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Derivitive of 2^x

  1. May 19, 2004 #1
    Hey, ive been tiring over this problem for a while....how do you get the derivitive of y=2^x using the definition of derivitive? Any help would be greatly appreciated.

    -Hackensack-
     
  2. jcsd
  3. May 19, 2004 #2

    AKG

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    Hmm... You mean from first principles?

    In case you don't already have it, there is a general rule for functions of the form:

    [tex]f(x)\ =\ a^{u(x)}[/tex]. That rule is:

    [tex]f'(x)\ =\ (a^{u(x)})[\ln(a)][u'(x)][/tex].

    I'll have to think about how one would go about doing this from first principles. See this site for proof of the exponential derivative rule.
     
  4. May 19, 2004 #3

    mathman

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    The simplest way to get the derivative is to use the equation:
    2x=exln2
    The derivative of the r.h.s is:
    ln2.exln2=ln2.2x

    To get the same result from first principles, expand the r.h.s. of the first equation in a power series, with x replaced by a (a=deltax), when you work on the difference quotient:
    (2x+a-2x)/a
     
  5. May 20, 2004 #4

    Zurtex

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    I realise this has been posted up two times but they actually both confused me so I'd like to see if I can help make it a little clearer. (a is a constant)

    [tex]y = a^x[/tex]

    [tex]\ln y = \ln \left( a^x \right)[/tex]

    [tex]\ln y = x \ln (a)[/tex]

    [tex]\frac{dy}{dx} \left( \frac{1}{y} \right) = \ln (a)[/tex]

    [tex]\frac{dy}{dx} = y \ln(a)[/tex]

    [tex]\frac{dy}{dx} = a^x \ln(a)[/tex]


    Hope that helps, whenever differentiating [itex]f(x)^{g(x)}[/itex] it can be tackled in the same way.
     
  6. May 20, 2004 #5

    HallsofIvy

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    If you really mean "using the definitionof the derivative then you would start this way:

    y(x+h)- y(x)= 2x+h- 2x= 2x(2h-1) so

    (y(x+h)- y(x))/h= 2x((2h- 1)/h)

    The derivative is the limit of that as h->0.

    Notice that the "2x" factor does not involve h while the "(2h-1)/h" factor does not involve x. The limit will be C2x where the constant C is the limit of (2h-1)/h. The hard part is proving that that limit is ln(2).
     
  7. May 20, 2004 #6
    :wink: (2^x) Lim ((1+1)^h -1)/h =
    h-->0
    =(2^x)Lim(1+h+h(h-1)/2!+h(h-1)(h-2)/3!+ - - - +(-1))/h=
    h-->0
    =(2^x)Lim(h+h(h-1)/2!+h(h-1)(h-2)/3!+ - - - -)/h =
    h-->0
    =(2^x)Lim(1+(h-1)/2!+(h-1)(h-2)/3!+ - - - -)
    h-->0
    =(2^x)(1-1/2!+2!/3!-3!/4!+ - + - + -)= (2^x)(1-1/2+1/3-1/4+1/5- + - +)

    =(2^x)(ln2). I've just finished HallsofIvy's work.
     
    Last edited: May 20, 2004
  8. May 21, 2004 #7
    implicit diffy.
     
  9. May 21, 2004 #8
    I imagine that using a series expansion would involve the derivative anyway. May as well use L'Hospital's if you can do that...

    cookiemonster
     
  10. May 21, 2004 #9

    HallsofIvy

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    If you were referring to wisky40's work, that was just an expansion using the binomial theorem, known long before calculus itself (and regularly used to find derivatives in the early days of calculus).
     
  11. May 21, 2004 #10

    Gokul43201

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    I haven't tried this but i imagine this {lim(2^h - 1)/h = ln(2), as h -->0} can also be proven inductively.
     
  12. May 21, 2004 #11

    Tom Mattson

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    What is the statement P(n)? Is it this:

    P(n): d(nx)/dx=(ln n)n2 for all integer n.

    If so, then the proof (if it can be done inductively) will only be good for integer bases, and therefore won't apply to, say, base e. Zurtex's post, on the other hand, covers all real bases.
     
  13. May 21, 2004 #12
    The binomial expansion is simply the taylor series of (x + 1)^n, is it not?

    cookiemonster
     
  14. May 22, 2004 #13

    HallsofIvy

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    Therefore what? A power series expansion of any function is its Taylor series. My point was that it doesn't necessarily "involve the derivative". The fact that the coefficients are related to the derivative means that, if one has another way of finding a power series, one can use that to find the derivative- the whole point of this thread.

    For example, it is easy to see that the Taylor's series (strictly speaking "McLaurin series") for f(x)= 1/(1- x) is 1+ x+ x2+ ... by using the sum of a geometric series (known long before the calculus) and from that deduce that f '(0)= 1, f"(0)= 2, and, in general, f(n)(0)= n!.
     
  15. Jun 15, 2004 #14
    the derivitive of 2^x(ln2)

    Can somebody explain how to find the n'th derivitive of f(x)=2^x. After getting the first derivitive finding the second derivitive is giving me alot of trouble.
     
  16. Jun 15, 2004 #15

    Gokul43201

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    [tex]\frac {d} {dx} 2^x (ln2) = (ln2) \frac {d} {dx} 2^x = (ln2)^2 2^x= \frac {d^2} {dx^2} 2^x[/tex]

    since ln2 is a constant, independent of x.

    This can be repeated any number of times to give :

    [tex] \frac {d^n} {dx^n} 2^x = (ln2)^n 2^x [/tex]
     
  17. Jun 30, 2004 #16

    you can do that?
     
  18. Jun 30, 2004 #17

    AKG

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    cair0

    Perhaps you misunderstood what Zurtex wrote:

    (dy/dx) * (1/y) = ln(a)
    dy/dx = y*ln(a)
     
  19. Jul 1, 2004 #18
    let y=2 ^ x
    then log y = log 2^x
    log y= x . log2

    now, u can easily do this
     
  20. Jul 23, 2004 #19

    mathwonk

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    I posted a complete answer to this question today in the thead "introducing logarithms"
     
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