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Proving the Product rule

  1. Dec 17, 2004 #1
    Hi all!
    I've been working with differentiation for quite some time now, and I know how to use the product rule.
    f(x) = u(x)*t(x)
    f'(x) = u'(x)*t(x) + t'(x)u(x)
    But what I don't understand is how one derives it... :cry:

    Oh, and one more thing.
    How do we know a number is irrational, like pi and e ?
    Like, how do we know that it has an infinite number of decimals?
    Is it because when you do the decimal operation you will eventually see that there's a sequence of decimals which repeats itself, or not? Something else perhaps?

    Thanks in advance.
    Last edited: Dec 17, 2004
  2. jcsd
  3. Dec 17, 2004 #2


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    Hi, it can be derived using the definition of the derivative. Start with:

    [tex]f'(a)=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}[/tex]

    Now use the fact that f(x)=u(x)*t(x). Do you want to try to proceed on your own?

    Proving a number like pi or e is irrational can be tricky. I think there's a post somewhere here where HallsOfIvy posted one of the easier proofs, I'll see if I can find it.

    edit-aha! https://www.physicsforums.com/showthread.php?t=8193 The method was originally by I. Niven though Halls is probably refering to an article of A. Parks, who generalized it. It's not short, but it's accessible after first year calculus.
    Last edited: Dec 17, 2004
  4. Dec 17, 2004 #3

    matt grime

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    That seems a remarkably uneasy* proof of the fact that e is irrational, which can be done from first principles merely from the definition of e.

    let e=p/q, where p and q have no common factors

    consider p(q-1)! - eq!

    the resulting number can be shown to be both integer and between 0,1 and not equal to either. contradiction, hence e is irrational. It's a proof in the first year of any maths course.

    irrationality of pi is harder.

    * in the sense of one has to prove more complicated theorems to use it
    Last edited: Dec 17, 2004
  5. Dec 17, 2004 #4


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    Pictures sometimes help (of course, sometimes they don't). Suppose you have a rectangle with a width of u and a height of t. How does the area change if you adjust u and t by small amounts? (Say, by an amount of Δu and Δt respectively)
  6. Dec 17, 2004 #5


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    Absolutely, but for pi it's decent. Getting e is a nice bonus, though it also gets e to the power of any non-zero rational for free too.

    I should have pointed out the simpler method for e, thank you. Though it's not p(q-1)! - eq! that will be an integer between 0 and 1, but q! times the terms in the infinite series for e after the qth term (maybe I'm misreading).
  7. Dec 17, 2004 #6

    matt grime

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    no, you're probably correct, few too many beers not helping my brain
  8. Dec 17, 2004 #7
    Let there be a y = uv, where u = g(x), v = q(x) and y = f(x)

    y = uv,

    y+dy = (u +du)(v+dv) = uv + udv + vdu + dudv

    then dy = udv +vdu + dudv,

    dudv moves toward zero thus,
    dy/dx = u(dv/dx) + v(du/dx) which is f'(x) = [g(x)]q'(x) + [q(x)]g'(x)
  9. Dec 17, 2004 #8
    i would use the mean-value theorem:
    [tex] f'(x) =\lim_{x\rightarrow y} \frac{f(x)-f(y)}{x-y}[/tex], [tex] g'(x) =\lim_{x\rightarrow y} \frac{g(x)-g(y)}{x-y}[/tex],

    for e i think the proof uses power series, for pi,
    let [tex]\pi = a/b[/tex] where a & b are positive integers, and let

    [tex]f(x) = \frac{x^n(a-bx)^n}{n!}[/tex]

    [tex]F(x) = f(x) - f^(2) (x) + f^(4) (x) - ... + (-1)^nf^(2n)(x)[/tex]

    since [tex]n!f(x)[/tex] has integer coefficients & terms in x whose degree is at least n, f(x) & its derivatives have integer values @ x=0, and also for [tex]x= \pi = a/b[/tex]

    (note that [tex]f(x) = \frac{b^n}{n!}(\pi - x)^n[/tex], so [tex]f(x) = f(\pi-x)[/tex] ).

    from calculus,
    [tex] \frac{d}{dx} [F'(x)sin(x) - F(x)cos(x)] = F''(x)sin(x) + F(x)sin(x) = f(x)sin(x)[/tex]

    and [tex]\int_{0}^{\pi} f(x)sin(x) dx = [F'(x)sin(x) - F(x)cos(x)]^\pi_0 = F(\pi) - F(0)[/tex], an integer, since the ith derivative of f evaluated at 0 & at [tex]\pi[/tex] are integers. but for [tex]0 < x < \pi[/tex],

    [tex]0 < f(x)sin(x) < \frac{\pi^n a^n}{n!}[/tex]

    so that the integral is positive, but arbitrarily small for large n, since (another result from calculus) [tex] \lim_{n\rightarrow \infinity} \frac{x^n}{n!} = 0[/tex],

    contradicting the fact that the integral above is an integer. so [tex]\pi[/tex] is irrational (that's ivan niven's excellent proof; i got it as a handout in a course i did last spring)

    it seems that most proofs of irrationality are by contradiction. the proofs that e, pi & square roots of a (non-square) numbers are all donce by pretending that they are rational, setting them to a/b & going from there
    Last edited: Dec 17, 2004
  10. Dec 18, 2004 #9
    Thanks for your help guys! I'll do my best to understand!
  11. Jan 4, 2005 #10


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    There are definitely more elegant derivations of the product rule. But here's one way:

    Let r(x)=p(x)q(x).

    Let s(x,h)=[p(x+h)-p(x)]/h, h is not 0
    Let t(x,h)=[q(x+h)-q(x)]/h, h is not 0


    write p(x+h) in terms of s(x,h) and q(x+h) in terms of t(x,h)


    Multiply out the numerator and simplify

    =limh->0 [h^2s(x,h)t(x,h) + p(x)ht(x,h) + hs(x,h)q(x)]/h
    =limh->0 h^2s(x,h)t(x,h)/h + limh->0 p(x)ht(x,h)/h + limh->0 hs(x,h)q(x)/h
    =limh->0 hs(x,h)t(x,h) + p(x)limh->0 t(x,h) + q(x)limh->0 s(x,h)

    Note that limh->0 t(x,h)=q'(x) and limh->0 s(x,h)=p'(x)

    =p'(x)q'(x)limh->0 h + p(x)q'(x) + q(x)p'(x)
    =0 + p(x)q'(x) + q(x)p'(x)
    =p(x)q'(x) + q(x)p'(x)
  12. Jan 5, 2005 #11


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  13. Jan 6, 2005 #12
    what about this:

    [tex] (fg)'(x) = \lim_{y \rightarrow x} \frac{ (fg)(y) - (fg)(x)}{y-x} = \lim_{y \rightarrow x} \frac{ f(y)g(y) - f(x)g(x)}{y-x} = \lim_{y \rightarrow x}\left( f(y) \frac{g(y) - g(x)}{y-x} + g(x) \frac{f(y) - f(x)}{y-x}\right) = f(x)g'(x)+f'(x)g(x)[/tex]
  14. Jan 6, 2005 #13


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    Sorry to disspoint you,but that's the traditional method of proving it.I was taught this method in school,when in the 11-th grade.
    It's nothing spectacular in it,just elegance.

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