Forskjell mellom versjoner av «Kvotient regel derivasjon-bevis»
Fra Matematikk.net
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| − | $$ | + | $f(x)= \frac{u(x)}{v(x)}, \quad f´(x)= \frac{u´(x) \cdot v(x) - u(x) \cdot v´(x)}{(v(x))^2}, \quad$ |
$f'(x)= \lim_{\Delta x \rightarrow0} \frac{\frac{u(x+\Delta x)}{v(x+ \Delta x)} - \frac{u(x)}{v(x)}}{\Delta x} \\ = \lim_{\Delta x \rightarrow0} \frac{u(x+\Delta x) \cdot v(x) - {u(x) \cdot v(x+ \Delta x)}}{\Delta x \cdot v(x+ \Delta x) \cdot v(x)} \\ = \lim_{\Delta x \rightarrow0} \frac{u(x+\Delta x) \cdot v(x)- u(x) \cdot v(x) - {u(x) \cdot v(x+ \Delta x) + u(x) \cdot v(x)}}{\Delta x \cdot v(x+ \Delta x) \cdot v(x)} $ | $f'(x)= \lim_{\Delta x \rightarrow0} \frac{\frac{u(x+\Delta x)}{v(x+ \Delta x)} - \frac{u(x)}{v(x)}}{\Delta x} \\ = \lim_{\Delta x \rightarrow0} \frac{u(x+\Delta x) \cdot v(x) - {u(x) \cdot v(x+ \Delta x)}}{\Delta x \cdot v(x+ \Delta x) \cdot v(x)} \\ = \lim_{\Delta x \rightarrow0} \frac{u(x+\Delta x) \cdot v(x)- u(x) \cdot v(x) - {u(x) \cdot v(x+ \Delta x) + u(x) \cdot v(x)}}{\Delta x \cdot v(x+ \Delta x) \cdot v(x)} $ | ||
Revisjonen fra 5. jun. 2015 kl. 16:52
$f(x)= \frac{u(x)}{v(x)}, \quad f´(x)= \frac{u´(x) \cdot v(x) - u(x) \cdot v´(x)}{(v(x))^2}, \quad$
$f'(x)= \lim_{\Delta x \rightarrow0} \frac{\frac{u(x+\Delta x)}{v(x+ \Delta x)} - \frac{u(x)}{v(x)}}{\Delta x} \\ = \lim_{\Delta x \rightarrow0} \frac{u(x+\Delta x) \cdot v(x) - {u(x) \cdot v(x+ \Delta x)}}{\Delta x \cdot v(x+ \Delta x) \cdot v(x)} \\ = \lim_{\Delta x \rightarrow0} \frac{u(x+\Delta x) \cdot v(x)- u(x) \cdot v(x) - {u(x) \cdot v(x+ \Delta x) + u(x) \cdot v(x)}}{\Delta x \cdot v(x+ \Delta x) \cdot v(x)} $
