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Derivative of 10^x using limit definition

  1. Jun 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Obtain the first derivative of 10x by the limit definition.


    2. Relevant equations
    f'(x)=limh->0 f(x+h)-f(x)/h


    3. The attempt at a solution
    f'(x)=limh->0 10x+h-10x/h
    I also know that h=1 as x approaches 0.

    Now, how do I make it so that you aren't dividing by h=0.
     
  2. jcsd
  3. Jun 29, 2008 #2

    HallsofIvy

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    10x+h= 10x10h so
    [tex]\frac{10^{x+h}- 10^x}{h}= 10^x\frac{10^h- 1}{h}[/tex]
    Now the question is, what is
    [tex]\lim_{h\rightarrow 0}\frac{10^h- 1}{h}[/tex]?
     
  4. Jun 29, 2008 #3
    To find
    [tex]\lim_{h\rightarrow 0}\frac{10^h- 1}{h}[/tex]
    I can't plug h=0 in because I would be dividing by 0. Do I plug in 1 since the limit as h approaches 0 is 1?
     
  5. Jun 29, 2008 #4

    nrqed

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    if you plug h = 0 then 10^h = 1. But you want the next order correction which you explect should be proportional to h. So you expect something of the form

    [tex] lim_{h\rightarrow 0} ~10^h = 1 + a h + \ldots [/tex]

    where a is some numerical value and the dots represent terms of higher powers in h. The problem is to find the value of a.

    Here is the trick: use that any number x may be written as [tex] e^{ \ln x} [/tex]. Then use what you know about rules for logs and then Taylor expand the exponential.
     
  6. Jun 29, 2008 #5
    I don't understand what this means. What is "Taylor expanding" and the "next order correction"?
     
  7. Jun 29, 2008 #6

    nrqed

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    Have you ever seen the relation
    [tex] e^\epsilon = 1 + \epsilon + \frac{\epsilon^2}{2} + \ldots [/tex] ?

    This is a Taylor expansion.

    Have you ever proved using the limit definition that the derivative of e^x is e^x? Then you must have used something similar to the above.

    If you haven't proved the e^x case and the expansion I wrote above is not familiar to you then I will let someone else help you because I don't see at first sight any other approach.


    EDIT: do you know what the limit as h goes to zero of [tex] \frac{e^h -1}{h} [/tex] gives? maybe you have been told this without proving it. If you know the result of this limit and are allowed to use it, then I can show you how to finish your problem. If not, I don't see how to help, unfortunately.


    Best luck!
     
  8. Jun 29, 2008 #7
    I've never proved e^x. Thank you for trying to help though.
     
  9. Jun 29, 2008 #8

    nrqed

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    Sorry. I can tell you that the limit as h goes to zero of 10^h is
    [tex] lim_{h \rightarrow 0} 10^h = lim_{h \rightarrow 0} e^{\ln 10^h} = lim_{h \rightarrow 0} e^{h \ln 10} \approx 1 + h \ln 10 [/tex]
    where I used an identity for logs and then I used the expansion of the exponential I mentioned earlier. Form this you can get the final answer of your question.


    Hopefully someone else will be able to find a way to show this result in some other way but I can't think of any!


    Best luck
     
  10. Jun 29, 2008 #9

    Redbelly98

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    Taylor expansion requires knowledge of what the derivative is. But we don't know the derivative, that is what we are supposed to find.
     
  11. Jun 29, 2008 #10

    Redbelly98

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    I don't know if this will be useful, but one might try using the definition of e:

    [tex]
    e = \lim_{N \rightarrow \infty} (1+\frac{1}{N})^N = \lim_{a \rightarrow 0} (1+a)^{1/a}
    [/tex]

    or

    [tex]
    e^A = \lim_{N \rightarrow \infty} (1+\frac{1}{N})^{NA} = \lim_{a \rightarrow 0} (1+a)^{A/a}
    [/tex]


    Also, the fact that

    [tex]
    10^h = e^{h \ln(10)}
    [/tex]
     
  12. Jun 29, 2008 #11

    HallsofIvy

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    I suspect this was given as a preliminary to the derivative of ex so the derivative of ex cannot be used. It is easy to see that the derivative of ax, for a any positive number, is a constant times ax. It is much harder to determine what that constant is! It's not too difficult to show that, for some values of a, that constant is less than 1 and, for some values of a, larger than 1. Define e to be the number such that that constant is 1. That is, define "e" by
    [tex]\lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1[/tex]
    As Redbelly98 said, 10h= eh ln(10) so
    [tex]\frac{10^h- 1}{h}= \frac{e^{h ln(10)}- 1}{h}[/itex]
    If we multiply both numerator and denominator of that by ln(10) we get
    [tex]ln(10)\left(\frac{e^{h ln(10)}-1}{h ln(10)}\right)[/tex]
    Clearly, as h goes to 0 so does h ln(10) so if we let k= h ln(10) we have
    [tex]ln(10)\left(\lim_{h\rightarrow 0}\frac{e^{h ln(10)}-1}{h ln(10)}\right)= ln(10)\left(\lim_{k\rightarrow 0}\frac{e^k- 1}{k}\right)[/tex]
    so the limit is ln(10) and the derivative of 10x is ln(10)10x.

    That is NOT something I would expect a first year calculus student to find for himself!
     
  13. Jun 29, 2008 #12

    dynamicsolo

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    Moreover, Taylor series are generally taught in second-semester calculus, while covering infinite sequences and series, while the limit

    [tex]
    \lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1
    [/tex]

    is often demonstrated or proven (if it is not simply stated without proof) in the first-semester course, shortly after having covered limits and while developing the rules of differentiation. I hardly expected that the OP would have seen Taylor series yet...

    I believe the proof given in textbooks usually revolves around the limit definition of e, which Redbelly98 gives in post #10.
     
  14. Jun 29, 2008 #13

    nrqed

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    Agreed. I should not have mentioned Taylor series. They seem so natural to me now that I tend to use them without even thinking about it.
    This is why I then asked the OP if he/she had seen the formula you wrote just above. I hope he/she has. Because if he/she has to go back to the limit definition and prove the above identity in order to solve the original question, this problem seems much more challenging than I would expect as an assignment problem at that level!!
     
  15. Jun 29, 2008 #14

    dynamicsolo

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    I suspect that OP's textbook presents the limit



    [tex]
    \lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1
    [/tex]

    somewhere in the chapter and that a student is just asked to recognize that they could apply it, in something like the manner Halls suggests in post #11...
     
  16. Jun 30, 2008 #15
  17. Jun 30, 2008 #16
    Sorry, the dereivative of 10^x is of course ln10(10^x) but the Math forum derivation of the derivative of e^x is still helpful
     
  18. Jun 30, 2008 #17
    More thoughts on this problem.

    f(x)=10^X=e^(x*ln10)

    f(x+h)= e^(ln10(x+h))=e^(ln10*x)*e^(ln10*h)

    Plugging into definitation of derivative and simplifying gives

    f'(x)= limit(h goes to 0) 10^x(10^h-1)/h

    tabulating the limit as h goes to 0 of (10^h-1)/h= ln10
     
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