(90)-高数专项练题8_08.2026考研数学高途王喆全程班_赠送2025课程_25考研数学（三）全年智达班_{2}--资料

第五章 多元函数微分学例5.1 设 f (x, y)  arcsin x2  y4 , 则下列选项正确的是( ). B (A) f (0,0)存在, f (0,0)存在 (B) f (0,0)不存在, f (0,0)存在 x y x y (C) f (0,0)不存在, f (0,0)不存在 (D) f (0,0)存在, f (0,0)不存在 x y x y fixd-flo arcsmIX) o fx I - was u 10 0) = - = . - o Y X-8 *T2 . flot)-floor ariswy f10 s - o 0) Me Mr = . I y -0 Yo = Y F    例5.2 设 f  x , y , f  x , y 均存在,则下列选项正确的是( C ). x 0 0 y 0 0   (A) lim f (x, y)存在 (B) f (x, y)在 x , y 处连续 0 0 xx 0     (C) lim f x, y 存在 (D) f (x, y)在U x , y 内有定义 0 0 0 xx 0 fix 40) fixo 40) fil Yo ex . - . F = fixto fixo = = . X-Vo d=a y 2ay + x + = = x yea pl -capsmo = 9 X = - Y 0 x+ = 1 yarctan ,( x, y)  (0,0),  例5.3 设 f (x, y)   x2  y2 则 f (x, y)在点  0, (x, y)  (0,0),  ( 0 , 0 ) ↑ C 处( ). (A)连续但不可微 (B)偏导数存在但不连续 (C)可微 (D)连续但偏导数不存在 = f M yar == 470 fixd-floor e Mu 0 - 0 fil 0 = = 0 . X o X o I flat - for Marim--0PR fil00) , my m ryl = I 470 Y& 82 dZ f(x .y) fix0) (fx 100.ox fy 000y) pu - - - + , Pa = P (x , y)+ 10 . 0) (M) + 10 . 01 X =2 I Ey Yarem -0 = Men & = p ①Y ( - Ju . arcta- = = (x . Y)+10. 0) x yz xty2 W . 41+10 . 0) x y2 Y + y I 1 ** Fr Wan Carin - ) Hu -dZ it = 0 ·: = 0 ( . Y1+10 . 0) ( . Y1+10 0) P . = 10 0) of S* . 2 f xy 例5.4 设函数 f  x, y    ext 2 dt，则  . 0 xy 1,1 : X-E *Ye 35 - : fixy ) n = / U s ** zru an un = 2 & t = - + - 382 y2 . = X=1 X=/ tesy / - u + = 57 -2 y oxa . y( (e2y ) e + = + y 1 = E =(e) ze = + = = ye - 2 f xy 例5.4 设函数 f  x, y    ext 2 dt，则  . 0 xy 1,1 of eX(xy) exyx 35 = : = X = · of exiy2 57 exy.3y + = 24 2x . st = Pe = Ex-ay 11 11 .↑ 2 2  u  u 例5.5 设函数u  u(x, y)满足方程   0及条件u(x,2x)  x, 2 2 x y u (x,2x)  x2,其中u  u(x, y)具有二阶连续偏导数,则u (x,2x) ( B ) x xx I 11 . 4x 4x 3x 3x (A) ； (B) ； (Ｃ) ； (Ｄ) ． 3 3 4 4 75XX Ux GET1 UNIX) III) IX 2x) : : . Xi 2 U(XIT : # 28 X I : UI 2X) X : . = , X U n'(x 2x) 42 2 -'n + (x 2x) 2 / . . · = I ! x * u (x 2x) = . x Ur 1 - : IX 2x) . = , 2ui(x2x) X U2(X2x) 1 x ↓ = = , ! Fiz #75 X U Hi Q Un (x2x) 2 2 x => (X 2x) + = . Un Fz8X5i Uz & => uni (2x) + (X , 2x1 . 