How much proof is needed in such paper (Maths related)? 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.
 A: The answer to your question depends on what you expect your readers to know in advance. Since you're writing about derivatives, they should know calculus.
Then the fact that $e^x$ is its own derivative is likely to be a better known prerequisite than the statement about logarithmic derivatives you want to use to prove it.
A: 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
$$\lim_{h \to 0}\left(\frac{\mathrm e^{h}-1}{h}\right)=1$$
A: For the record, I agree with @EthanBolker's answer above.
With regards to the question of whether a proof of the fact exists without assuming the derivative of $e^x$ is $e^x$, the answer is yes.
https://en.wikipedia.org/wiki/Logarithmic_derivative
https://proofwiki.org/wiki/Derivative_of_Natural_Logarithm_Function
The equality follows from the chain rule, and hence depends on a proof that $$ {d \over dx} \ln x = {1 \over x}$$
One can prove this, for instance, using the fundamental theorem of calculus and the fact that the antiderivative of $\frac{1}{x}$ is $\ln x$.
