f '(x0) = limh→0 \(\frac{f(x_0+h)) - f(x_0)}{h}\) = limh→0 [ \(\frac{1}{h}\) · ( f(x0+h) - f(x0) ) ]
f(x) = 1/x für x0= -2
f '(-2) = limh→0 [ \(\frac{1}{h}\) ·(\(\frac{1}{-2+h}\) - \(\frac{1}{-2}\)) ] = limh→0 [ \(\frac{1}{h}\) ·( \(\frac{1}{-2+h}\) + \(\frac{1}{2}\) ) ]
Brüche in der (...) auf Hauptnenner bringen:
= limh→0 [ \(\frac{1}{h}\) · \(\frac{2+(-2+h)}{2 · (-2+h)}\) ] = limh→0 [ \(\frac{1}{h}\) · \(\frac{h}{-4+2h}\) ]
jetzt kann man h wegkürzen:
= limh→0 \(\frac{1}{-4+2h}\) = \(\frac{1}{-4}\) = -1/4
Gruß Wolfgang