: Mar 16, 2007
  8. Mar 16, 2007 #7
    Well, I read one chapter a head so I figured out how to do it using the power rule.. :-/ it's a way easier that way.
     
  9. Mar 16, 2007 #8
    My comp. is being stupid so I can't write the whole thing =\ I had a test once where we had to do a lot of those derivatives without the shortcuts. U can solve it that way by multiplying both the numerator and the denominator by [tex]\sqrt{x+h}+\sqrt{x}\\[/tex]

    edit: yay my comp is working lets give this another try

    [tex]
    \[\lim_{h \to 0}\frac{x+h+\sqrt{x+h}-x-\sqrt{x}}{h}\]
    & =\]\lim_{h \to 0}\frac{h+\sqrt{x+h}-\sqrt{x}}{h}\]
    & =\[\lim_{h \to 0}\frac{h(\sqrt{x+h}+\sqrt{x})+x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\]
    & =\[\lim_{h \to 0}\frac{h(\sqrt{x+h}+\sqrt{x})+h}{h(\sqrt{x+h}+\sqrt{x})}\]
    & =\[\lim_{h \to 0}1+\frac{1}{\sqrt{x+h}+\sqrt{x}}\]
    & =\[1+\frac{1}{2\sqrt{x}}\]
    [/tex]

    Damn I suck at this.
     
    Last edited: Mar 16, 2007
  10. Mar 16, 2007 #9
    ok, I'll have to practice that one. :) thanks
     
  11. Mar 16, 2007 #10
    U can also use the addition rules of limits to calculate derivatives for x and sqrt x separately and then add them.
     
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