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Proof that the derivative of e^x is e^x.

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Differential calculus on Khan Academy: Limit introduction, squeeze theorem, and epsilon-delta definition of limits.

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34 thoughts on “Proof: d/dx(e^x) = e^x | Taking derivatives | Differential Calculus | Khan Academy

  1. You can use the definition of a derivative to solve this which would make the proof seem more valid.

  2. The last step is wrong
    U r multiplying e^x on both sides but on the RHS u have to multiply it out of the derivative that is d/dy. e^x.1/e^x.that step is wrong

  3. so the video used d/dx (e^x) in a chain rule to prove that d/dx(e^x) = e^x. I don't think that would be considered valid proof since ur using the result with chain rule to prove the result.

  4. So I spent an hour proving to myself that the derivative of lnx=1/x.. Then I was gonna try the derivative of e^x and I got lazy.. Now I wish I did it alone because it looks so easy.. But I have just one question.. Isn't d/dx(e^x) it's own term ? In a proof, you can't assume the right answer in it, you must prove the right answer. If we don't know the derivative of e^x is e^x yet, why are you saying that (e^x(d/dx(e^x)))/e^x=e^x. I'm only in algebra II, so I could really be wrong here.. But does anyone else feel the same ? I'm a bit confused..

  5. Derivative of e^x is the y coordinate
    Derivative of 1/2*x^2 is the x coordinate
    I think they should be equally celebrated

  6. I have a way of finding the derivative of e^x without using the chain rule or properties of natural logs.

    d/dx(e^x)= lim(h->0) [e^(x+h)-e^x]/h = e^x. lim(h->0) [e^h-1]/h = e^x.lim(h->0) [1+h-1]/h=e^x

    Proof for e^h=1+h :

    Consider two functions "e^h" and "log_{1+h} {e^h^2.(1+h)}"

    It is easy to see that both these functions approach 1 as h approaches 0.

    So for values of h very very close to 0, we can say that:
    e^h = log_{1+h} {e^h^2 * (1+h)}
    = log_{1+h} {1+h} + log_{1+h} {e^h^2}
    = 1 + h^2/log_e{1+h}
    = 1 + h/[log_e{1+h}]/h
    = 1+h/log_e [(1+h)^(1/h)] –(1)

    Now for values of h close to 0, (1+h)^(1/h) ~ 2.71828… = e –(2)

    So putting (2) in (1): e^h = 1+h/log_e{e} = 1+h

  7. if d/dx lne(e^x)=1/e^x and d/dx e^x=1/e^x then
    d/dx lne(e^x) = d/dx e^x
    doesn't that mean that lne(e^x) = e^x
    lne(e)*x=e^x
    x=e^x

  8. When I try and find the integral from 0-1 for this function, i run into the issue of receiving -infinity for 0 with the way i integrated it. So i figure im doing something wrong.

  9. The poof goes like this:

    1. Take the derivative of ln(e^x) by first using log rules to show ln(e^x) = x. d/dx (x) = 1.

    2. Now that you know that, go through taking the derivative of ln(e^x) a second time. This time, you use the chain rule.

    3. AFTER using chain rule, you get:
    (1/e^x) * ( d/dx(e^x) ) = d/dx( ln(e^x) ) = 1 [From Step 1]
    Or simply:
    (1/e^x) * ( d/dx(e^x) ) = 1

    Multiply both sides by e^x to get rid of the fraction, and you're done.

  10. I understand the chain rule but how did you make the jump at 2:43
    Also when you multiplied both sides by e^x, you can't because 1/(e^x) is being derived, or so I thought?

  11. Two things I don't understand, I thought we were trying to prove d/dx(e^x), but in this video he starts by doing it as d/dx(ln(e^x)). I'm assuming ln(e^x) = e^x?
    Next he's doing the chain rule to ln(e^x). First step is easy, taking the derivative of the inside which is e^x. Then the next step, which is supposed to be the exponent – 1 (n – 1), he gets 1/e^x instead. He got that from taking the derivative from lnx = 1/x… but I don't understand how that correlates.

  12. I'm pretty sure you went wrong somewhere here, you differentiated ln(e^x) to get 1, then differentiated it a different way to get 1/e^x, now assuming that makes sense you say that they must equal each other so 1/e^x=1 so 1=e^x you cant just scribble out the 1. a way to prove it is to say on the curve of e^x we can choose two points, p and p+h where h is the difference in the x axis between the two points so dy/dx((e^(p+h)-e^p)/h) as h gets closer to 0 it becomes effectively e^x

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