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Major problem in differentiation from first principles

  1. Apr 6, 2012 #1
    I am trying to differentiate the functions xn, eax and ln(ax) from first principles. I have successful in all three, but here's my problem. In finding the limit in each problem, you need to first Taylor expand to remove Δx from the denominator. But the very process of Taylor expansion uses differentiation to find its coefficients. So, it doesn't help if I am trying, for instance, trying to differentiate eax from first principles using Taylor expansion as I have used the derivative of eax itself in its differentiation.

    Any help on this would be great!
     
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  3. Apr 6, 2012 #2

    chiro

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    Hello failexam.

    I know there is a trick you need to do for exponential function but I can't remember it.

    For x^n, use the binomial theorem (or the generalized binomial if n is not an integer). For integer values you can expand the expression using factorial representations and then collect terms to get the final answer.

    Here is the identity:

    http://en.wikipedia.org/wiki/Binomial_theorem

    I just did a quick search for e^x and I got this:

    http://www.phsmath.org/BC/PDF%27s/topics/dex.pdf [Broken]

    For d/dx ln(x) I know you can use high school methods to prove this, but again this was a very long time ago.
     
    Last edited by a moderator: May 5, 2017
  4. Apr 6, 2012 #3

    HallsofIvy

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    Yes, using the Taylor series expansion is NOT "first principles". For xn, at least, you can use, instead, the "binomial theorem":
    [tex](x+ h)^n= \sum_{i= 0}^n \begin{pmatrix}n \\ i\end{pmatrix}h^i x^{n-i}[/tex]
    so so that
    [tex](x+ h)^n- x^n= \sum_{i= 1}^n \begin{pmatrix}n \\ i\end{pmatrix}h^i x^{n-i}[/tex]
    and then
    [tex]\frac{(x+ h)^n- x^n}{h}= \sum_{i= 1}^n \begin{pmatrix}n \\ i\end{pmatrix}h^{i-1}x^{n-i}[/tex]
    Since every term except i= 1 has an "h", the limit as h goes to 0 is just that term:
    [tex]\begin{pmatrix}n \\ 1\end{pmatrix}x^{n- 1}= nx^{n-1}[/tex]

    As far as ex is concerned, ex+h= exeh so that
    [tex]\frac{e^{x+h}- e^x}{h}= \frac{e^xe^h- e^x}{h}= \frac{e^x(e^h- 1)}{h}[/tex]
    so that
    [tex]\lim_{h\to 0}\frac{e^{x+h}- e^x}{h}= e^x \left(\lim_{h\to 0}\frac{e^h- 1}{h}\right)[/tex]

    So the problem becomes finding the limit
    [tex]\lim_{h\to 0}\frac{e^h- 1}{h}[/tex]
    and there are many ways to do that depending on exactly how you are defining the number "e". In fact, one way is to define h as the number that makes that limit 1!

    Since there is no simple property of ln(a+ b), I don't think there is any reasonable way to find the derivative of ln(x) by "first principles" rather than using the fact that ln(x) is the inverse function to ex.
     
  5. Apr 6, 2012 #4

    chiro

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  6. Apr 6, 2012 #5

    HallsofIvy

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    Perfectly good but NOT "first principles". That does precisely what I said- uses the fact that ln(x) is inverse to ex and that the derivative of ex is ex.
     
  7. Apr 6, 2012 #6


    Well, that ln x is inverse to ex still could be considered first principles, just as using theorems related to limits, continuity and stuff. I propose the following

    using only continuity of functions and their limits and basic properties of logarithms and their inverse functions and also the fact

    that [itex]\lim_{x\to ∞}\left(1+\frac{1}{f(x)}\right)^{f(x)}=e\,\,[/itex] for any function f(x) s.t. [itex]\lim_{x\to ∞}f(x)=∞[/itex]:

    [itex]\frac{\ln (x+h)-\ln x}{h}=\ln\left(1+\frac{h}{x}\right)^{1/h}=\ln\left[\left(1+\frac{1}{x/h}\right)^{x/h}\right]^{1/x}\to \ln e^{1/x}=\frac{1}{x}[/itex] .

