Kvotient regel derivasjon-bevis: Forskjell mellom sideversjoner
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Bevis: | Bevis: | ||
$f'(x)= \lim_{\Delta x \rightarrow0} \frac{\frac{u(x+\Delta x)}{v(x+ \Delta x)} - \frac{u(x)}{v(x)}}{\Delta x} \\ = \lim_{\Delta x \rightarrow0} \frac{u(x+\Delta x) \cdot v(x) - {u(x) \cdot v(x+ \Delta x)}}{\Delta x \cdot v(x+ \Delta x) \cdot v(x)} \\= \lim_{\Delta x \rightarrow0} \frac{u(x+\Delta x) \cdot v(x)- u(x) \cdot v(x) - {u(x) \cdot v(x+ \Delta x) + u(x) \cdot v(x)}}{\Delta x \cdot v(x+ \Delta x) \cdot v(x)} \\ = \lim_{\Delta x \rightarrow0} ( \frac{u(x+\Delta x)- u(x)}{\Delta x}$ | $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)} \\ = \lim_{\Delta x \rightarrow0} ( \frac{u(x+\Delta x)- u(x)}{\Delta x} \cdot \frac{v(x)}{v(x+\Delta x) \cdot v(x)} - $ | ||
Sideversjonen fra 5. jun. 2015 kl. 17:10
Vi har:
$f(x)= \frac{u(x)}{v(x)}, \quad f´(x)= \frac{u´(x) \cdot v(x) - u(x) \cdot v´(x)}{(v(x))^2}, \quad f´(x)= \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$
Bevis:
$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)} \\ = \lim_{\Delta x \rightarrow0} ( \frac{u(x+\Delta x)- u(x)}{\Delta x} \cdot \frac{v(x)}{v(x+\Delta x) \cdot v(x)} - $