| 2 | 3√x2 | |||
f(x) = ( | −√3)( 4x 3√x + | ) najpierw moge przekształcic te funkcje do | ||
| √x | 3x |
| 1 | ||
prostszej postaci i otrzymam (2*x−1/2 −√3)(4 * x4/3 + | * x−1/3) skorzystam | |
| 3 |
| 20 | 5 | 16√3 | √3 | |||||
f'(x) = | x−1/6− | x−11/6− | x1/3 + | x−4/3 | ||||
| 3 | 9 | 3 | 9 |
| 1 | ||
(2x−1/2 − √3)(4x4/3 + | x−1/3) = | |
| 3 |
| 2 | √3 | |||
= 8x5/6 + | x−5/6 − 4√3x4/3 − | x−1/3 = u | ||
| 3 | 3 |
| 5 | 2 | 5 | 4 | |||||
u' = 8* | x−1/6 + | (− | )x−11/6 − 4√3* | x1/3 − | ||||
| 6 | 3 | 6 | 3 |
| √3 | 1 | |||
*(− | )x−4/3 = | |||
| 3 | 3 |