2 = - X 2BPE i > u Un Hi : U X = = ex . Hid ui) : => . (X 2x) + (x 2x) 2 X , . · = - 0 ul * 0 : Y . = - 3 2z   例5.6 求z  f sin x,cos y,ex y 的 ,其中 xy f 具有二阶连续偏导数. ficosx fsey = + 1(SMX) X f z = 2(10sy) 82 fis". **Y) (fir" -smy) e + los X = . Y)] 5xzY # >(e* ** ** (fs' fase ** ) fo +s <-smy) + . + fir SmycoSX fis &M Y fir ** fas" ex+ Y) = - . + losX - . Smy , e ! ++Y fs e +  例5.7 设u  f xyz, g(2 y  z) ,其中 f 具有连续三阶连续偏导数, g 二阶可  3u X 导,试求 . I (XYz) xyz (fil fir filu f = z)Y , , 2(9(24 + u fi yz - = Z fiz * (fixz fir g 21 fi x y2 fi29: fiz 4z = + . : . + = . + 24z + y 34 (fil-xy fis g xyz fi + 2xyz + - = . 2x2y77 . (fixy fi 9) 29yz + fizg" fingzy + . + Yz + . . (fix fig) + f + · 2 +fi 42" 19 xyz ! xyz) fill &xyzfil 2197-yzfic + + 29 + - . + ! fir f (29" 29 7 z9 + +z + + +例5.8 设函数 f (u,v)可微，z  z(x, y)由方程(x  1)z  y2  x2 f (x  z, y) 确定，则dz  . (0,1) ↑ X=0 4 = SXGE = z = . EF L , -c dIXfNx -z 41] d[(xellz] => = . f(x dX Xdf(x => zd(x + ) + (x + )dz - 24my = - z , y) . + -z , 4) => zdX + (x + 1)dz - 24dy = f(x zy) . 2xax + y · (fidix -z) + f2 dy) 4 y=. x X =0 zz/ , = dy + dz - 2ay = 0 = dz)(a)) ax +2dy = -例5.9 设 y  g(x, z)，z  z(x, y)由方程 f (x  z, xy)  0所确定，其中 f ， dz g具有一阶连续的偏导数，求 . dx < x f(x - 9(z)) 35 y 91z) = -z = 0 = = , - / F(xz) f(x-z (xzl) xg = , fix =fi + =_ fi f2-X ! 21 + 9 . .例5.9 设 y  g(x, z)，z  z(x, y)由方程 f (x  z, xy)  0所确定，其中 f ， dz g具有一阶连续的偏导数，求 . dx 13 9(z) = =(y = CT Y YI) Z z(x) = = , f(x xy) -z = 0 , di 24 25 X an g 9 . = + = & dX fi (l - fr( - ) · + + X 8 # =↑ ↑ 例5.10 设 f (x, y, z)  x2z3e y ，其中z  z(x, y)由方程 x3  y3  z3  3xyz 确定的隐函数，试求u(x, y)  f (x, y, z(x, y))在点(1,0)处的全微分 du . (1,0) Xe7 zix) f(x zix1) 41 Y ux = = . , , . 2 ! z xie1 32:820 2x e + . = . &X OX &* + el #J #XS = +, Y = z / X = = 0 . =& X y z3 F(xy z) + + 3xyz . = - 35 F 347 - * - = - Y = - G 0 1) 32 3XY . . - 3y2 07 E 3xz (2 - = - = + 2 , 0 . 1) 322 - 3xy = Bo x 1 Y z = * -11 : = = 0 + , . = 0 13 . . &( * x0 0 = -14 . = 5 2 all = - , 1) ,0 . /( . du %) = - Jdx = 2 dy .例5.11 设二元函数F(x, y)具有二阶连续的偏导数，且F(x , y )  0， 0 0 F(x , y )  0，F(x , y )  0.