Forskjell mellom versjoner av «Produktregel derivasjon - bevis»

Fra Matematikk.net
Hopp til:navigasjon, søk
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} $
+
$f´(x)= \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)v(x + \Delta x)- u(x)v(x)}{\Delta x}  \\  = \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)v(x + \Delta x)- u(x)v(x+\Delta x) +u(x)v(x+\Delta x )u(x)v(x)}{\Delta x}  \\ \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)v(x + \Delta x)- u(x)v(x+\Delta x) +u(x)v(x+\Delta x )u(x)v(x)}{\Delta x}$

Revisjonen fra 11. jun. 2015 kl. 03:38

$f(x)=u(x) \cdot v(x) \quad f´(x)= u´(x) \cdot v(x) + u(x) \cdot v´(x) \quad f´(x)= \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$

Bevis:

$f´(x)= \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)v(x + \Delta x)- u(x)v(x)}{\Delta x} \\ = \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)v(x + \Delta x)- u(x)v(x+\Delta x) +u(x)v(x+\Delta x )u(x)v(x)}{\Delta x} \\ \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x)v(x + \Delta x)- u(x)v(x+\Delta x) +u(x)v(x+\Delta x )u(x)v(x)}{\Delta x}$