Hallo,
\( \begin{aligned} f(x) &=-\frac{1}{8} x^{3}+\frac{3}{8} x^{2}+\frac{9}{8} x-1 \\[10pt] f(x+h) &=-\frac{1}{8}(x+h)^{3}+\frac{3}{8}(x+h)^{2}+\frac{9}{8}(x+h)-1 \end{aligned} \)
\( \begin{aligned}=&-\frac{1}{8}\left(x^{3}+3 x^{2} h+3 x h^{2}+h^{3}\right)+\frac{3}{8}\left(x^{2}+2 x h+h^{2}\right) \\+& \frac{9}{8} x+\frac{9}{8} h-1 \\[8pt]=&-\frac{1}{8} x^{3}-\frac{3}{8} x^{2} h-\frac{3}{8} x h^{2}-\frac{1}{8} h^{3}+\frac{3}{8} x^{2}+\frac{3}{4} x h+\frac{3}{8} h^{2} \\ &+\frac{9}{8} x+\frac{9}{8} h-1 \end{aligned} \)
\( f(x)=-\frac{1}{8} x^{3}+\frac{3}{8} x^{2}+\frac{9}{8} x-1 \)
\(f'(x)=\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}\)
\( \begin{aligned} f^{\prime}(x) &=\lim \limits_{h \rightarrow 0} \frac{-\frac{3}{8} x^{2} h-\frac{3}{8} x h^{2}-\frac{1}{3} h^{3}+\frac{3}{4} x h+\frac{3}{8} h^{2}+\frac{9}{8} h}{h} \\ &=\lim \limits_{h \rightarrow 0} \frac{h\left(-\frac{3}{8} x^{2}-\frac{3}{8} x h-\frac{1}{3} h^{2}+\frac{3}{4} x+\frac{3}{8} h+\frac{9}{8}\right)}{h} \\ &=-\frac{3}{8} x^{2}+\frac{3}{4} x+\frac{9}{8} \end{aligned} \)
Gruß, Silvia
