| −15 | ||
y= | ||
| 4(x−3)4 |
| 15 | −10 | −4 | ||||
Wyszło mi : | − | − | ||||
| (x−3)5 | (x−3)4 | (x−3)3 |
| x8 | ||
y= | ||
| 8(1−x2)4 |
| 64x7 − 64x9 − x8 | ||
Wyszło mi : | ||
| 64(1−x2)5 |
| √2x2 − 2x + 1 | ||
y= | ||
| x |
| 1 | ||
ale z tego co widze jednak nie. | ||
| 2 * √2x2 − 2x + 1 |
| f '(x) | ||
[√f(x)] ' = | ||
| 2√f(x) |
| 15 | 1 | 15 | ||||
f(x) = − | * | = − | *( x −3)−4 | |||
| 4 | ( x − 3)4 | 4 |
| 15 | 15 | |||
f '(x) = − | *(−4)*( x −3)−5 = 15*( x −3)−5 = | |||
| 4 | (x −3)5 |
| 10 | 1 | |||
− | ||||
| 3(x−3)3 | 2(x−3)2 |