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Homework Help: Find the derivative, but using the definition thereof

  1. Jan 25, 2010 #1
    1. Prove that the derivative x(-2/3) is equal to (-2/3)x(-5/3) using the definition of a derivative "limit as h approaches zero of [(f(x+h) - f(x))]/h".



    2. Relevant equations

    f(x) = x(-2/3)

    limit as h approaches zero of [(f(x+h) - f(x))]/h

    sorry I couldn't type out the full equation, I hope that everyone understands what this is...




    3. So first I substituted everything in

    lim as h approaches zero of [f(x+h) - f(x)]/h


    then I multiplied the numerator by the conjugate root. After doing so, I then expanded (x + h)4 = x4+4hx3+6h2x2+4h3x+h4

    then i plugged it in and got 1 over the cubed root of x4+4hx3+6h2x2+4h3x+h4 MINUS 1 over the cubed root of x4.

    Now remember, we're taking the limit of this whole thing (line above) divided by H, as h approaches zero.

    that's where I don't know what to do...where do I go from there? Algebraically, I can't really simplify or cancel anything...

    It's 3 am and I've been working on this question for over an hour, and I'm dead tired, so I do apologize if my post has broken any rules, or my attempt at a solution is not detailed enough...I simply do not have the energy to show more work than i already have. I would scan the work i've done and upload it, but my scanner doesn't work on this computer (64bit OS).

    If you cannot answer the question or help because of that, then I do apologize for wasting anyones time. My basic question is, how do I prove that only using the definition of a derivative?

    Again, thanks so much in advance, and I apologize for my lack of detail and perhaps a very unorganized post.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 25, 2010 #2
    My reaction would be "huh?" unfortunately. The way to approach this question is by direct substitution of f(x) into the limit expression. So,
    [tex]
    \lim_{h\rightarrow 0} \frac{f (x + h) - f (x)}{h} = \lim_{h\rightarrow 0} \frac{(x + h)^{-\frac{2}{3}} - x^{-\frac{2}{3}}}{h}
    [/tex]

    Next, expand using the binomial expansion:
    [tex]
    (x + h)^{-\frac{2}{3}} = x^{-\frac{2}{3}} (1 + \frac{h}{x})^{-\frac{2}{3}} = x^{-\frac{2}{3}} (1 + (-\frac{2}{3})(\frac{h}{x}) + \frac{(-\frac{2}{3})(-\frac{5}{3})}{2} (\frac{h}{x})^2 + ...)
    [/tex]
    Try to continue by yourself first before viewing the spoiler below.







    This reduces the first limit expression to
    [tex]
    \lim_{h\rightarrow 0} x^{-\frac{2}{3}} ((-\frac{2}{3})(\frac{1}{x}) + \frac{(-\frac{2}{3})(-\frac{5}{3})}{2} (\frac{h}{x^2}) + ...) = (-\frac{2}{3}) x^{-\frac{5}{2}}
    [/tex]
     
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