Izvod funkcije
Pravila diferenciranja

Izvod funkcije $f(x)=\ln(x)$

Odredimo izvod $f(x)=\ln{x}$ u proizvoljnoj $x_{0}>0$:
Posmatrajmo $g(x_{0})$: $$g(x_{0}, \Delta x)=\frac {\Delta f(x_{0})}{\Delta x}=\frac{\ln{(x_{0}+\Delta x)}-\ln{x_{0}}}{\Delta x },$$ $f$ u $x_{0}$:

This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

$$f'(x_{0})=\lim_{\Delta x\rightarrow 0}g(x_{0}, \Delta x)=$$ $$=\lim_{\Delta x\rightarrow 0}\frac {\Delta f(x_{0})}{\Delta x}=\lim_{\Delta x\rightarrow 0}\frac{\ln{(x_{0}+\Delta x)}-\ln{(x_{0})}}{\Delta x }=$$ $$=\lim_{\Delta x\rightarrow 0}\frac {\ln(1+\frac{\Delta x}{x})}{\frac{\Delta x}{x}}\cdot \frac{1}{x}=\frac{1}{x},$$ jer je $\lim_{t \rightarrow 0}\frac{ln(1+t)}{t}=1.$