Forskjell mellom versjoner av «Produktregel derivasjon - bevis»
Fra Matematikk.net
| (5 mellomliggende revisjoner av samme bruker vises ikke) | |||
| Linje 3: | Linje 3: | ||
Bevis: | Bevis: | ||
| − | $f´(x)= \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)v(x + \Delta x)- u(x)v(x)}{\Delta x} \\ = \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)v(x + \Delta x)- u(x)v(x+\Delta x) +u(x)v(x+\Delta x )u(x)v(x)}{\Delta x} \\ = \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)v(x + \Delta x) | + | $f´(x)= \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)v(x + \Delta x)- u(x)v(x)}{\Delta x} \\ = \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)v(x + \Delta x)- u(x)v(x+\Delta x) +u(x)v(x+\Delta x ) - u(x)v(x)}{\Delta x} \\ = \lim_{\Delta x \rightarrow 0} \frac{(u(x + \Delta x)-u(x))v(x+\Delta x) +u(x)(v(x+\Delta x )-v(x))}{\Delta x} \\ = \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)- u(x)}{\Delta x}v(x+\Delta x) + u(x) \frac{v(x+\Delta x )-v(x)}{\Delta x} \\ = u´(x)v(x) + u(x)v´(x)$ |
Nåværende revisjon fra 11. jul. 2015 kl. 11:20
$f(x)=u(x) \cdot v(x) \quad f´(x)= u´(x) \cdot v(x) + u(x) \cdot v´(x) \quad f´(x)= \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$
Bevis:
$f´(x)= \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)v(x + \Delta x)- u(x)v(x)}{\Delta x} \\ = \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)v(x + \Delta x)- u(x)v(x+\Delta x) +u(x)v(x+\Delta x ) - u(x)v(x)}{\Delta x} \\ = \lim_{\Delta x \rightarrow 0} \frac{(u(x + \Delta x)-u(x))v(x+\Delta x) +u(x)(v(x+\Delta x )-v(x))}{\Delta x} \\ = \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)- u(x)}{\Delta x}v(x+\Delta x) + u(x) \frac{v(x+\Delta x )-v(x)}{\Delta x} \\ = u´(x)v(x) + u(x)v´(x)$
