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Derivatives of e^x

  1. Jun 13, 2010 #1
    1.

    f(x)= ax^2 = ae^TR, nez
    Prove f '(x) = a(n) x^(n-1)


    2.
    n does not equal 0


    3.
    I don't even understand it
     
    Last edited: Jun 13, 2010
  2. jcsd
  3. Jun 13, 2010 #2

    HallsofIvy

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    Re: Prove

    I don't understand it either- go back and check the problem again!

    In particular, "F(x)= ax^2 = ae^TR" makes no sense- it appears to be saying that F(x) is equal to two different functions. The only way I could make sense of it is if they are the same function: x^2= e^(ln(x^2))= e^(2 ln(x)). How is "TR" defined?

    In any case "f'(x)= a(n)x^(n-1)" makes no sense because there was no mention of either "f" or "n" before. Assuming you mean "F" and "f" to be the same function (a very bad practice- capital letters and small letters are different symbols and should represent different variables) there still was n "n" in the original formulation.

    Of course, it is true that if f(x)= ax^n, not x^2, then f'(x)= a(n x^(n-1)). That is normally proved by using induction on n or the binomial theorem, long before the derivatives of e^x and ln(x) are introduced. But if you do have those, then you could argue that f(x)= ax^n= a e^(n ln(x)) so that f'(x)= a e^(n ln(x))*(n/x)= a(x^n)(n/x)= a(nx^(m-1)).
     
  4. Jun 13, 2010 #3
    Re: Prove

    f(x)= ax^2 = aeTR (Not sure if its exponential), neZ
    prove f '(x) = a(n)x^(n-1)
    n does not equal 0
    sorry about the mistake

    This was for a pre-calculus class but the teacher stated that it was in introductory calculus, so i assume that it is before the derivatives of e^x.

    Btw do you know what type of problem this is so I can look it up in the book?
     
    Last edited: Jun 13, 2010
  5. Jun 13, 2010 #4
    Re: Prove

    I am confused on how you got:
    F' (x) = ae ^( n ln(x)) * (n/x) = a(x^n)(n/x) then to a(nx ^(n-1))
     
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