    DonAntonio
     
  8. Apr 6, 2012 #7
    It depends on how your book is defining the exponential function and the logarithm function. For example, Rudin defines the exponentiol function as: [itex]exp(x) := \sum_{n=0}{\infty} \frac{x^n}{n!}[/itex]. Then, you are allowed to use this with out having to define this as a power series representation for the exponential function.
     
  9. Apr 6, 2012 #8
    Thank you very much for your reply.

    I need to mention here that the binomial theorem is a special case of the Taylor series, so again we are back to square one, right?

    As for ex, that derivation cannot be first principles as we are not using the basic fundamental definition of differentiation, right?

    I shall be very happy if you reply.
     
    Last edited by a moderator: May 5, 2017
  10. Apr 6, 2012 #9
    Thank you so much for contributing to so many of my questions before! I would like to mention that the binomial theorem is a special case of the Taylor series, so we are back to square one, aren't we?

    As for the second derivation (that for ex) , I would be grateful to you if you mention a link which gives a detalied look at how to find that limit.

    Again, thank you very much!
     
  11. Apr 6, 2012 #10

    HallsofIvy

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    It is incorrect to say that the binomial formula is a "special case" of the Taylor's series.

    Every polynomial or power series could be written in terms of a Taylor's series but that does not make it a "Taylor's series". A "Taylor's series" is a series that is specifically derived from the Tayor's formulas- the fact that result is, say, a polynomial does NOT make that polynomial as power series. For example, if a problem asked for the "Taylor's series about x= 2" for [itex]f(x)= x^2- 3x+ 5[/itex], we can calculate that f(2)= 4- 6+ 5= 3, f'(x)= 2x- 3, so that f'(2)= 4- 3= 1, f''(x)= 2 so f''(2)= 2, and all other derivatives are identically 0. The "Taylor's series about x= 2" for f is [itex](2/2!)(x- 2)^2+ 1/1!(x- 2)+ 3[/itex]. That does NOT mean that we can just say "[itex](x- 2)^2+(x- 2)+ 3[/itex]" is a "Taylor's series".

    The binomial formula, which is purely algebraic, predates both Calculus and the "Taylor's series"
     
  12. Apr 6, 2012 #11

    Office_Shredder

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    The binomial theorem CAN be derived via Taylor series, and in fact the expansion is the Taylor series of an appropriate polynomial, but you don't have to do it that way.

    The formula
    [tex] (x+y)^n = \sum_{k=0}^n {n \choose k} x^k y^{n-k} [/tex]
    for an integer value of n and any numbers x and y can be derived purely combinatorially. Let's see for example how to get (x+y)3:
    The possible terms in the expansion are x3, x2y, xy2 and y3. When we distribute it as (x+y)(x+y)(x+y) and multiply everything together, every term is gotten by:choosing one of x or y from the first x+y, choosing one of x or y from the second x+y, choosing one of x or y from the third x+y, and multiplying these things together. The number of ways to get x3 is the number of ways to choose an x from each of the three (x+y)s, which is exactly 3 choose 3. So the coefficient of x3 is 3 choose 3.

    The number of ways to get an x2y is the number of ways to choose 2 x's and one y from the three (x+y) parts, which is 3 choose 2 (choose two places to take an x from, and take y's from the rest). The number of ways to get x2y is 3 choose 2, so the coefficient of x2y is 3 choose 2.

    The number of ways to get an xy2 is the number of ways to pick one x and 2 y's from the three (x+y)s. To count this, we see that there are 3 choose 1 ways to pick the x+y term you take the x from, and then take y's from the rest of them. So the coefficient of xy2 is 3 choose 1.

    Similarly you get the coefficient for y3

    There was nothing special about the number 3 here, in general given (x+y)(x+y)....(x+y), the number of ways you can get xk yn-k is n choose k: pick k (x+y)'s to take an x from, and take a y from the rest of them
     
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