Proof of the derivative of e^x: A Step-by-Step Proof and Explanation

In this video, we will use the limit definition of the derivative, also known as the First Principle, to thoroughly prove what the derivative of e^x is. To do this, we will utilize Euler's number, and I will provide a detailed explanation so that you can easily grasp the concept. If there are any parts that you find difficult to understand or if you have any questions, please feel free to leave a comment.

If you're eager to explore more differentiation techniques for various functions, click the link below.

Stay tuned for a variety of upcoming differentiation videos. Additionally, I recommend watching the easy integration videos using the Box Rule below, as they will complement your learning experience.

Also, there are various other videos available on my channel and more will be uploaded continuously, so feel free to visit and check them out.

#derivatives #differentiation #mathwithalex

[MadScientyst]

This is what's called a 'rigorous proof' folks.
Brilliant exposition & well done bro....a new subscriber!

[nektariosorfanoudakis]

Much better proof than the one I'd found when I was 18, where I used the definition of exp as a limit of a sequence and used Bernoulli's inequality and sandwich theorem to find the final limit and I'm not sure if I did it correctly anymore either. 😂

[tw]

Half expected the proof to be circular. Glad it wasn't. Nice

[Matematicand]

A really ellegant proof! Congratulations!

[uggupuggu]

Why didnt you let

e = (1+h)^1/h as h goes to 0 and

e^h =(1+h)

thus the final limit = h/h which is 1

[gokaytaspnar]

I rewrite lim h -> inf (1 + x/h)^h using binomial expansion
1 + (hC1) * 1 * 1/h + (hC2) * 1 * 1/h^2 ...

And found that e^x = 1 + x/1! + x^2/2! ...
Without using it's derivative

[darcash]

This was a fun proof, but I remember when I first tried to prove it, I stopped at e^x *
lim h —> 0 (e^h - 1)/h. I saw that evaluating it became 0+/0+, and 0-/0- from each side. At this point, I realized these were approaching zero at the same speed, I guess you could say. Like how x^2 approaches infinity faster than x bc it has a higher order. It seemed logical to me to say that the remaining part was equal to 1 for this reason. Is there a mathematical term for this, being of the same magnitude/order I guess you might call it?

[abdouab]

The euler limit was proven by bringing the ln in and use l'hopital rule, but how do we know the ln dereivetive, isn't related with exp derevetive ?

[pauselab]

Another way to define e is to first prove that e^x is never 0 in real numbers. Then show that the wronskian of any 2 function that is it’s derivative is always defined and zero. So there is a unique function up to a factor such that f’=f. Call that function e^x. Then Taylor series to prove that e is a real number.

[randomname]

This is a circular proof. The limit definition of e derives from the Taylor series of e^x, meaning it relies on the fact that the derivative of e^x is e^x. You essentially proved that if the derivative of e^x is e^x (a requirement for the limit) then the derivative of e^x is e^x (what you tried proving). The truth is there is no actual way to prove this, because that is the definition of e.

[squeezy]

Damn that's some neat writing!

I always thought of this derivative being a kind of definition, like if you consider the function b^x for any base b, we know that its derivative is proportional to the function itself:

Let f(x) = b^x (for well defined functions, let's say b > 0)
Using the limit definition of the derivative we get:
f'(x) = lim h -> 0 (b^(x+h) - b^x)/h
= lim h -> 0 (b^x * b^h - b^x)/h
= lim h -> 0 (b^x(b^h - 1))/h
= b^x * lim h -> 0 (b^h - 1)/h
= K b^x (where K is some constant of proportionality defined by the h limit above)

Now to determine K for any base (remembering that we're trying to avoid circular reasoning), we don't really have a better way than just plugging in really small values of h. For example let b = 2, the corresponding limit value seems to approach ~0.693 if you let h get really small (which we now know to be ln(2) but pretend we don't know this). With b = 3 you get K ~= 1.099 - so suppose there exists a number between 2 and 3, let's call it 'e', where this K value is exactly 1 (i.e. lim h -> 0 (e^h - 1)/h = 1) and we have defined e in the following way:

e is the number such that the function f(x) = e^x has derivative e^x (i.e. f(x) = f^(n)(x) for all positive integers n) and we can use numerical methods to find that e ~= 2.718.

I'm not exactly sure how this relates to the definition you used but I'm 100% sure there's a connection - great video and let me know what you think!

[griffgruff]

If one accepts the power series of e^x as being valid, then the identity is obtained by simply taking the derivative wrt x.

[avonbarksdale]

That limit being equal to e is not obvious to me. Even if it is the definition it could be interesting to learn more about that specific property.

[jan-willemreens]

... also known as " FIRST PRINCIPLES " ...

[BedrockBlocker]

Not bad, to note us that we used continuity of the exponential function and it's inverse here.
I find it a little confusing that you use an "x" as multiplication operator, but overall a neat proof.

[InnerView_]

The question is related to the variables h and n. Is it legal to interchange variables in the middle of a mathematical proof, and substituted? The expression was originally about h approaching zero, and then it became about n approaching zero. Is that allowed?

[nuraka]

Isn't e = lim n--> + infinity (1+1/n)^n ?

[Niloctronic]

At 2:45, I didn't understand how could you move the limit operator to inside the logarithm's function. Why was that allowed ? Thanks for the video.

[pauselab]

it's by definition of e. not exactly a reason why that is the case, it just happens to be that number. if you wanted to find say a function whose derivative is itself, you can see that that function has to be an exponential function. then you can just define the exponential function e^x as being the one. then you try to figure out what is e numerically by using taylor series

[davannaleah]

From the series expansion of e^x, for very small x, we approximate e^x as 1+x...remember this is not the limit! as x->0. So (e^(x+h) - e^x)/h becomes e^x(e^h - 1)/h -> e^x(1+h - 1)/h -> e^x(h)/h == e^x! QED