Differenzialrechnung
Steigung durch Limes
Wir kennen zwar die normale Ableitung, die man eigentlich benutzt, um die Steigung und Krümmung einer Funktion zu ermitteln, doch gibt es auch noch einen weiteren Weg, auf dem die Ableitungen basieren. Wir errechnen den Steigungswert für einen Punkt anhand des Limes.
f(x) = x2
| f(x) |
= |
x2 |
| fI(x) |
= |
lim
h -> 0 |
f(x + h) - f(x)
x + h - x |
| fI(x) |
= |
lim
h -> 0 |
f(x + h) - f(x)
h |
| fI(x) |
= |
lim
h -> 0 |
(x + h)² - x²
h |
| fI(x) |
= |
lim
h -> 0 |
x² +2xh + h² - x²
h |
| fI(x) |
= |
lim
h -> 0 |
2xh + h²
h |
| fI(x) |
= |
lim
h -> 0 |
2x + h |
| fI(x) |
= |
lim
h -> 0 |
2x |
|
|
f(x) = x3
| f(x) |
= |
x3 |
| fI(x) |
= |
lim
h -> 0 |
f(x + h) - f(x)
x + h - x |
| fI(x) |
= |
lim
h -> 0 |
f(x + h) - f(x)
h |
| fI(x) |
= |
lim
h -> 0 |
(x + h)³ - x³
h |
| fI(x) |
= |
lim
h -> 0 |
x³ + 3x² · h + 3x · h² + h³ - x³
h |
| fI(x) |
= |
lim
h -> 0 |
3x² · h + 3x · h² + h³
h |
| fI(x) |
= |
lim
h -> 0 |
(3x² + 3x · h + h²) |
| fI(x) |
= |
lim
h -> 0 |
3x² |
|
|
f(x) = 3x2 + 1
| f(x) |
= |
3x2 + 1 |
| fI(x) |
= |
lim
h -> 0 |
f(x + h) - f(x)
h |
| fI(x) |
= |
lim
h -> 0 |
3(x + h)² + 1 - (3x² + 1)
h |
| fI(x) |
= |
lim
h -> 0 |
3(x² + 2xh + h²) + 1 - 3x² - 1
h |
| fI(x) |
= |
lim
h -> 0 |
3x² + 6xh + 3h² + 1 - 3x² - 1
h |
| fI(x) |
= |
lim
h -> 0 |
6xh + 3h²
h |
| fI(x) |
= |
lim
h -> 0 |
(6x + 3h) |
| fI(x) |
= |
lim
h -> 0 |
6x |
|
|
f(x) = 4x3 - x
| f(x) |
= |
4x3 - x |
| fI(x) |
= |
lim
h -> 0 |
f(x + h) - f(x)
h |
| fI(x) |
= |
lim
h -> 0 |
4(x + h)³ - (x + h) - (4x³ - x)
h |
| fI(x) |
= |
lim
h -> 0 |
4(x³ + 3x²h + 3xh² + h³) - (x + h) - (4x³ - x)
h |
| fI(x) |
= |
lim
h -> 0 |
4x³ + 12x²h + 12xh² + 4h³ - x - h - 4x³ + x
h |
| fI(x) |
= |
lim
h -> 0 |
12x²h + 12xh² + 4h³ - h
h |
| fI(x) |
= |
lim
h -> 0 |
(12x² + 12xh + 4h² - 1) |
| fI(x) |
= |
lim
h -> 0 |
12x² - 1 |
|
|
21.04.2005
