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$(\ln x)'=\dfrac{1}{x}$

$f(x)=\ln x$;

$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{\ln(x+h)-\ln x}{h}$

But due to the properties of logarithms [??]

$\dfrac{\ln(x+h)-\ln x}{h}=\dfrac{1}{h}\ln\left(\dfrac{x+h}{x}\right)=\ln\left(1+\dfrac{h}{x}\right)^{\frac{1}{h}}=\dfrac{1}{x}\ln\left(1+\dfrac{h}{x}\right)^{\frac{x}{h}}$

When taking the limit, which does not affect $\frac{1}{x}$, we get

$\dfrac{1}{x}\ln e= \dfrac{1}{x}$

by the definition of the number $e$ [??]