Kvotient regel derivasjon-bevis: Forskjell mellom sideversjoner

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Bevis:   
Bevis:   


f(x)=limΔx0u(x+Δx)v(x+Δx)u(x)v(x)Δx=limΔx0u(x+Δx)v(x)u(x)v(x+Δx)Δxv(x+Δx)v(x)=limΔx0u(x+Δx)v(x)u(x)v(x)u(x)v(x+Δx)+u(x)v(x)Δxv(x+Δx)v(x)=limΔx0(u(x+Δx)u(x)Δxv(x)v(x+Δx)v(x)
$f'(x)= \lim_{\Delta x \rightarrow0} \frac{\frac{u(x+\Delta x)}{v(x+ \Delta x)} - \frac{u(x)}{v(x)}}{\Delta x} \ = \lim_{\Delta x \rightarrow0} \frac{u(x+\Delta x) \cdot v(x) - {u(x) \cdot v(x+ \Delta x)}}{\Delta x \cdot v(x+ \Delta x) \cdot v(x)} \= \lim_{\Delta x \rightarrow0} \frac{u(x+\Delta x) \cdot v(x)- u(x) \cdot v(x) - {u(x) \cdot v(x+ \Delta x) + u(x) \cdot v(x)}}{\Delta x \cdot v(x+ \Delta x) \cdot v(x)}  \ = \lim_{\Delta x \rightarrow0} ( \frac{u(x+\Delta x)- u(x)}{\Delta x} \cdot \frac{v(x)}{v(x+\Delta x) \cdot v(x)} - \frac{v(x+\Delta x)- v(x)}{\Delta x} \cdot \frac{u(x)}{v(x+\Delta x) \cdot v(x)}$

Sideversjonen fra 5. jun. 2015 kl. 17:12

Vi har:

f(x)=u(x)v(x),f´(x)=u´(x)v(x)u(x)v´(x)(v(x))2,f´(x)=limΔx0f(x+Δx)f(x)Δx

Bevis:

f(x)=limΔx0u(x+Δx)v(x+Δx)u(x)v(x)Δx=limΔx0u(x+Δx)v(x)u(x)v(x+Δx)Δxv(x+Δx)v(x)=limΔx0u(x+Δx)v(x)u(x)v(x)u(x)v(x+Δx)+u(x)v(x)Δxv(x+Δx)v(x)=limΔx0(u(x+Δx)u(x)Δxv(x)v(x+Δx)v(x)v(x+Δx)v(x)Δxu(x)v(x+Δx)v(x)