Wiki-Quellcode von Lösung Ableitungsregeln entdecken und begründen
Zuletzt geändert von akukin am 2025/08/03 21:25
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author | version | line-number | content |
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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_1m_2)x+(m_2b_1+b_2) | ||
23 | \end{align} | ||
24 | {{/formula}} | ||
25 | |||
26 | ))) | ||
27 | ))) | ||
28 | 1. (((Die Ableitung an der Stelle {{formula}}x_0{{/formula}} berechnet sich mit Hilfe des Differenzialquotienten durch | ||
29 | {{formula}}f'(x_0)=\lim\limits_{x\rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0}{{/formula}} | ||
30 | |||
31 | 1. (((Summenfunktion: | ||
32 | |||
33 | {{formula}} | ||
34 | \begin{align} | ||
35 | f'(x_0)&=\lim\limits_{x\rightarrow x_0} \frac{(m_1+m_2)x+(b_1+b_2)-((m_1+m_2)x_0+(b_1+b_2))}{x-x_0}\\ | ||
36 | &=\lim\limits_{x\rightarrow x_0} \frac{(m_1+m_2)x+(b_1+b_2)-(m_1+m_2)x_0-(b_1+b_2)}{x-x_0} \\ | ||
37 | &=\lim\limits_{x\rightarrow x_0} \frac{(m_1+m_2)x-(m_1+m_2)x_0}{x-x_0} \\ | ||
38 | &=\lim\limits_{x\rightarrow x_0} \frac{(x-x_0)(m_1+m_2)}{x-x_0} \\ | ||
39 | &=\lim\limits_{x\rightarrow x_0} \left((m_1+m_2)\frac{x-x_0}{x-x_0}\right) \\ | ||
40 | &=m_1+m_2 | ||
41 | \end{align} | ||
42 | {{/formula}} | ||
43 | |||
44 | Somit ist {{formula}}f'(x)=m_1+m_2{{/formula}}. | ||
45 | ))) | ||
46 | 1. (((Vielfachenfunktion: | ||
47 | |||
48 | {{formula}} | ||
49 | \begin{align} | ||
50 | f'(x_0)&=\lim\limits_{x\rightarrow x_0} \frac{(am_1)x+ab_1-((am_1)x_0+ab_1)}{x-x_0}\\ | ||
51 | &=\lim\limits_{x\rightarrow x_0} \frac{(am_1)x+ab_1-(am_1)x_0-ab_1}{x-x_0}\\ | ||
52 | &=\lim\limits_{x\rightarrow x_0} \frac{(am_1)x)-(am_1)x_0}{x-x_0}\\ | ||
53 | &=\lim\limits_{x\rightarrow x_0} \frac{am_1(x-x_0)}{x-x_0}\\ | ||
54 | &=am_1 | ||
55 | \end{align} | ||
56 | {{/formula}} | ||
57 | |||
58 | Somit ist {{formula}}f'(x)=am_1{{/formula}}. | ||
59 | ))) | ||
60 | 1. (((Produktfunktion: | ||
61 | |||
62 | {{formula}} | ||
63 | \begin{align} | ||
64 | f'(x_0)&=\lim\limits_{x\rightarrow x_0} \frac{m_1m_2x^2+(m_1b_2+m_2b_1)x+b_1b_2-(m_1m_2x_0^2+(m_1b_2+m_2b_1)x_0+b_1b_2)}{x-x_0}\\ | ||
65 | &=\lim\limits_{x\rightarrow x_0} \frac{m_1m_2x^2+(m_1b_2+m_2b_1)x+b_1b_2-m_1m_2x_0^2-(m_1b_2+m_2b_1)x_0-b_1b_2}{x-x_0}\\ | ||
66 | &=\lim\limits_{x\rightarrow x_0} \frac{m_1m_2x^2-m_1m_2x_0^2+(m_1b_2+m_2b_1)x-(m_1b_2+m_2b_1)x_0}{x-x_0}\\ | ||
67 | &=\lim\limits_{x\rightarrow x_0} \frac{m_1m_2(x^2-x_0^2)+(m_1b_2+m_2b_1)(x-x_0)}{x-x_0}\\ | ||
68 | &=\lim\limits_{x\rightarrow x_0} \frac{m_1m_2(x-x_0)(x+x_0)+(m_1b_2+m_2b_1)(x-x_0)}{x-x_0}\\ | ||
69 | &=\lim\limits_{x\rightarrow x_0} (m_1m_2(x+x_0)+(m_1b_2+m_2b_1)) \\ | ||
70 | &=m_1m_2 2x_0+(m_1b_2+m_2b_1)\\ | ||
71 | \end{align} | ||
72 | {{/formula}} | ||
73 | |||
74 | Somit ist {{formula}}f'(x)=2m_1m_2 x+(m_1b_2+m_2b_1){{/formula}}. | ||
75 | ))) | ||
76 | 1. (((Verkettung: | ||
77 | |||
78 | {{formula}} | ||
79 | \begin{align} | ||
80 | f'(x_0)&=\lim\limits_{x\rightarrow x_0} \frac{(m_1m_2)x+(m_2b_1+b_2)-((m_1m_2)x_0+(m_2b_1+b_2))}{x-x_0}\\ | ||
81 | &=\lim\limits_{x\rightarrow x_0} \frac{(m_1m_2)x+m_2b_1+b_2-(m_1m_2)x_0-m_2b_1-b_2}{x-x_0}\\ | ||
82 | &=\lim\limits_{x\rightarrow x_0} \frac{(m_1m_2)x-(m_1m_2)x_0}{x-x_0}\\ | ||
83 | &=\lim\limits_{x\rightarrow x_0} \frac{(m_1m_2)(x-x_0)}{x-x_0}\\ | ||
84 | &=m_1m_2 | ||
85 | \end{align} | ||
86 | {{/formula}} | ||
87 | |||
88 | Somit ist {{formula}}f'(x)=m_1m_2{{/formula}}. | ||
89 | ))) | ||
90 | ))) |