Proof d/dx (ln x)

Based on Khanacademy way of proofing $$\ \frac{d}{dx}\left ln(x) = \frac{1}{x} $$ which one of my student showed it to me ... I have another way of proofing this. Almost similar but using imaginary variable.

$$ Proof :$$
Let $$ f(x) = ln(x) $$ then, $$ \frac{ln(x+ \Delta h) - ln(x)}{\Delta h} \,\,= \frac{ln\big( \frac{x+ \Delta h}{x}\big)}{\Delta h} $$ Now we add the imaginary $$ {\color{blue} x}. $$
$$={\color{blue} x}\, ln \frac{\big (\frac{x+ \Delta h}{x} \big )}{{\color{blue} x} \Delta h} \,\, = \frac{1}{x} \, \bigg (\frac{x}{\Delta h} \, ln \bigg (\frac{x + \Delta h}{x} \bigg ) \bigg )$$

$$=\frac{1}{x} \, \bigg ( ln \big( 1 + \frac{1}{\frac{x}{\Delta h}} \big ) \bigg )^{\frac{x}{\Delta h}} \, = \frac{1}{x} \big( (e^{0})=1 \big) \, = \frac{1}{x}$$. This completes the proof