Ved hjelp av definisjonen for den deriverte:

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:

$f'(x)= \lim_{\Delta x \rightarrow0} \frac{\frac{u(x+\Delta x)}{v(x+ \Delta x)} - \frac{u(x)}{v(x)}}{\Delta x} \\ = \lim_{\Delta x \rightarrow0} \frac{u(x+\Delta x) \cdot v(x) - {u(x) \cdot v(x+ \Delta x)}}{\Delta x \cdot v(x+ \Delta x) \cdot v(x)} \\= \lim_{\Delta x \rightarrow0} \frac{u(x+\Delta x) \cdot v(x)- u(x) \cdot v(x) - {u(x) \cdot v(x+ \Delta x) + u(x) \cdot v(x)}}{\Delta x \cdot v(x+ \Delta x) \cdot v(x)} \\ = \lim_{\Delta x \rightarrow0} ( \frac{u(x+\Delta x)- u(x)}{\Delta x} \cdot \frac{v(x)}{v(x+\Delta x) \cdot v(x)} - \frac{v(x+\Delta x)- v(x)}{\Delta x} \cdot \frac{u(x)}{v(x+\Delta x) \cdot v(x)} ) \\ = u´(x) \cdot \frac{v(x)}{v(x) \cdot v(x)} - v´(x) \cdot \frac{u(x)}{v(x) \cdot v(x)} \\ = \frac{u´(x) \cdot v(x) - u(x) \cdot v´(x)}{(v(x))^2}$

Eller ved hjelp av logaritmeregler:

$f(x)= \frac uv \quad \quad u>0, \quad v>0 \\ (\ln f(x))´ = \frac 1u \cdot u´- \frac 1v \cdot v´= \frac{u´v-vú}{uv}$


$( \frac uv)´ = (e^{\ln \frac uv})´ = e^{\ln \frac uv} \cdot \frac{u´v-vú}{uv} =\frac uv \cdot \frac{u´v-v`u}{uv} = \frac{uv´- v`u}{v^2}$


Derivasjonsregler