TITLE: How much proof is needed in such paper (Maths related)?
QUESTION [5 upvotes]: I'm writing a paper (report) regarding  Euler's Number $\space e \space$ (even though he didn't discover it).
Within this paper, I show that:
$${d\over dx} {e^x} = {e^x}$$
**NOTE: ** This is not what the whole paper is about.
However, the proof uses the fact that:
$$\space {d\over dx}\ln f(x) = {f'(x)\over f(x)}$$
Do I need to prove this first?
Or can I just leave it as prerequisite knowledge before reading the paper?
**P.S. - ** I understand this isn't exactly a Maths question in a sense. But it is Maths related and it takes a Mathematician to answer.

REPLY [4 votes]: You haven't really proven anything. You have used something that needs proving, to "prove" that the derivative of $\mathrm e^x$ is itself. It's really just a circular argument.
I would suggest that you prove it from first principles:
$$\mathrm f'(x) = \lim_{h \to 0} \left(\frac{\mathrm f(x+h)-\mathrm f(x)}{h}\right)$$
In your case, $\mathrm f(x) = \mathrm e^x$ and so
\begin{eqnarray*}\frac{\mathrm d}{\mathrm d x}\mathrm e^x &=& \lim_{h \to 0}\left(\frac{\mathrm e^{x+h}-\mathrm e^x}{h}\right) \\ \\
&=& \lim_{h \to 0}\left(\frac{(\mathrm e^{h}-1)\mathrm e^x}{h}\right) \\ \\
&=& \mathrm e^x \cdot \lim_{h \to 0}\left(\frac{\mathrm e^{h}-1}{h}\right) 
\end{eqnarray*}
To prove your statement from first principles you need only show that
$$\lim_{h \to 0}\left(\frac{\mathrm e^{h}-1}{h}\right)=1$$