Den deriverte til en funksjon i et punkt er den momentant veksten i punktet.
Vi regner dette slik som vi regner gjennomsnittlig vekst men avstanden mellom de to x-verdiene er veldig liten.
Vi sier da at delta x går mot null
$\Delta x \to 0$
$$ f’(x)=\lim_{\Delta\to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$$
\begin{align}f(x)&=x^2\\f(x+\Delta x)&=(x+\Delta x)^2\\&=x^2+2x\Delta x+(\Delta x)^2\\f’(x)&=\lim_{\Delta x\to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}\\&=\lim_{\Delta x\to 0} \frac{x^2+2x\Delta x+(\Delta x)^2-x^2}{\Delta x}\\&=\lim_{\Delta x\to 0} \frac{2x\Delta x+(\Delta x)^2}{\Delta x}\\&=\lim_{\Delta x\to 0} \frac{\Delta x(2x+\Delta x)}{\Delta x}\\&=\lim_{\Delta x\to 0} (2x+\Delta x)\\&=2x\end{align}
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\begin{align}f(x)&=x^2\\f(x+\Delta x)&=(x+\Delta x)^2\\&=x^2+2x\Delta x+(\Delta x)^2\\f’(x)&=\lim_{\Delta x\to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}\\&=\lim_{\Delta x\to 0} \frac{x^2+2x\Delta x+(\Delta x)^2-x^2}{\Delta x}\\&=\lim_{\Delta x\to 0} \frac{2x\Delta x+(\Delta x)^2}{\Delta x}\\&=\lim_{\Delta x\to 0} \frac{\Delta x(2x+\Delta x)}{\Delta x}\\&=\lim_{\Delta x\to 0} (2x+\Delta x)\\&=2x\end{align}
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