Đạo hàm của $f(x)$ tại điểm $x_o$ kí hiệu $f'(x_o)$ và:
$f'(x_o) = \mathop{\lim}\limits_{x \to x_o}\dfrac{f(x) - f(x_o)}{x - x_o}$
Xét $f(x) = \ln x$ ta được:
$(\ln x_o)' = \mathop{\lim}\limits_{x \to x_o}\dfrac{\ln x- \ln x_o}{x - x_o}$
$= \mathop{\lim}\limits_{x \to x_o}\dfrac{\ln(x_o + (x - x_o))- \ln x_o}{x - x_o}$
Đặt $\Delta x = x - x_o$
$\Rightarrow (\ln x_o)' = \mathop{\lim}\limits_{\Delta x \to 0}\dfrac{\ln(x_o + \Delta x)- \ln x_o}{\Delta x}$
$= \mathop{\lim}\limits_{\Delta x \to 0}\dfrac{\ln\left(\dfrac{x_o + \Delta x}{x_o}\right)}{\Delta x}$
$= \mathop{\lim}\limits_{\Delta x \to 0}\dfrac{1}{\Delta x}.\ln\left(1 + \dfrac{\Delta x}{x_o}\right)$
$= \mathop{\lim}\limits_{\Delta x \to 0}\ln\left(1 + \dfrac{\Delta x}{x_o}\right)^{\dfrac{1}{\Delta x}}$
Đặt $\dfrac{\Delta x}{x_o} = h$
$\Rightarrow \Delta x = hx_o$
$\Rightarrow \dfrac{1}{\Delta x} = \dfrac{1}{hx_o}$
Do $\Delta x \to 0$
Nên $h \to 0$
Ta được:
$(\ln x_o)' = \mathop{\lim}\limits_{h \to 0}\ln\left(1 + h\right)^{\dfrac{1}{hx_o}}$
$= \mathop{\lim}\limits_{h \to 0}\ln\left[\left(1 + h\right)^{\dfrac{1}{h}}\right]^{\dfrac{1}{x_o}}$
$= \dfrac{1}{x_o}.\mathop{\lim}\limits_{h \to 0}\ln\left[\left(1 + h\right)^{\dfrac{1}{h}}\right]$
$= \dfrac{1}{x_o}.\ln\left[\mathop{\lim}\limits_{h \to 0}\left(1 + h\right)^{\dfrac{1}{h}}\right]$
$= \dfrac{1}{x_o}.\ln e$
$= \dfrac{1}{x_o}.1 = \dfrac{1}{x_o}$
Vậy ta được công thức tổng quát:
$(\ln x)' = \dfrac{1}{x}$
