Izvod funkcije
Pravila diferenciranja

Izvod funkcije $f(x)=\sin{x}$

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

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$$f'(x_{0})=\lim_{\Delta x\rightarrow 0}g(x_{0},\Delta x)=$$ $$=\lim_{\Delta x\rightarrow 0}\frac {\Delta f(x_{0})}{\Delta x}=\lim_{\Delta x\rightarrow 0}\frac{\sin{(x_{0}+\Delta x)}-\sin{x_{0}}}{\Delta x }=$$ $$=\lim_{\Delta x\rightarrow 0}\frac {2\sin{\frac{\Delta x}{2}}\cos{\frac{2x_{0}+\Delta x}{2}}}{\Delta x}$$ Koristeći poznati rezultat $\lim_{t \rightarrow 0}\frac{\sin t}{t}=1$ i neprekidnost kosinusne funkcije, dobijamo da je $$f'(x)=\lim_{\Delta x \rightarrow 0}\frac{f(x_{0}+\Delta x)-f(x)}{\Delta x}=\cos{x_{0}}.$$