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Homework Help: Simple Single Variable Derivative Question

  1. Oct 28, 2005 #1
    wow! I am having a major brain fart here .. I'm in multivariable calc but I need to take the derivative of a single variable function using the definition of derivative. The function is:

    f(x)=abs(x)^(3/2) and I need to find the derivative when a=0

    so by the def of derivative, the derivative equals:

    lim (h->0) of [abs(a+h)^(3/2)-abs(a)^3/2]/h

    I know I'm not allowed to just "plug in" a=0 to get the derivative at this point .. but how do I do it? I can't combine the absolute value terms because of the triangle inequality (for ex. abs(-6)-abs(3)=3 but abs(-6-3)=9)

    help! I feel really stupid right now. :grumpy:
     
  2. jcsd
  3. Oct 28, 2005 #2

    quasar987

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    why can't you plug a=0 ?:grumpy:
     
  4. Oct 28, 2005 #3
    I thought when you were using definition of derivitive, you aren't allowed to plug in for a until you have it simplified? Am I wrong? I could be .. it's been a year since I've had to work with definition of derivative.
     
  5. Oct 28, 2005 #4
    There's no such rule. "a" is a placeholder for a constant; place your constant in there to get the derivative there. Leave it as "a" to get an expression for the derivative (if it exists) at all values of "a". :smile:
     
  6. Oct 28, 2005 #5

    benorin

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    use the *other* definition of limit:
    f '(a) = lim (x->a) of [abs(x)^(3/2)-abs(a)^3/2]/(x-a)

    then factor the top as a difference of cubes.
     
  7. Oct 28, 2005 #6

    HallsofIvy

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    You cannot plug in h= 0 because that would give 0/0. But if you are finding the derivative at a given x= a, you certainly can plug in that value of a before you start. You are looking for
    [tex]lim_{h\rightarrow0}\frac{h^{\frac{3}{2}}}{h}[/tex]
    That should be easy.
     
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