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Proof d/dx e^x=e^x using substitution

  1. Jun 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Proof d/dx e^x=e^x, use e=limit (1+1/h)^h h->infinity

    Show how that implies d/dx e^x=e^x

    t

    2. Relevant equations



    3. The attempt at a solution
    Ive tried using chain rule--wasnt accepted
    Also, I did e=(1+h)^[1/h]-->e^h=1+h, then reduced to e^x, still didnt accept

    How can I prove it using limit substitution? He told me to google finding limits by substitution then he said e^h-1=U
     
  2. jcsd
  3. Jun 29, 2012 #2

    Mentallic

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    So have you tried taking the derivative of ex in the limit form?

    [tex]\frac{d}{dx}e^x=\frac{d}{dx}\lim_{h\to \infty}\left(1+\frac{1}{h}\right)^{hx}[/tex]
     
  4. Jun 29, 2012 #3
    Ive attempted, but its a limit inside a limit?
     
  5. Jun 29, 2012 #4

    Mentallic

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    Aren't you allowed to use the derivative rules? [tex]\frac{d}{dx}a^{cx}=c\ln(a)a^{cx}[/tex]
     
  6. Jun 29, 2012 #5
    Exactly what i did when he first said, its obvious, but he wants mathematical reasoning using limits..
     
  7. Jun 29, 2012 #6
    [tex](e^x)'=\lim_{\delta\to 0}\dfrac{e^{x+\delta}-e^{x}}{\delta}=e^{x}\lim_{\delta\to 0}\dfrac{e^{\delta}-1}{\delta}[/tex]
    Now you want to show
    [tex]\lim_{\delta\to 0}\dfrac{e^{\delta}-1}{\delta}=1[/tex]
    And you are given that
    [tex]\lim_{h\to \infty}\left(1+\frac{1}{h}\right)^{h}=e[/tex]
    That's very close. Don't forget the hint you are given: letting u=e^h-1.
     
    Last edited: Jun 29, 2012
  8. Jun 29, 2012 #7

    Mentallic

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    Hmm... ok...

    Can you clarify what he meant by e^h-1=U?

    What about if we tried to just go back to the basics. So we'll use any positive number a,

    [tex]\frac{d}{dx}a^x=\lim_{h\to 0}\frac{a^{x+h}-a^x}{h}[/tex]

    [tex]=\lim_{h\to 0}\frac{a^xa^h-a^x}{h}[/tex]

    [tex]=a^x\lim_{h\to 0}\frac{a^h-1}{h}[/tex]

    Now, we're interested in the case where the limit is equal to 1,

    [tex]\lim_{h\to 0}\frac{a^h-1}{h}=1[/tex]

    [tex]\lim_{h\to 0}a^h-1=\lim_{h\to 0} h[/tex]

    [tex]\lim_{h\to 0}a^h =\lim_{h\to 0}1+h[/tex]

    Can you see where this is heading?

    edit: klondike beat me to it.
     
  9. Jun 29, 2012 #8
    uh-oh, this doesn't sound right.
     
  10. Jun 29, 2012 #9
    Put that in my "attempts" he says that e^h ≠ (1+h) because its a ≈ and not =, I even told him under a certain domain "its true" he wont accept lol..
     
  11. Jun 29, 2012 #10
    So what exactly should I be substituting??
     
  12. Jun 29, 2012 #11

    Mentallic

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    Actually, yeah, you're right. We should be starting off with the limit expression for e.
     
  13. Jun 29, 2012 #12
    let [tex]\frac{1}{h}=e^{\delta}-1 \quad;; h\to \infty \quad as \quad \delta \to 0[/tex] then write δ in terms of h, and plug them back in, and you are all set.

     
    Last edited: Jun 29, 2012
  14. Jun 29, 2012 #13
    Im a bit confused here. Where are you getting the (1/h)=e^δ−1 from?
     
  15. Jun 29, 2012 #14
    Because e^δ−1 goes zero as δ goes zero. You want some quantity goes zero as something goes zero or, more conveniently as something goes infinity as you are given a known limit as its independent variable goes infinity.

    Hence, it's mostly conveniently to set the quantity to 1/h as h approach infinity. It satisfies both condition and closely resemble what you are given.
    You can, of course first set y=e^δ−1 as well.


     
  16. Jul 2, 2012 #15
    how do I put δ in terms of h..where am I plugging it in??
    EDIT: SORRY I realized h=1/(e^δ - 1)

    do I insert that h into the e= limit (1+ 1/h)^h ??

    thanks
     
  17. Jul 2, 2012 #16
    [tex]e^{\delta}-1=\frac{1}{h}[/tex]
    [tex]\delta=ln(1+\frac{1}{h})[/tex]
    And put these 2 into it. And recall that aln(x)=ln(x^a). I'm sure you will figure it out.
     
  18. Jul 3, 2012 #17
    mhm Im doing this:

    lim e^δ−1/ δ =1

    e^δ - 1= 1/h
    δ=ln(1+ 1/h)

    so I get: limit [(1/h)]/ [ln(1+(1/h))] h-->infinity

    Im getting 1
     
  19. Jul 3, 2012 #18
    Now go back to check post #6. Isn't that what you're trying to prove?

     
  20. Jul 3, 2012 #19
    ohhhhhh so 1=e^x holy ****!
     
  21. Jul 3, 2012 #20
    well 1 implies it..wow so cool
     
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