f'(x0) = limh→0 \(\frac{f(x_0+h)) - f(x_0)}{h}\)
f(x) = -3x2 für x0= 1
f'(1) = limh→0 \(\frac{-3·(1+h)^2 - (-3)}{h}\) = limh→0 \(\frac{-3·(1+2h+h^2)+3}{h}\)
= limh→0 \(\frac{-3-6h-3h^2+3}{h}\) = limh→0 \(\frac{h·(-6-3h)}{h}\)
= limh→0 \(\frac{-6-3h}{h}\) = -6
Gruß Wolfgang