Bài tập toán cao cấp part 9 pdf

9.2. Vi phˆan cu ’ a h`am nhiˆe ` ubiˆe ´ n 127 hay l`a f(x +∆x, y +∆y) ≈ f(x, y)+ ∂f ∂x (M)∆x + ∂f ∂y (M)∆y (9.8) Cˆong th´u . c (9.8) l`a co . so . ’ d ˆe ’ ´ap du . ng vi phˆan t´ınh gˆa ` nd´ung. Dˆo ´ i v´o . i h`am c´o sˆo ´ biˆe ´ n nhiˆe ` uho . n2tac˜ung c´o cˆong th´u . ctu . o . ng tu . . . 9.2.3 C´ac t´ınh chˆa ´ tcu ’ a vi phˆan Dˆo ´ iv´o . i c´ac h`am kha ’ vi f v`a g ta c´o: (i) d(f ± g)=df ±dg; (ii) d(fg)=fdg + gdf, d(αf)=αdf, α ∈ R; (iii) d  f g  = gdf − fdg g 2 , g =0; (iv) Vi phˆan cˆa ´ p1cu ’ a h`am hai biˆe ´ n f(x, y)bˆa ´ tbiˆe ´ nvˆe ` da . ng bˆa ´ t luˆa . n x v`a y l`a biˆe ´ ndˆo . clˆa . p hay l`a h`am cu ’ a c´ac biˆe ´ ndˆo . clˆa . p kh´ac. 9.2.4 Vi phˆan cˆa ´ p cao Gia ’ su . ’ h`am w = f(x, y) kha ’ vi trong miˆe ` n D. Khi d ´o vi phˆan cˆa ´ p1 cu ’ a n´o ta . idiˆe ’ m(x, y) ∈ D tu . o . ng ´u . ng v´o . i c´ac sˆo ´ gia dx v`a dy cu ’ a c´ac biˆe ´ ndˆo . clˆa . pdu . o . . cbiˆe ’ udiˆe ˜ nbo . ’ i cˆong th´u . c df = ∂f ∂x dx + ∂f ∂y dy. (9.9) O . ’ d ˆa y , dx =∆x, dy =∆y l`a nh ˜u . ng sˆo ´ gia t`uy ´y cu ’ abiˆe ´ ndˆo . clˆa . p, d´o l`a nh˜u . ng sˆo ´ khˆong phu . thuˆo . c v`ao x v`a y.Nhu . vˆa . y, khi cˆo ´ di . nh dx v`a dy vi phˆan df l`a h`am cu ’ a x v`a y. Theo di . nh ngh˜ıa: Vi phˆan th ´u . hai d 2 f (hay vi phˆan cˆa ´ p 2) cu ’ a h`am f(x, y)ta . idiˆe ’ m M(x, y)du . o . . cdi . nh ngh˜ıa nhu . l`a vi phˆan cu ’ avi phˆan th ´u . nhˆa ´ tta . idiˆe ’ m M v´o . i c´ac diˆe ` ukiˆe . n sau dˆay: (1) Vi phˆan df l`a h`am chı ’ cu ’ a c´ac biˆe ´ ndˆo . clˆa . p x v`a y. 128 Chu . o . ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe ` ubiˆe ´ n (2) Sˆo ´ gia cu ’ a c´ac biˆe ´ ndˆo . clˆa . p x v`a y xuˆa ´ thiˆe . n khi t´ınh vi phˆan cu ’ a f  x v`a f  y du . o . . c xem l`a b˘a ` ng sˆo ´ gia d ˆa ` u tiˆen, t ´u . cl`ab˘a ` ng dx v`a dy. T`u . d´o d 2 f(M)= ∂ 2 f(M) ∂x 2 dx 2 +2 ∂ 2 f ∂x∂y (M)dxdy + ∂ 2 f ∂y 2 (M)dy 2 (9.10) trong d´o dx 2 =(dx) 2 , dy 2 =(dy) 2 v`a ta xem c´ac da . o h`am riˆeng hˆo ˜ n ho . . pb˘a ` ng nhau. Mˆo . t c´ach h`ınh th´u . c d ˘a ’ ng th ´u . c (9.10) c´o thˆe ’ viˆe ´ tdu . ´o . ida . ng d 2 f =  ∂ ∂x dx + ∂ ∂y dy  2 f(x, y) t´u . c l`a sau khi thu . . chiˆe . n ph´ep “b`ınh phu . o . ng” ta cˆa ` ndiˆe ` n f(x, y) v`ao “ˆo trˆo ´ ng”. Tu . o . ng tu . . d 3 f =  ∂ ∂x dx + ∂ ∂y dy  3 f(x, y) = ∂ 3 f ∂x 3 dx 3 +3 ∂ 3 f ∂x 2 ∂y dx 2 dy +3 ∂ 3 f ∂x∂y 2 dxdy 2 + ∂ 3 f ∂y 3 dy 3 , v.v Mˆo . t c´ach quy na . p ta c´o d n f(x, y)= n  k=0 C k n ∂ n f ∂x n−k ∂y k dx n−k dy k . (9.11) Trong tru . `o . ng ho . . pnˆe ´ u w = f(t, v),t= ϕ(x, y),v= ψ(x, y) th`ı dw = ∂f ∂t dt + ∂f ∂v dx (t´ınh bˆa ´ tbiˆe ´ nvˆe ` da . ng !) d 2 w = ∂ 2 f ∂t 2 dt 2 +2 ∂ 2 f ∂t∂v dtdy + ∂ 2 f ∂v 2 dv 2 + ∂f ∂t d 2 t + ∂f ∂v d 2 v. (9.12) 9.2. Vi phˆan cu ’ a h`am nhiˆe ` ubiˆe ´ n 129 9.2.5 Cˆong th´u . c Taylor Nˆe ´ u h`am f(x, y)l`an +1 lˆa ` n kha ’ vi trong ε-lˆan cˆa . n V cu ’ adiˆe ’ m M 0 (x 0 ,y 0 )th`ıdˆo ´ iv´o . idiˆe ’ mbˆa ´ tk`yM(x, y) ∈Vta c´o cˆong th´u . c Taylor f(x, y)=f(x 0 ,y 0 )+ 1 1!  f  x (x 0 ,y 0 )(x − x 0 )+f  y (x 0 ,y 0 )(y −y 0 )  + 1 2!  f  xx (x 0 ,y 0 )(x − x 0 ) 2 +2f  xy (x 0 ,y 0 )(x −x 0 )(y − y 0 ) + f  yy (x 0 ,y 0 )(y −y 0 )  + ···+ 1 n! m  i=0 C i n ∂ n f(x 0 ,y 0 ) ∂x n−i ∂y i (x −x 0 ) n−i (y − y 0 ) i + 1 (n + 1)! n  i=0 ∂ n+1 f(ξ,η) ∂x n−i ∂y i (x − x 0 ) n−i (y −y 0 ), (9.13) trong d ´o ξ = x 0 + θ(x − x 0 ), η = y 0 + θ(y −y 0 ), 0 <θ<1. hay l`a f(x, y)=f(x 0 ,y 0 )+ 1 1! df (x 0 ,y 0 )+ 1 2! d 2 f(x 0 ,y 0 )+ + 1 n! d n f(x 0 ,y 0 )+R n+1 , = P n (x, y)+R n+1 (9.14) trong d´o P n (x, y)go . il`adath´u . c Taylor bˆa . c n cu ’ a hai biˆe ´ n x v`a y, R n+1 l`a sˆo ´ ha . ng du . .Nˆe ´ ud ˘a . t ρ =  ∆x 2 +∆y 2 th`ı (9.14) c´o thˆe ’ viˆe ´ tdu . ´o . ida . ng f(x, y)=P n (x, y)+0(ρ),ρ→ 0, o . ’ dˆay R n+1 = o(ρ) l`a phˆa ` ndu . da . ng Peano. 130 Chu . o . ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe ` ubiˆe ´ n 9.2.6 Vi phˆan cu ’ a h`am ˆa ’ n Theo di . nh ngh˜ıa: biˆe ´ n w du . o . . cgo . i l`a h`am ˆa ’ ncu ’ a c´ac biˆe ´ nd ˆo . clˆa . p x, y, , t nˆe ´ un´od u . o . . cchobo . ’ iphu . o . ng tr`ınh F (x,y, ,w)=0 khˆong gia ’ id u . o . . cd ˆo ´ iv´o . i w. D ˆe ’ t´ınh vi phˆan cu ’ a h`am ˆa ’ n w ta lˆa ´ y vi phˆan ca ’ hai vˆe ´ cu ’ aphu . o . ng tr`ınh (xem nhu . dˆo ` ng nhˆa ´ tth´u . c) rˆo ` it`u . d´ot`ımdw.Dˆe ’ t´ınh d 2 w ta cˆa ` n lˆa ´ y vi phˆan cu ’ a dw v´o . ilu . u´yr˘a ` ng dx v`a dy l`a h˘a ` ng sˆo ´ , c`on dw l`a vi phˆan cu ’ a h`am. Ta c˜ung c´o thˆe ’ thu d u . o . . c vi phˆan dw b˘a ` ng c´ach t´ınh c´ac da . o h`am riˆeng: w  x = − F  x (·) F  w (·) ,w  y = − F  y (·) F  w (·) , rˆo ` ithˆe ´ v`ao biˆe ’ uth´u . c dw = ∂w ∂x dx + ∂w ∂y dy + ···+ ∂w ∂t dt, v.v C ´ AC V ´ IDU . V´ı du . 1. T´ınh vi phˆan df nˆe ´ u 1) f(x, y)=xy 2 ,2)f(x, y)=  x 2 + y 2 . Gia ’ i. 1) Ta c´o f  x =  xy 2   x = y 2 ,f  y =  xy 2 )  y =2xy. Do d´o df (x, y)=y 2 dx +2xydy. 2) Ta t´ınh c´ac da . o h`am riˆeng: f  x = x  x 2 + y 2 ,f  y = y  x 2 + y 2 · 9.2. Vi phˆan cu ’ a h`am nhiˆe ` ubiˆe ´ n 131 Do d´o df = x  x 2 + y 2 dx + y  x 2 + y 2 dy = xdx + ydy  x 2 + y 2 ·  V´ı du . 2. T´ınh df (M 0 )nˆe ´ u f(x, y, z)=e x 2 +y 2 +z 2 v`a M 0 = M 0 (0, 1, 2). Gia ’ i. Ta c´o df (M)= ∂f ∂x (M)dx + ∂f ∂y (M)dy + ∂f ∂z (M)dz, M = M(x, y, z). Ta t´ınh c´ac da . o h`am riˆeng ∂f ∂x =2xe x 2 +y 2 +z 2 ⇒ ∂f ∂x (M 0 )=0, (v`ı x =0) ∂f ∂y =2ye x 2 +y 2 +z 2 ⇒ ∂f ∂y (M 0 )=2e 5 , ∂f ∂z =2ze x 2 +y 2 +z 2 ⇒ ∂f ∂z (M 0 )=4e 5 . T`u . d ´o df (M 0 )=2e 5 dy +4e 5 dz.  V´ı d u . 3. T´ınh dw ta . idiˆe ’ m M 0 (−1, 1) nˆe ´ u w = f(x + y 2 ,y+ x 2 ). Gia ’ i. C´ach 1. T´ınh c´ac da . o h`am riˆeng cu ’ a h`am f(x, y) theo x v`a theo y rˆo ` i´apdu . ng cˆong th´u . c (9.9). T`u . v´ıdu . 4, mu . c 9.1 ta c´o ∂f ∂x (M 0 )=f  t (0, 2) − 2f  v (0, 2) ∂f ∂y (M 0 )=2f  t (0, 2) + f  v (0, 2) t = x + y 2 ,v= y + x 2 v`a do d´o df (M 0 )=  f  t (0, 2) − 2f  v (0, 2)  dx +2  2f  t (0, 2) + f  v (0, 2)  dy. 132 Chu . o . ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe ` ubiˆe ´ n C´ach 2. ´ Ap du . ng t´ınh bˆa ´ tbiˆe ´ nvˆe ` da . ng cu ’ a vi phˆan cˆa ´ p1. Ta c´o t = x + y 2 ⇒ dt = dx +2ydy, v = y + x 2 ⇒ dv =2xdx + dy. Do d ´o df (M 0 )= ∂f ∂t (0, 2)dt + ∂f ∂v (0, 2)dv = f  t (0, 2)[dx +2ydy]+f  v (0, 2)[2xdx + dy] =  f  t (0, 2) − 2f  v (0, 2)  dx +  2f  t (0, 2) + f  v (0, 2)  dy.  V´ı d u . 4. 1) Cho h`am f(x, y)=x y . H˜ay t`ım vi phˆan cˆa ´ p hai cu ’ a f nˆe ´ u x v`a y l`a biˆe ´ ndˆo . clˆa . p. 2) T`ım vi phˆan cˆa ´ p hai cu ’ a h`am f(x + y, xy)nˆe ´ u x v`a y l`a biˆe ´ n d ˆo . clˆa . p. Gia ’ i. 1) T`u . v´ıdu . 2, 1) v`a cˆong th ´u . c (9.10) ta c´o d 2 f = ∂ 2 f ∂x 2 dx 2 +2 ∂ 2 f ∂x∂y dxdy + ∂ 2 f ∂y 2 dy 2 , trong d´o ∂ 2 f ∂x 2 = y(y − 1)x y−2 , ∂ 2 f ∂y 2 = x y (lnx) 2 , ∂ 2 f ∂x∂y = x y−1 (1 + ylnx) v`a do d ´o d 2 f = y(y − 1)x y−2 dx 2 + x y−1 (1 + ylnx)dxdy + x y (lnx) 2 dy 2 . 2) Ta viˆe ´ t h`am d˜a cho du . ´o . ida . ng u = f(t, v), trong d ´o t = x + y, v = xy. 9.2. Vi phˆan cu ’ a h`am nhiˆe ` ubiˆe ´ n 133 1 + C´ach I. T´ınh c´ac da . o h`am riˆeng rˆo ` i ´ap du . ng (9.10). Ta c´o: ∂f ∂x = f  t (x + y,xy)+f  v (x + y,xy) · y, ∂f ∂y = f  t (x + y,xy)+f  v (x + y,xy) · x, ∂ 2 f ∂x 2 = f  tt + f  tv y + f  tv y + f  vv y 2 = f  tt +2yf  tv + y 2 f  vv , ∂ 2 f ∂x∂y = f  tt + f  tv x + f  tv y + f  vv xy + f  v = f  tt +(x + y)f  tv + xyf  vv + f  v , ∂ 2 f ∂y 2 = f  tt + f  tv x + f  tv x + f  vv x 2 = f  tt +2xf  tv + x 2 f  vv . Thˆe ´ c´ac d a . o h`am riˆeng t`ım du . o . . c v`ao (9.10) ta thu du . o . . c d 2 f =(f  tt +2yf  tv + y 2 f  vv )dx 2 +2(f  tt +(x + y)f  tv + xyf  vv + f  v )dxdy +(f  tt +2xf  tv + x 2 f  vv )dy 2 . 2 + C´ach II. Ta c´o thˆe ’ thu du . o . . ckˆe ´ t qua ’ n`ay nˆe ´ ulu . u´yr˘a ` ng v´o . i t = x + y ⇒ dt = dx + dy v`a v = xy → dv = xdy + ydx v`a t`u . d ´o d 2 t = d(dx + dy)=d 2 x + d 2 y =0 (v`ı x v`a y l`a biˆe ´ nd ˆo . clