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Query concerning derivative of e^x

  1. May 16, 2009 #1
    I've been studying calculus and have always been confused about the property of e^x.
    "e is the unique number such that e^x is equal to its derivative."

    I haven't ever really understood why, but have figured it would pop up some time later. Unfortunately, it still hasn't popped up, so I've decided to ask. How is e^x the derivate of itself?

    I was mucking around with numbers a bit and came up with a couple of things:

    Asumming e^x is its derivative:
    [tex]y = e^{x^{2}+2}[/tex]
    [tex]\frac{dy}{dx} = e^{x^{2}+2}[/tex]
    However (using the chain rule):
    [tex]\frac{dy}{dx} = 2xe^{x^{2}+2}[/tex]
    Which isn't the same thing...

    Also, while I'm on the topic, is there a way to calculate NP derivatives such as: [tex]2^{x}[/tex]?

    I am sorry if this seems basic, or is commonly asked, but I just can't seem to figure it out.
  2. jcsd
  3. May 16, 2009 #2


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    Hi Kaimyn! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    2x = (eln(2))x = ex.ln(2), so (2x)' = ln(2).2x :smile:

    (and since ln(e) = 1, e is the only number with (ex)' =ex)
  4. May 16, 2009 #3
    The derivative of ex with respect to x is ex. If x is a function of t (ie., x = t2) and you want the derivative with respect to t of the whole expression, then of course you have to use the chain rule.

    One common method is to take a logarithm (it doesn't have to be the natural logarithm) of both sides and use implicit differentiation (ie., use the fact that if A(x) = B(x) and A is differentiable, then A'(x) = B'(x)).
  5. May 16, 2009 #4


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    You just said ""e is the unique number such that e^x is equal to its derivative."
    Given that, it is impossible that [tex]e^{x^2}[/tex] is also its own derivative!
    Why would you assume that?

  6. May 16, 2009 #5


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    Both tiny-tim and slider showed how to get the derivative of ex from the derivative of ln(x). Here's how to get it directly.

    If f(x)= ax, for any positive number a, then
    [tex]\frac{f(x+h)- f(x)}{h}=\frac{a^{x+h}- a^x}{h}= \frac{a^xa^h- a^x}{h}= a^x\frac{a^h- 1}{h}[/itex]

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

    In other words, the dependence on x can be taken out of the limit: since that limit does not depend on x, it is a constant.

    So, for any positive real number, a, the derivaive of ax is a constant time ax.

    It is not too difficult to see that if a= 2, say, then that limit must be less than 1 or that if a= 3, it is larger than 1. Thus, there exist a value of a between 2 and 3 such that
    [tex]\lim_{h\rightarrow 0}\frac{a^h- 1}{h}= 1[/itex]
    and we define e to be that number.
    Last edited by a moderator: May 16, 2009
  7. May 16, 2009 #6
    Assuming you can find the derivative of ln(x) straight from the defintion, you can use the chain rule to find the derivative of e^x.

    [tex] \frac{d(ln(e^x))}{dx} = \frac{d(x)}{dx} = 1 [/tex]

    but [tex]\frac{d(ln(e^x))}{dx} = \frac{d(ln(u))}{du}\frac{d(e^x)}{dx} = \frac{1}{e^x}\frac{d(e^x)}{dx} [/tex]

    therefore, [tex] \frac{1}{e^x}\frac{d(e^x)}{dx} = 1 [/tex]

    or [tex] \frac{d(e^x)}{dx} = e^x [/tex]
  8. May 16, 2009 #7
    Thank you for all your helpful replies.

    Thanks, now it actually makes sense.

    Now I have "proof".

    Another mathematical proof that makes sense.

    Thanks again for all the helpful replies.
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