| x2+y2 | ||
Mam policzyć pochodne cząstkowe f(x,y)= | ||
| xy |
Prosiłbym o rozpisanie
| f(x) | ||
policz teraz pochodną z : | ||
| g(x) |
| (x+A2)'*AX − (X+A2)(Ax)' | (1+0)*AX − (x+A2)*(0*1) | |||
f(x)g(x)' = | = | = | ||
| (Ax)2 | Ax2 |
| AX | ||
| AX2 |
| (2x+2A)*AX−(x+A2)*A1 | ||
f(x)g(x)= | = 2x2A+2A2x−Ax+A3AX2 = porażka | |
| (AX)2 |
| 2x*(xy) − (x2 + y2)*y | ||
Pliczę Ci pochodną po x: f'x = | = ... | |
| (xy)2 |
| 2x*Ax − (x2 + A2)A | ||
... w podanym przeze mnie schemacie: f' = | ||
| (Ax)2 |
| 2x2y−(x2y+y3 | |
= x2y + y3x2y2 = y(x2+y2)x2y2 = x2+y2yx2 = | |
| x2y2 |
| 2x2y − x2y − y3 | x2y − y3 | x2 − y2 | ||||
.... = | = | = | =
| |||
| x2y2 | x2y2 | x2y |
| 1 | y | ||
− | ....czyli masz dobrze.. ![]() | ||
| y | x2 |
| x2 + y2 | ||
f(x, y) = | ||
| xy |
| df(x, y) | d | x2 + y2 | |||
= | ( | ) = | |||
| dx | dx | xy |
| d | x2 | d | y2 | |||||
= | ( | ) + | ( | ) = | ||||
| dx | xy | dx | xy |
| 1 | d | d | 1 | ||||
= | (x) + y | ( | ) = | ||||
| y | dx | dx | x |
| 1 | y | |||
= | − | = | ||
| y | x2 |
| x2 − y2 | ||
= | ||
| x2y |
| df(x, y) | d | x2 + y2 | |||
= | ( | ) = | |||
| dy | dy | xy |
| d | x2 | d | y2 | |||||
= | ( | ) + | ( | ) = | ||||
| dy | xy | dy | xy |
| d | 1 | 1 | d | ||||
= x | ( | ) + | (y) = | ||||
| dy | y | x | dy |
| x | 1 | |||
= − | + | = | ||
| y2 | x |
| y2 − x2 | ||
= | ||
| xy2 |