多元函数微分学部分习题解析.pdf

õ ¼© Æ SK ‰ !ÀJK 1§ f (x, y) = xy + (y − 1) arcsin x , K f ′ (x, 1) = ( ) y x µǑ f (x, 1) = x, ¤ fx (x, 1) = 1 y x 2, u = z , K du (1,1,1) = ( ) 1 (ln y −ln z) µu = e x u (1, 1, 1) = 0, u (1, 1, 1) = 1, u (1, 1, 1) = −1, ¤ du = dy − dz x y z 3, ¼ f (x, y) 3 : (x , y ) ?  5Ÿ §( ·K 0 0 A) ë Y  3 7^‡ £ £B)  3 ë Y ©^‡ £C)  3 Œ © 7^‡ (D) Œ ë Y  3 ©^‡ µD ( 2 ∂u 2 ∂u 4 § u = u(x, y) Œ §  y = x §k u(x, y) = 1 9 = x, K  y = x (x = 0) § = ( ), ∂x ∂y A. 1 , B. − 1 , C.0 D.1 2 2 ∂u 1 2 2 2 1 2 1 µd ∂x = x, =⇒ u = 2 x + ϕ(y) 5  u(x, x ) = 1 =⇒ ϕ(x ) = 1 − 2 x , =⇒ ϕ(y) = 1 − 2 y ¤ u(x, y) = 1 x2 + 1 − 1 y ,=⇒ ∂u = − 1 2 2 ∂y 2  2 sin(x y) 5. f (x, y) =  xy , xy = 0 , K f ′ (0, 1) = ( ) x  x, xy = 0 A.0, B.√1, C. 2D. Ø 3 f (0, 1) = lim f (x,1)−f (0,1) = lim sin x2 = 1 µ x 2 x→0 x x→0 x 6. |^C þ O† u = x, v = y Œò  x ∂z + y ∂z = z z Ǒ#  ( ) x ∂x ∂y A√.u ∂z = z, B.v ∂z = z, C.u ∂z = z, D.v ∂z = z ∂u ∂v ∂v ∂u