Proof of the Derivative of e^x ( Definition of Derivative)

Using the definition of derivative to prove the derivative of e^x.

[Lucas-nudu]

Thanks man, I want to add, without using L'hopital rule (since we are working with the very foundation of derivatives, and not with its aplications) how to prove that limit:
(e^h-1)/h it can be written as:
((e^h)-e^0)/lne^h) and we write all as logaritms:
[ln(e^e^h)-ln(e^e^0)]/ln e^h
we regroup the upper term:
[ln(e^e^h)/(e^e^0)]/ln e^h
and it follows:
[ln(e^(e^h-e^0))]/ln e^h
both logarithms tends to ln e^0 and before they reach 0/0 they can cancel each other and you get 1 as a result

[TheNetkrot]

it´s not a proof you are proposing that the division e^h - 1/h is equal to one and that is not proving anything. You can't give yourself this ratio is one unless you have stated what you are going to proof.

[kishoremoorthy]

Is there a proof for that "special limit definition" if so can you please tell me :)

[potatooup]

3:27 this is done using l'hopital rule which basically says that the lim f(x)/g(x) is equal to the lim f'(x)/g'(x) so you cannot use it to prove a function's (in this case e^x's) derivative

[davidkippy]

Aren't you assuming the final answer already when you take that limit? Surely it needs L'Hopital rule

[zafe_]

What he has done is enough tbh...the derivative of e^x is (e^h - 1)/h X e^x which he wrote at 2:50 and that derivative can also be written as lne X e^x which he didn't write. Now lne=1 so you are left with the derivative being 1 X e^x = e^x. hope that helped.

[johnnolen]

Pardon me for saying so but your so called proof here is not a proof, it's an exercise in circular reasoning. The lim (h-->0) {[e^h - 1] ÷ h} = 1 is what you were supposed to show but instead you just assumed it was true.

[navhari]

this proof is not specific to e^x it could be used for any number. Surely using this proof you could prove the derivative of any number to the power x is equal to the that number to the power x.

[tabotcharlesbessong]

Thank very much sir you just solved my problem

[benthomas]

Can someone please explain why e^x is e^any constant is 0? Wouldn't
saying e^x is e^x mean whatever constant you plug in for x the slop is
the value of that function?

[peterlohnes]

sigh (hey I'm just going to skip an important part)

[shawnclifford]

lim (e^h - 1)/h, as h ---> 0
Using L'Hopital rule
lim d/dh(e^h - 1)/d/dh(h), as h ---> 0
lim e^h(1)/1, as h ---> 0
lim e^h, as h ---> 0
Then, substitute h to 0.
lim e^0, as h ---> 0
lim 1, as h ---> 0
(n^0 = 1, n≠0)

[ryanzheng]

This proof is not correct, as it uses d(e^x)/dx = e^x

[tahaelturki]

I'm Sorry this proof is completely wrong

[Baandarxx]

WOW THIS
DISLIKE sir sorry I have an assignment to do

[martinwashington]

To convince yourself that the limit he just assumed was 1. Express e^(h) as its infinite series I.e. sum from 0 to inf of h^(n) /n! After simplying and then taking the limit youll be left with a constant 1. Youre welcome.

[benthomas]

Can someone please explain why e^x is e^any constant is 0? Wouldn't
saying e^x is e^x mean whatever constant you plug in for x the slop is
the value of that function?