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Using the limit definition (f(x+h)-f(x)\h

  1. Oct 10, 2006 #1
    I have to differentiate some problems using the limit definition (f(x+h)-f(x)\h), but Iam having some trouble on a couple.

    1. Square root of 30 is 0 because its a constant correct?

    2. R(t) = 5t^(-3/5) <---- I have tried this problem many times with the definition, but I cant come up with the correct solution and I dont know why!

  2. jcsd
  3. Oct 10, 2006 #2
    1. correct

    2. rewrite this as [tex] 5\frac{d}{dt} \frac{1}{t^{\frac{3}{5}}} [/tex].

    So: [tex] \frac{1}{\sqrt[5]{(t+h)^{3}}} - \frac{1}{\sqrt[5]{t^{3}}} [/tex]. Subtract the fractions. Then rationalize the numerator. take the limit, and then multiply by 5.
    Last edited: Oct 11, 2006
  4. Oct 11, 2006 #3


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    1. Wow, I hate it when people misuse terminology so badly!!

    NO!!! [itex]\sqrt{30}[/itex] is NOT 0!!!. It is about 5.4! Of course, it is true that the derivative of the function [itex]f(x)= \sqrt{30}[/itex], since that is a constant function, is 0.
  5. Oct 11, 2006 #4
    hmm iam still confused.. sorry
    not about square root of 30.. about the next problem
  6. Oct 11, 2006 #5
    courtigrad, you surely meant the derivative wrt. t, didn't you? ;)
  7. Oct 11, 2006 #6
    yeah, just a typo. my bad. also i notice people type "courtigrad" instead of "courtrigrad".
  8. Oct 11, 2006 #7
    the trouble is: what do you multiply the fraction by to rationalize the numerator.

    [tex] t^{\frac{3}{5}} - (t+h)^{\frac{3}{5}} [/tex] multiplied by something to get [tex] t - (t+h) = -h [/tex]

    For example, [tex] t^{\frac{1}{2}} - (t+h)^{\frac{1}{2}}(t^{\frac{1}{2}} + (t+h)^{\frac{1}{2}}) = -h [/tex]
    Last edited: Oct 11, 2006
  9. Oct 11, 2006 #8
    Alright I got it.. Thanks Guys!
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