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.