| df | f(Xo+Δx,Yo)−f(Xo,Yo) | ||
(Xo,Yo)=lim(Δx→0) | |||
| dx | Δx |
| x+y | ||
a) f(x,y)= | (1,1) | |
| x |
| df | (1+Δx,1)−f(1,1) |
| |||||||||
(1,1)=lim(Δx→0) | =lim(Δx→0) | ||||||||||
| dx | Δx | Δx |
| |||||||||||
= U{ | }−2 }{Δx} (Δx powinna byc pod całym tym ułamkiem jak we | ||||||||||
|
| df | Δy | ||
=lim(Δy→0) | =1 | ||
| dy | Δy |
| x3+y3 | ||
b) f(x,y)= | dla (x,y)≠0 i 0 dla (x,y)=0 | |
| √x2+y2 |
| df | Δx2 | ||
=lim(Δx→0) | =0 | ||
| dx | Δx |
| df | Δy2 | ||
=lim(Δy→0) | =0 | ||
| dy | Δy |
| xy2 | ||
c) f(x,y,z)= | (0,1,1) | |
| z |
| df | |
=lim(Δx→0)ΔxΔx=1 | |
| dx |
| df | |
=lim(Δy→0)0Δy=0 ![]() | |
| dy |
| df | |
=lim(Δz→0)0Δz=0 ![]() | |
| dz |
| df | √11+Δx−1 | ||
=lim(Δx→0) | = i nie wiem jak tu obl ta granice.. | ||
| dx | Δx |
| df | 1+Δy−1 | ||
=lim(Δy→0) | =1 | ||
| dy | Δy |
| df | √1+Δz−1 | ||
=lim(Δz→0) | = ![]() | ||
| dz | Δz |