Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivative of y=(1+1/x)^x

  1. Nov 30, 2006 #1
    How can can prove that the derivative of y=(1+1/x)^x is always positive?
    Thank you
     
  2. jcsd
  3. Nov 30, 2006 #2
    y=(1+1/x)^x
    Let u=1+1/x,y=u^x
    y=u^x
    ln y=x ln u
    y'/y=x/u
    dy/du=[x/(1+1/x)](1+1/x)^x
    =x(1+1/x)^(x-1)
    du/dx=-1/x^2
    dy/dx=(dy/du)(du/dx)
    =x(1+1/x)^(x-1) (-1/x^2)
    =-(1+1/x)^(x-1) /x
     
  4. Dec 1, 2006 #3
    y=(1+1/x)^x
    Let u=1+1/x,y=u^x
    y=u^x
    ln y=x ln u
    Then how come y'/y=x/u ?
    I think the derivative of y will be calculated as follow:
    y'/y=(x ln u)' = x*(1/u)u' + ln u (derivative of a product)
    ==> y' = y[x/(1+1/x)* (-1/x^2) +ln (1+1/x)]
    y' = y [ln(1+1/x)-1/(x+1)]
    y' = (1+1/x)^x [ln(1+1/x)-1/(x+1)]

    The problem here is that I can not prove y' always positive
     
  5. Dec 1, 2006 #4

    radou

    User Avatar
    Homework Helper

    Your derivative is correct. Now, you should look at the domain and image of this function.
     
  6. Dec 1, 2006 #5
    it's not. at x = 0 it's undefined, for instance
     
  7. Dec 1, 2006 #6
    Yes, I should avoid the value x=0. Except from that, let say, for all x >0, why the derivative is always positive?
     
  8. Dec 1, 2006 #7

    radou

    User Avatar
    Homework Helper

    So? Graph the function and it will be pretty obvious that y'(x) > 0, for every x from [tex](0, +\infty)[/tex].
     
  9. Dec 1, 2006 #8
    because the question asked about the function without any constraints on x's domain.

    part of the problem people have while tackling questions like this is they don't emphasise the domain, as you pointed out yourself

    also a graph may make it "pretty obvious", but this isn't a proof
     
  10. Dec 1, 2006 #9

    radou

    User Avatar
    Homework Helper

    Well, constraints on domains are self understood. It wouldn't make any sence to talk about values of the function on an interval where it isn't defined, would it?
     
  11. Dec 1, 2006 #10
    depends how far into analysis you get. undefined points are crucial and interesting at some levels
     
  12. Dec 1, 2006 #11
    Thank you all for discussing. I have graphed the curve and the derivative will leads to plus zero (+0) when x --> infinity. So in the domain of 0<x<inf, the derivative must be positive. But infact that is not a proof.
     
  13. Dec 1, 2006 #12
    have you tried the binomial theorem?

    edit: done some calculating and that doesn't seem to help. i only have pencil and paper here, lol
     
    Last edited: Dec 1, 2006
  14. Dec 1, 2006 #13

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The domain of this function is [itex](-\infty, -1) U (0, \infty)[/itex], not just x> 0.
     
  15. Dec 1, 2006 #14
    being really pedantic: the domain is any set you (or the questioner) chooses so long as the rule is defined on that set

    a function is defined by the rule and the domain

    i still haven't found a neat way to do proof though
     
  16. Dec 2, 2006 #15
    y' = (1+1/x)^x [ln(1+1/x)-1/(x+1)]
    We make it : y' = A*B
    Where A= (1+1/x)^x and B=ln(1+1/x/)-1/(x+1)

    Because A>0 (every x>0), so we check the state of B.
    The derivative of B: B'= [ln(1+1/x)-1/(x+1)]'=...=1/(x+1)[1/(x+1)-1/x] <0 with all the values of x>0. So B is a contravariant function.
    We also have the fact that when x--> inf the B leads to zero, so all the values of B in the domain x>0 must be positive.
    Then y' = A*B also is positive.
     
  17. Dec 2, 2006 #16

    Gib Z

    User Avatar
    Homework Helper

    Anyone noticed the limit of this function as it approaches infinity is e? Just some interesting info for you all..
     
  18. Dec 4, 2006 #17
    Yes, and the function y=(1+a/x)^x will approaches e^a when x goes to infinity. Can you prove that?
     
  19. Dec 5, 2006 #18

    Gib Z

    User Avatar
    Homework Helper

    There is no need to prove that, it is true by definition. The challenge would be to prove different definitions of [itex]e[/itex] are equivalent.
     
  20. Dec 5, 2006 #19

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It is true by one definition of e. If, for example, you define
    [tex]ln(x)= \int_1^x \frac{1}{t}dt[/tex]
    and then define exp(x) to be the inverse function to ln(x), you would need to prove that
    [tex]\lim_{x\rightarrow \infty}\left(1+ \frac{a}{x}\right)^x= exp(a)[/tex]
     
  21. Dec 5, 2006 #20
    To simplify the question, prove that (1+1/x)^x < 0 for x < 0 and x =/= 0 and x=/= -1. For x < -1, (1+1/x)^x takes a positive value. If we define ln -|x| = - ln |x|, then the derivative is negative for x < -1. It is not always positive.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Derivative of y=(1+1/x)^x
  1. Area of y=x^2 at x=1 (Replies: 17)

  2. Y = (-1)^x (Replies: 3)

Loading...