Differenzenquotient von 1/x^2

Ich hoffe, du kennst den allgemeinen Differenzielqoutienten noch

${\displaystyle \underset{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_1\to {\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}{\mathrm{lim<mspace\; width="0.12em"></mspace>}}\left(\frac{\mathrm{f<mspace\; width="0.12em"></mspace>}\left({\mathrm{x<mspace\; width="0.12em"></mspace>}}_1\right)-\mathrm{f<mspace\; width="0.12em"></mspace>}\left({\mathrm{x<mspace\; width="0.12em"></mspace>}}_0\right)}{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_1-{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}\right)}$

Es gilt: ${\displaystyle {\mathrm{x<mspace\; width="0.12em"></mspace>}}_1={\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+\mathrm{h<mspace\; width="0.12em"></mspace>}}$

${\displaystyle \underset{\mathrm{h<mspace\; width="0.12em"></mspace>}\to 0}{\mathrm{lim<mspace\; width="0.12em"></mspace>}}\left(\frac{\mathrm{f<mspace\; width="0.12em"></mspace>}({\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+\mathrm{h<mspace\; width="0.12em"></mspace>})-\mathrm{f<mspace\; width="0.12em"></mspace>}\left({\mathrm{x<mspace\; width="0.12em"></mspace>}}_0\right)}{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+\mathrm{h<mspace\; width="0.12em"></mspace>}-{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}\right)=\underset{\mathrm{h<mspace\; width="0.12em"></mspace>}\to 0}{\mathrm{lim<mspace\; width="0.12em"></mspace>}}\left(\frac{\mathrm{f<mspace\; width="0.12em"></mspace>}({\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+\mathrm{h<mspace\; width="0.12em"></mspace>})-\mathrm{f<mspace\; width="0.12em"></mspace>}\left({\mathrm{x<mspace\; width="0.12em"></mspace>}}_0\right)}{\mathrm{h<mspace\; width="0.12em"></mspace>}}\right)}$

${\displaystyle \underset{\mathrm{h<mspace\; width="0.12em"></mspace>}\to 0}{\mathrm{lim<mspace\; width="0.12em"></mspace>}}\left(\frac{\frac{1}{{({\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+\mathrm{h<mspace\; width="0.12em"></mspace>})}^2}-\frac{1}{{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}^2}}{\mathrm{h<mspace\; width="0.12em"></mspace>}}\right)}$

${\displaystyle \underset{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_1\to {\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}{\mathrm{lim<mspace\; width="0.12em"></mspace>}}\left(\frac{\frac{{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}^2-{({\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+\mathrm{h<mspace\; width="0.12em"></mspace>})}^2}{{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}^2*{({\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+\mathrm{h<mspace\; width="0.12em"></mspace>})}^2}}{\mathrm{h<mspace\; width="0.12em"></mspace>}}\right)}$

${\displaystyle \underset{\mathrm{h<mspace\; width="0.12em"></mspace>}\to 0}{\mathrm{lim<mspace\; width="0.12em"></mspace>}}\left(\frac{{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}^2-{({\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+\mathrm{h<mspace\; width="0.12em"></mspace>})}^2}{{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}^2*{({\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+\mathrm{h<mspace\; width="0.12em"></mspace>})}^2*\mathrm{h<mspace\; width="0.12em"></mspace>}}\right)}$

${\displaystyle \underset{\mathrm{h<mspace\; width="0.12em"></mspace>}\to 0}{\mathrm{lim<mspace\; width="0.12em"></mspace>}}\left(\frac{-2{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0*\mathrm{h<mspace\; width="0.12em"></mspace>}-{\mathrm{h<mspace\; width="0.12em"></mspace>}}^2}{{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}^2*{({\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+\mathrm{h<mspace\; width="0.12em"></mspace>})}^2*\mathrm{h<mspace\; width="0.12em"></mspace>}}\right)}$

${\displaystyle \underset{\mathrm{h<mspace\; width="0.12em"></mspace>}\to 0}{\mathrm{lim<mspace\; width="0.12em"></mspace>}}(-\frac{\mathrm{h<mspace\; width="0.12em"></mspace>}(2{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+\mathrm{h<mspace\; width="0.12em"></mspace>})}{{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}^2*{({\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+\mathrm{h<mspace\; width="0.12em"></mspace>})}^2*\mathrm{h<mspace\; width="0.12em"></mspace>}})}$

${\displaystyle \underset{\mathrm{h<mspace\; width="0.12em"></mspace>}\to 0}{\mathrm{lim<mspace\; width="0.12em"></mspace>}}(-\frac{(2{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+\mathrm{h<mspace\; width="0.12em"></mspace>})}{{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}^2*{({\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+\mathrm{h<mspace\; width="0.12em"></mspace>})}^2})}$

${\displaystyle =(-\frac{(2{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+0)}{{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}^2*{({\mathrm{x<mspace\; width="0.12em"></mspace>}}_0+0)}^2})=-\frac{2{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}{{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}^4}=-\frac{2}{{{\mathrm{x<mspace\; width="0.12em"></mspace>}}_0}^3}}$

${\displaystyle \mathrm{f<mspace\; width="0.12em"></mspace>}\prime \left(\mathrm{x<mspace\; width="0.12em"></mspace>}\right)=-\frac{2}{{\mathrm{x<mspace\; width="0.12em"></mspace>}}^3}}$

Ich hoffe, das konnte dir helfen

lg Dravo5