Wiki-Quellcode von Lösung Ableitungsregeln entdecken und begründen
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| author | version | line-number | content |
|---|---|---|---|
| 1 | (%class=abc%) | ||
| 2 | 1. (((1. Summenfunktion {{formula}}f(x)=f_1(x)+f_2(x)=m_1x+b_1+m_2x+b_2=(m_1+m_2)x+(b_1+b_2){{/formula}} | ||
| 3 | 1. Vielfachenfunktion {{formula}}f(x)=a\cdot f_1(x)=a\cdot (m_1x+b_1)=(am_1)x+ab_1{{/formula}} | ||
| 4 | 1. (((Produktfunktion | ||
| 5 | |||
| 6 | {{formula}} | ||
| 7 | \begin{align} | ||
| 8 | f(x)&=f_1(x)\cdot f_2(x) \\ | ||
| 9 | &=(m_1x+b_1)\cdot(m_2x+b_2)\\ | ||
| 10 | &=m_1m_2x^2+m_1b_2x+m_2b_1x+b_1b_2\\ | ||
| 11 | &=m_1m_2x^2+(m_1b_2+m_2b_1)x+b_1b_2 | ||
| 12 | \end{align} | ||
| 13 | {{/formula}} | ||
| 14 | |||
| 15 | ))) | ||
| 16 | 1. (((Verkettung | ||
| 17 | |||
| 18 | {{formula}} | ||
| 19 | \begin{align} | ||
| 20 | f(x)&=f_2(x)\circ f_1(x)=f_2(f_1(x))=f_2(m_1x+b_1) \\ | ||
| 21 | &=m_2(m_1x+b_1)+b_2 \\ | ||
| 22 | &=(m_2m_1)x+(m_2b_1+b_2) | ||
| 23 | \end{align} | ||
| 24 | {{/formula}} | ||
| 25 | |||
| 26 | ))) | ||
| 27 | ))) |