Izvod funkcije
Pravila diferenciranja

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

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

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$$f'(x_{0})=\lim_{\Delta x\rightarrow 0}g(x_{0}, \Delta x)=$$ $$=\lim_{\Delta x\rightarrow 0}\frac {\Delta f(x_{0})}{\Delta x}=\lim_{\Delta x\rightarrow 0}\frac{\sqrt{x_{0}+\Delta x}-\sqrt{x_{0}}}{\Delta x }=$$ $$=\lim_{\Delta x\rightarrow 0}\frac {(x_{0}+\Delta x)-x_{0}}{\Delta x(\sqrt{x_{0}+\Delta x} + \sqrt{x_{0}})}=$$ $$=\lim_{\Delta x\rightarrow 0}\frac {1}{\sqrt{x_{0}+\Delta x}+\sqrt{x_{0}}}=\frac{1}{2\sqrt{x_{0}}}$$ Na , za $ x_{0}=1$ je$f'(1)=\frac{1}{2}$ pa krive $f(x)=\sqrt{x}$ u tački $(1,1)$ ima jednačinu $y-1=\frac {1}{2}(x-1).$