若一元函数 y  y(x)是由方程F(x, y)  0 x 0 0 y 0 0 所确定的在点( x , y )附近的隐函数，则 0 0 x 0 是函数 y  y(x)的极小值点 的一个充分条件是( B ). (A)F (x , y )  0 (B)F (x , y )  0(C)F (x , y )  0 (D)F (x , y )  0 xx 0 0 xx 0 0 yy 0 0 yy 0 0 Ear day : EHE Y = TEXO 8 = 20 & dX dX X=Xo x=Xo um - F(x - Y) = 0 , ELIX , Y = MX) . Ex Fy Stx Y => + = 0 X= Xo = 40 , , => Ex (40 40) Fy (0.40) Exilx x(x + . = 0 => = 0 = xo =vo# Fy . #EX Y = YI) , YM 0 + = , , F* Fy an ax (Fix Fyy. M Fy + + < x(x0 Yo . + + =0 - x day FyNo Fix Yo) Yo) => (0 . + . = O x=Xo day # Yol (Xo . = = - ax #y 40) (o >0 . = so/Af Ex E ( Yo) = Y =NXEA . 0 xxo -例5.12 设函数z  (1 e y )cos x  ye y ,则函数z  f (x, y)（ ） （A）无极值点 （B）有有限个极值点 （C）有无穷多个极大值点 （D）有无穷多个极小值点 Ex -(Hellsmx 0 = X= kx = = zy el10X el -y e) (10X y) e) = - . = - 1 - . = 0 EkE Y (K 2) EKEl SEE (KE Of . - , , . Z (l e) cosX Ziy -e4 Smx Zy ! (10sx = - + . , = . = e - - y) E kEBAf (kX -2) ACO B 0 Cco AC-BcO DEEBI . = . . , EKESBAJ (K O) AdO B (20 AG-Bo LACO = 0 , , , , FR :例5.13 设 f (x, y)  kx2  2kxy  y2 在点(0,0)处取得极小值，求 k 的取值 范围. fi 3 5 - : = 2kX + 24y [ => fi 2xx + 2 = fin fin fix 2 = 2 = 2 = AC-B" D >0 24/ 4k 4k0 & - => ockal As o 2 > o ACB ② At P4k -4k 04 =0 =0 = k = =/ f(xy) y f(0 At : OKI) k =0 = 0 = . 0 , >0 f(x x q2 (x flo k = 1 At - y) = + 2xy + = + % = ,例5.13 设 f (x, y)  kx2  2kxy  y2 在点(0,0)处取得极小值，求 k 的取值 范围. flo 35 =: . % = o f(x y k* ky" ky y . y) = kX + 2kxy + = + 2kxy + - + y 4)y2 k(x C + + - = # ( #T GOzKelAt fixy30 flo S ,0 , = 0 . C-K)% 0例5.13 设 f (x, y)  kx2  2kxy  y2 在点(0,0)处取得极小值，求 k 的取值 范围. f fNM) f10 0 * E 94 *Li 33 =: >, % = , . 10 )") &10 A = = OKE例5.14 设曲线C : x2  y2  xy  3，求曲线 C 上的点到(1,1)距离的最 大值. CF E-E(x) 34 (1-1) E 3) : 2 At+C d = P /H x(x y * FIXM N (+ + + -xy = + - , F 2(X + 1) + 2xX + xy = 0 = & Fy 2(y+ ) + 2xy + NX = 0 = = y #x xy = x + + - 3 20F 2(X + 1) + 2xX + xy = 0 Q = & Fy 2(y+ ) + 2xy + NX = 00 = #x = y 20 xy = x + + - 3 0 - 2X - zy + xX - xy = 0 = 2(x - y) + x(X - y) = 0 - y)(2 => (x - + x) = 0 #x casel x= #F - X = 1 y = 1 x= +, 4 . , , AJ(X0 y / PY XiX case. x == = x + = - y y = X = 2 = X=+ =2 : S d(1 . 1) = 0 dH ,+1 = 0 d(2 . 1) = 3 d( +,2) = 3 .例5.15 求函数z  x2  y2  2x  y在区域D : x2  y2  1上的最大值与最 小值. X Ex f 2X +2 =0 = X = = ( z) - . & t zy = 24 + = 0 = y = - , #TTEDIA # X &XS tElo 22] ItX = E = . cosE SmE = + + 2cost + Sit = 1 + 2cost + Sit 1 + 5 Sk(t +4) = Hl-5 H5