| √x−1 | ||
f(x) = | ||
| x |
Tak w ramach przypomnienia co i jak
√1−x
| ||||||||
f'(x)= | ||||||||
| x2 |
Myśl dalej
| √1 − x | ||
y = | ||
| x |
| u | ||
y = | u = √w w = 1 − x v = x | |
| v |
| u | u'v − uv' | |||
y' = [ | ]' = | = *) | ||
| v | v2 |
| 1 | ||
u' = [√w]' = | * w' = **) | |
| 2√w |
| 1 | −1 | |||
**) = | * (−1) = | |||
| 2√1 − x | 2√1 − x |
| 2 − x | ||||||||||||
*) = | = − | ||||||||||||
| x2 | 2x2√1 − x |
| 2 − x | ||
y' = − | ||
| 2x2√1 − x |
| u | ||
y' = − | u = 2 − x v = 2wz w = x2 z = √t t = 1 − x | |
| v |
| u | u'v − uv' | |||
y'' = [− | ]' = − | = *) | ||
| v | v2 |
| 1 | ||
z' = [√t]' = | * t' = ***) | |
| 2√t |
| 1 | 1 | |||
***) = | * −1 = − | |||
| 2√1 − x | 2√1 − x |
| x2 | x(5x − 4) | |||
**) = 2(2x√1 − x − | ) = − | |||
| 2√1 − x | √1 − x |
| |||||||||||
*) = − | = | ||||||||||
| (2x2√1 − x)2 |
| −3x(x − 4) − 8 | ||
= − | ||
| 4x3√(1 − x)3 |
Dziadku dziękuję