How much proof is needed in such paper (Maths related)?

Question

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.

Answer 1

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.

Answer 2

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 $\ln x$.

Answer 3

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 $$\lim_{h \to 0}\left(\frac{\mathrm e^{h}-1}{h}\right)=1$$