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Before you say you can't use the derivative of a function to prove the derivative of a function, just please see the proof.

I created it myself, and was wondering whether this could applied to any function and/or whether there is a generalization.

For this proof, the only additional knowledge you need to know is the integral of $\frac1x$

Proof:

$\frac d{dx}e^x=\lim_{h\to0}\frac{e^{x+h}-e^x}{h}$

L'Hospital's rule states that:

$\lim_{h\to0}\frac{e^{x+h}-e^x}{h}$= $\lim_{h\to0}\frac{\frac d{dh}e^{x+h}-e^x}{\frac d{dh}h}$

The derivative of h is 1 so:

$\lim_{h\to0}\frac{e^{x+h}-e^x}{h}$= $\lim_{h\to0}\frac d{dh}(e^{x+h}-e^x)$

Factor out the right hand side and replace in the derivative from earlier for the right hand side limit to get this:

$\frac d{dx}e^x$= $\lim_{h\to0}\frac d{dh}(e^x(e^h-1))$

Move the $e^x$ out of the derivative on the right hand side as the derivative is in respect to h, not x

$\frac d{dx}e^x$= $\lim_{h\to0}e^x(\frac d{dh}(e^h-1))$

This bit is slightly trickier - the derivative of $e^h-1$ in respect to h is just the derivative of $e^h$, and then when you apply the limit it will be just some constant.

If one denotes $e^x$ as $y$, then that constant will be $y'(0)$, and we get this differential equation:

$y'=y\times y'(0)$ or for simplicity (we will call that constant C)

$\frac{dy}{dx}=y\times C$

Re-arrange to get:

$dy \frac1y = Cdx$

Integrate both sides to get

$ln(y) = Cx$

Substitute in $y = e^x$

$ln(e^x)=Cx$

$x =Cx$

$1=C$

Substitute this into the original differential equation and we have our answer:

$\frac{dy}{dx}=y\times 1$

or in other words:

$\frac{d}{dx}e^x=e^x$

Can this proof work on other differntiable functions and/or is there a generalization.

Also, tell me what you think of it.

Bernard
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Jakub Skop
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    You don't need L'Hôpital's rule, if $f(x) = e^x$, then:

    $$f'(x) = \lim_{h\to 0}\frac{e^{x+h}-e^x}{h}= e^x\lim_{h\to 0}\frac{e^{h}-1}{h} = f'(0) f(x).$$

     – Ennar Jan 05 '20 at 09:53
  • I know - this is an alternative proof – Jakub Skop Jan 05 '20 at 09:54
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    It's not alternative, you just use more than you need: $\lim_{h\to 0} \frac{e^h-1}{h} = f'(0)$ by definition, no need for L'Hôpital. Even more so, you are assuming that the derivative is continuous. – Ennar Jan 05 '20 at 09:59
  • Oh, I see what you mean- but how did you get to that? – Jakub Skop Jan 05 '20 at 09:59
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    $f'(c) = \lim_{x\to c}\frac{f(x)-f(c)}{x-c}$, so let $x = h$ and $c = 0$. – Ennar Jan 05 '20 at 10:02
  • Thanks, I didn't realise that - but I guess it still works either way. – Jakub Skop Jan 05 '20 at 10:05
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    But the question is, how do you know that $e^x$ is differentiable and has derivative continuous at $0$? – Ennar Jan 05 '20 at 10:07
  • Well, I don't - I guess this is a sort of auxillary proof. Do you know of any way to prove $e^x$ is differentiable and continuous at 0? Actually - correct me if I'm wrong, but since the limit after differentiating the original quotient exists, then the derivative of $e^x$ is continuous at 0, is it not? – Jakub Skop Jan 05 '20 at 11:08
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    It all comes down to showing that $\lim_{h\to 0} \frac{e^h-1}h$ exists and is equal to $1$. From there you get $(e^x)' = e^x$ and hence, $e^x$ is infinitely many times differentiable. But this depends on definition of exponential function that you are using. If you want serious consideration of your proof, you are missing these details: definition and properties that you can prove from definition. Then we can judge whether your proof is satisfactory or not. – Ennar Jan 05 '20 at 11:33
  • If I were to treat that limit as a limit and not as a constant as I did, would that then eliminate the problem of continuity that I had experienced? – Jakub Skop Jan 05 '20 at 12:57
  • It is a constant, if you can show that the limit exists. Combine that with my first comment and you are done. Take a look at this question, for example. – Ennar Jan 05 '20 at 13:25
  • I understand what you mean, but I wanted this to be a proof that somehow doesn't use that fact - all the other proofs I've seen use that. I wanted it to be different - by using that it is just the same old proof just slightly re-written.However, I've seen that this method of proving using L'Hopitals rule does not work on $e^x$ without a few assumptions. Are there any other functions that do not require these assumptions? – Jakub Skop Jan 05 '20 at 13:28
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    Everybody uses that fact to avoid being circular. What you are doing with L'Hôpital's rule is the following: assuming that $f$ is differentiable and $\lim_{x\to c} f'(x)$ exists, then $\lim_{x\to c}\frac{f(x)-f(c)}{x-c} = \lim_{x\to c} f'(x)$. But $f'(c) = \lim_{x\to c}\frac{f(x)-f(c)}{x-c}$ by definition. The point is, using L'Hôpital's rule to prove differentiability is circular. It does show something interesting though, look at this Wiki section. – Ennar Jan 05 '20 at 14:21

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