What is the precise relation between the definition of $e$ from compound profit and the exponential function As in I understand the relation between compound profit and exponentiation:
20% interest after $n$ years is given by $1.2^n$
and I understand the derivations of $e$ as the limit as$ n\rightarrow \infty$ of $(1+1/n)^n$
and as the value such that $d/dx(a^x)=a^x$
I just don't understand the exact mathematical link between the two.
Is it just conincedence that the constant derived from continuous interest's exponential is equal to its derivative or is there an rigorous mathematical link between the two.
 A: As you stated, we know $$e=\lim_{m\to \infty}(1+\frac1m)^m$$
Which represents the maximum (limit) amount when a bank applies an instant compound interest of 100% with an initial amount of $1$ dollar. Let $n=\frac1m$, then we have:
$$e=\lim_{n\to0}(1+n)^\frac1n$$
As $m\to \infty$, $n\to0$
Now, take the derivative of $f(x)=e^x$ by definition:
$$f^´(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h=\lim_{h\to0}\frac{e^{x+h}-e^x}h=\lim_{h\to0}\frac{e^xe^h-e^x}h=e^x\lim_{h\to0}\frac{e^h-1}h$$
Let $y=e^h-1 \Rightarrow h=ln(1+y)$. As $h\to0$, $y\to0$. Substituting:
$$f^´(x)=e^x\lim_{y\to0}\frac y{ln(1+y)}=e^x\lim_{y\to0}\frac 1{\frac 1y ln(1+y)}=e^x\lim_{y\to0}\frac 1{ln(1+y)^\frac 1y }=e^x\frac 1{ln\lim_{y\to0}(1+y)^\frac 1y}=e^x\frac 1{lne}=e^x$$
The exponential function is defined by the constant $e$, which comes from instant compound interest. This definition of $e$ is the reason that allows the derivative of the exponential function be the exponential function itself.
I hope this answered your question
