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Help! can someone differentiate e^(nx)

  1. Nov 30, 2003 #1
    can someone differentiate e^(nx) where n is any integer. i think is equal to n*e^(nx).

    please show the proof, thanx.
    :smile: :smile: :smile:
  2. jcsd
  3. Dec 1, 2003 #2
    Well the proof is very big, so i just give u an idea as to how to go abt proving:

    [tex]d(e^{(nx)})/dx = ne^{nx}[/tex]

    now increment y by a small value such that:

    [tex] y + \Delta y = e^{(n(x + \Delta x))}[/tex]

    Divide the whole term by y such that:

    [tex] (\Delta y)/y = [[e^{(n(x + \Delta x))}]/(e^{(nx)})] - 1[/tex] {We are putting [tex]d(e^{nx})/dx = n*e^{nx}[/tex] on the right hand side of the eqtn.}

    take the one on the other side to get:

    [tex] \Delta y/ y = [e^{(n(x + \Delta x))}- e^{(nx)}]/[e^{(nx)}] [/tex]

    now apply the limits [tex]\Delta y \longrightarrow 0[/tex] to bot sides and evaluate the limit to get your answer.

    Last edited: Dec 1, 2003
  4. Dec 1, 2003 #3


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    Or, assuming you don't actually need to go back to the original definition of the derivative: apply the chain rule.

    To differentiate y= enx, let u= nx.

    Then y= eu so dy/du= eu

    (If you don't know that then you will need to go back to definitions- in particular exactly how you are defining ex.)

    du/dx= n, of course, and

    dy/dx= (dy/du)(du/dx)= eu(n)= n enx.
  5. Dec 1, 2003 #4
    thanx guys. i think the product rule can also use to differentiate
    e^(nx), since e^(nx)=e^(n)*e^(x). but that will require much more time.
  6. Dec 1, 2003 #5
    this is not correct, the product rule does not apply here.
  7. Dec 1, 2003 #6
    Hmm . .. I don't think that is quite legal since e^n is a constant, the product rule only applies if you have two functions.
    you could alternately take the log of both sides which may look something like:

    differentiate with respect to x
    and since y= e^nx

    Of course I used the chain rule now on the left hand side, now
  8. Dec 1, 2003 #7
    The product rule applies fine for a constant term, since a constant is a perfectly good function. It does not apply the way he said though; his exponential relation is wrong.

  9. Dec 1, 2003 #8
    It's mightily unnecessary as one term will automatically go to zero, and yes I missed his algebraic mistake
  10. Dec 2, 2003 #9
    sorry about that mistake, what i meant was that,the product rule can be use repeatly to differentiate e^(nx).

    eg. differentiate e^(3x)

    e^(2x) = e^(x)*e^(x)
    [e^(2x)]'=e^(x)*[e^(x)]' + e^(x)*[e^(x)]' = e^(x)*e^(x) + e^(x)*e^(x)
    [e^(2x)]'= 2e^(2x)

    e^(3x) = e^(2x)*e^(x)
    [e^(3x)]'=e^(x)*[e^(2x)]' + e^(2x)*[e^(x)]' "sub [e^(2x)]'= 2e^(2x)"
    [e^(3x)]'=e^(x)*(2e^(2x)) + e^(2x)*e^(x)
    [e^(3x)]'=2e^(3x) + e^(3x)
  11. Dec 2, 2003 #10


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    Okay, you recognize that enx is NOT enex so that the a single application of the product rule does not work.

    Yes, you can use the product rule repeatedly (in a proof by induction) but why?

    What's wrong with using the chain rule as I suggested?
  12. Dec 2, 2003 #11


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    Actually, this is a more interesting problem that I first thought- there are several different ways of doing it.

    The "obvious" way (to me anyway) is to note that the derivative of ex is ex and use the chain rule:
    denx/dx= ex(d(nx)/dx)= nenx.

    But we can also write enx= (en)x and use the fact that dax/dx= (ln a) ax:
    d((en)x)/dx= ln(en)(en)x= nenx.

    Or do it the other way around: enx= (ex)n) and use the power rule (together with the chain rule and the derivative of ex):
    d((en)x)/dx= n((ex)n-1)(ex)= n(ex)n= nenx
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