f ( x ) = x 3 + x − 2 {\displaystyle f(x)=x^{3}+x-2}
f ′ ( x ) = lim h → 0 f ( x 0 + h ) − f ( x 0 ) h {\displaystyle f'(x)=\lim _{h\to 0}{\frac {f(x_{0}+h)-f(x_{0})}{h}}}
f ′ ( x ) = lim h → 0 ( x 0 + h ) 3 + ( x 0 + h ) − 2 − x 0 3 − x 0 + 2 h {\displaystyle f'(x)=\lim _{h\to 0}{\frac {(x_{0}+h)^{3}+(x_{0}+h)-2-x_{0}^{3}-x_{0}+2}{h}}}
f ′ ( x ) = lim h → 0 ( x 0 + h ) 3 + ( x 0 + h ) − x 0 3 − x 0 h {\displaystyle f'(x)=\lim _{h\to 0}{\frac {(x_{0}+h)^{3}+(x_{0}+h)-x_{0}^{3}-x_{0}}{h}}}
f ′ ( x ) = lim h → 0 x 0 3 + 3 x 0 2 h + 3 x 0 h 2 + h 3 − x 0 3 − x 0 h {\displaystyle f'(x)=\lim _{h\to 0}{\frac {x_{0}^{3}+3x_{0}^{2}h+3x_{0}h^{2}+h^{3}-x_{0}^{3}-x_{0}}{h}}} (ausmultiplizieren oder pascalsches dreieck)
f ′ ( x ) = lim h → 0 3 x 0 2 h + 3 x 0 h 2 + h 3 − x 0 h {\displaystyle f'(x)=\lim _{h\to 0}{\frac {3x_{0}^{2}h+3x_{0}h^{2}+h^{3}-x_{0}}{h}}}
(ausmultiplizieren oder pascalsches dreieck)
