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Differentiate from first principles

  1. Oct 31, 2013 #1
    1. The problem statement, all variables and given/known data

    Differentiate from first principles with respect to x: x^n (where n belongs to the natural numbers).


    2. Relevant equations

    f'(x) = Lim x→0 [f(x+h) - f(x)]/h

    3. The attempt at a solution

    f'(x) = Lim x→0 [f(x+h) - f(x)]/h
    = Lim x→0 [(x+h)^n - x^n]/h

    I need some help simplifying this.
     
  2. jcsd
  3. Oct 31, 2013 #2

    UltrafastPED

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    [(x+h)^n - x^n] ... use the binomial expansion theorem to get the first few terms of (x+h)^n:

    x^n + n x^(n-1) h^1 + (n|2) x^(n-2) h^2 ... where (n|2) is the number of combinations of n items taken 2 at a time.

    Then simplify, and take the limit.
     
  4. Oct 31, 2013 #3
    I don't have a very good understanding of the binomial theorem. I'm not sure what this means: "where (n|2) is the number of combinations of n items taken 2 at a time" or how it helps me.

    My only experience of the binomial theorem has been making the h into a 1 thus giving me (x+h)^n = [(1/h)^n].(x/h + 1)^n where n has always been given to me.
     
  5. Oct 31, 2013 #4

    UltrafastPED

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  6. Oct 31, 2013 #5
    I tried the above and just came out with x^n-1. I'm not sure where to obtain the n I need.

    I'm having some serious issues with [sin(x+h) - sin(x)]/h and {[1/(x+h^1/2)]-[1/(x^1/2)]}. I think the second of which can be solved with (a-b)(a+b) = a^2 - b^2 or am I completely wrong?
     
  7. Oct 31, 2013 #6

    Mark44

    Staff: Mentor

    Show us what you did when you expanded (x + h)n using the binomial theorem.
     
  8. Oct 31, 2013 #7
    I've just managed to do it. I expanded (x+h)^n, subtracted x^n and divided by h. Substituting 0 in for h I am left with just one term which had no h after the division, which was (n 1)x^n-1 which I now realise gives me n.x^n-1.

    But I am still unsure about the others I mentioned above. I tried just about every identity I could to get the sine one to work, unless I missed something. I have lim θ→0 Sinθ/θ = 1 but I'm not sure how to make it helpful.
     
    Last edited: Oct 31, 2013
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