How to find the derivative of ln x?

Consider f(x) = lnx and using the first principles work out
${\displaystyle f\prime \left(x\right)=limh\to 0\frac{f(x+h)-f\left(x\right)}{h}}$
= ${\displaystyle limh\to 0\frac{\ln (x+h)-\ln x}{h}}$
=${\displaystyle limh\to 0\frac{\ln \left(\frac{x+h}{x}\right)}{h}}$
= ${\displaystyle limh\to 0\frac{\ln (1+\frac{h}{x})}{h}}$
= ${\displaystyle limh\to 0\frac{1}{x}\frac{\ln (1+\frac{h}{x})}{\frac{h}{x}}}$
= ${\displaystyle \frac{1}{x}}$ ${\displaystyle limh\to 0\frac{\ln (1+\frac{h}{x})}{\frac{h}{x}}}$
= ${\displaystyle \frac{1}{x}}$, because ${\displaystyle [limh\to 0\frac{\ln (1+\frac{h}{x})}{\frac{h}{x}}=1]}$

The "how" depends on the definition you are using for ln x.
${\displaystyle \ln x={\int }_1^x\frac{1}{t}dt}$ and the result is immediate useing the Fundamental Theorem of Calculus.
If you start with a definition of ${\displaystyle e^x}$ and find its derivative, then you probably defined ln x by:
${\displaystyle y=\ln x}$ ${\displaystyle \iff }$ ${\displaystyle x=e^y}$
Differentiate implicitely:
${\displaystyle \frac{d}{dx}\left(x\right)=\frac{d}{dx}\left(e^y\right)}$
${\displaystyle 1=e^y\frac{dy}{dx}}$
Hence, ${\displaystyle \frac{dy}{dx}=\frac{1}{e^y}=\frac{1}{x}}$