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y
∂xf(x, y) =
2 |xy| sign(xy) ∂yf(x,
x
y) =
2 sign(xy)
|xy|
y
xy = 0 Df(x, y) =
2 sign(xy) dx +
|xy|
x
2 sign(xy) dy.
|xy|
x
∂xf(x, y) = −√
∂yf(x, y) = −√
y
|2−x2−y2 | |2−x2−y2 |
R 2 x 2 + y 2 = 2
x
y
Df(x, y) = −
dx − dy.
|2 − x2 − y2 | |2 − x2 − y2 |
v = (1/ √ 2, 1/ √ 2, 0) v (0, 0, 0)
f(x, y, z) = (2x − 3y + 4z) cos(xyz)
f
∂xf(x, y, z) = 2 cos(xyz) − yz(2x − 3y + 4z) sin(xyz)
∂yf(x, y, z) = −3 cos(xyz) − xz(2x − 3y + 4z) sin(xyz)
∂zf(x, y, z) = 4 cos(xyz) − xy(2x − 3y + 4z) sin(xyz).
R 3 R 3
∂f
∂u (0, 0, 0) = Df(0, 0, 0)u = ∂xf(0,
√
2
0, 0)ux + ∂yf(0, 0, 0)uy + ∂zf(0, 0, 0)uz = −
2 .
f(x, y) = x3y x6 (x, y) = 0 f(0, 0) = 0
+ y2 f(x, y) = log(1 + 3y3 )
x2 + y2 (x, y) = 0 f(0, 0) = 0
sin y +
f(x, y) =
|x| log(1 + y2 )
(x, y) = 0 f(0, 0) = 0
x 2 + y 2
f(x, y) = arctan(x2 + y 2 )
x 2 + y 2
(x, y) = 0 f(0, 0) = 0
γ(t) = (t, t 3 ) (0, 0) t → 0 +
lim f(γ(t)) = lim
t→0 t→0
t6 1
= = 0 = f(0, 0).
2t6 2
(0, 0)
v = (vx, vy)
vx = 0 vy = 0
f((0, 0) + t(vx, vy)) − f(0, 0) f(t(vx, vy)) 1
∂vf(0, 0) = lim
= lim
= lim
t→0
t
t→0 t
t→0 t
= lim
t→0
tv 3 xvy
t 4 v 6 x + v 2 y
= 0.
t 4 v 3 xvy
t 6 v 6 x + t 2 v 2 y
(0, 0)
(0, 0)