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Definition of a derivative

  1. Sep 30, 2004 #1
    okay... could someone show me how to find the derivative of 5/(x^2/5) using ONLY the definition of a derivative( (f(x+h)-f(x))/h as h approaches 0), and not using any of the rules or proving the rules

    strictly algebraicly...(probably spelled that wrong)
     
  2. jcsd
  3. Sep 30, 2004 #2
    i get it to (5( (x)^2/5 - (x+h)^2/5) )/((x^2+xh)^2/5)
    but cant get rid of the radicals... =/
     
  4. Sep 30, 2004 #3
    After about an hour (I was about to just give up) of messing around I had an idea. I needed to know if I could factor [tex]x^5 - y^5[/tex] and it turns out that I can!

    For your reference

    [tex]x^5-y^5 = (x-y)(x^4 + x^3y + x^2y^2 +xy^3 + y^4)[/tex]

    what you have is
    [tex]
    5\lim_{h\rightarrow 0} \frac {x^\frac {2} {5} - (x+h)^\frac {2} {5}} {h(x^2 + 2hx)^\frac {2} {5}} [/tex]

    Now the trick to getting rid of that h in the denominator is to figure out how you could go about adding the two parts of the numerator.

    Start by noticing that the numerator is a difference of squares

    [tex]
    x^\frac {2} {5} - (x+h)^\frac {2} {5}
    [/tex]

    Which of course you can factor out to

    [tex]
    (x^\frac {1} {5} + (x+h)^\frac {1} {5})(x^\frac {1} {5} - (x+h)^\frac {1} {5})
    [/tex]

    Now all you have got to do is figure out how to get ride of that nasty little [tex]h[/tex] in the denominator. That is not so easy unless you multiply the entire quotient by 1. That is, multiply by

    [tex]
    \frac {x^\frac {4} {5} + x^\frac {3} {5}(x+h)^\frac {1} {5} + x^\frac {2} {5}(x+h)^\frac {2} {5} + x^\frac {1} {5}(x+h)^\frac {3} {5} + (x+h)^\frac{4} {5}} {x^\frac {4} {5} + x^\frac {3} {5}(x+h)^\frac {1} {5} + x^\frac {2} {5}(x+h)^\frac {2} {5} + x^\frac {1} {5}(x+h)^\frac {3} {5} + (x+h)^\frac{4} {5}}
    [/tex]

    Which, if you will recall from our earlier results, we can easily multiply the [tex] (x^\frac {1} {5} - (x+h)^\frac {1} {5})[/tex] with the [tex]x^\frac {4} {5} + x^\frac {3} {5}(x+h)^\frac {1} {5} + x^\frac {2} {5}(x+h)^\frac {2} {5} + x^\frac {1} {5}(x+h)^\frac {3} {5} + (x+h)^\frac{4} {5}[/tex] and that will get rid of the fifth root and thus will give us a nice and neat numerator from which you can factor out the h. If you have gotten this far the rest is cake.

    Regards
     
    Last edited: Sep 30, 2004
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