$\ln(x)$, $e^{x}$ and $\int \frac{1}{x}dx$ relationship My math professor told me that $\int_1^x \frac{1}{t} dt$ is $\ln(x)$ by the definition; so far so good.
But how/why does $\ln(x)$ ($\int_1^x\frac{1}{t} dt$: by defintion) coincide with the inverse of $e^{x}$? Thanks!
 A: If $$f(x) = \int_1^x \dfrac{dt}t$$ and $$g'(x) = g(x)$$ with $g(0) = 1$, then we can show that $f(x) = g^{-1}(x)$.
Setting $t = g(y)$, we get that $dt = g'(y) dy$. When $y=0$, we get that $t=1$ and when $y = g^{-1}(x)$, we get that $t=x$.
Hence,
$$f(x) = \int_{0}^{g^{-1}(x)} \dfrac{g'(y) dy}{g(y)} = \int_{0}^{g^{-1}(x)} dy = g^{-1}(x)$$
A: I disagree/disapprove of any math teacher that says this is definition.  Its not.  No one can assign more than one definition to a single concept without somewhere proving the equivalence.
You want to talk definition?  One of the Bernoulli family, while working on problems of bank account growth and compounding interest problems, was able to show that a bank account that grows with 100% APR, compounded continuously, converged to a value a little over 2.7 times that of the principle.  HE showed that $\lim_{n\to \infty} (1+\frac{1}{n})^n = e$.
THIS is definition. The most fundamental and principle of equalities. This is the historical origin, the chronological first, and the basis of all other properties which followed, and which were proved equivalent.
So now youve established the existence of e.  Taking $e^x$ is a trivial algebraic concept. Taking example from the interest growth problems, $e^x$ is the balance of a bank account with a principle of 1 after $x$ years.
Enter logarithms: If $e^m = n $ then (by definition of the logarithm) we have $m = \log_e(n)$.
Then, we can say $\log_e(n) = \ln(n)$.  This is nomenclature. Its just a standard and simpler way of writing a frequently appearing logarithm. This equivalence is not proven, its definition. But its not a mathematical coincidence, its simply an arbitrated truth. It was assigned and invented for simplicity.
Proving $\frac{d}{dx} \ln(x) = \frac{1}{x}$ is done via the limit definition of the derivative.  Here, let us prove the derivative of any arbitrary logarithm with respect ot its argument x.
$\frac{d}{dx} \log_a(x) = \lim_{h\to 0}\frac{\log_a(x+h)-\log_a(x)}{h}$
$\frac{d}{dx} \log_a(x) = \lim_{h\to 0}\frac{\log_a(\frac{x+h}{x})}{h}$
$\frac{d}{dx} \log_a(x) = \lim_{h\to 0}\frac{1}{h}\log_a(1+\frac{h}{x})$
Letting $h=x/n$ then as $h\to 0$ we have $n\to\infty$. The substitution creates:
$\frac{d}{dx} \log_a(x) = \lim_{n\to \infty}\frac{n}{x}\log_a(1+\frac{1}{n})$
Factoring out constants and continuous functions independent of $n$:
$\frac{d}{dx} \log_a(x) = \frac{1}{x}\log_a\lim_{n\to \infty}(1+\frac{1}{n})^n$
By the ACTUAL definition of e, we have
$\frac{d}{dx} \log_a(x) = \frac{1}{x}\log_a(e)$
This is a general truth for any log.  If you let $a=e$ then $\frac{d}{dx} \ln(x) =\frac{1}{x}\log_e(e)  = \frac{1}{x}$. It is because of this that $e$ holds special importance to calculus.
And by the general theorem of calculus, we also have the integral, $\ln(x) = \int_1^x \frac{1}{t} dt$.  This is just the reverse rule.
Proving that $\frac{d}{dx}e^x =e^x$ is as easy as letting $y=e^x$ and evaluating $\frac{d}{dx}\ln(y) = \frac{d}{dx}x$ using the chain rule.  It falls out kind of trivially.
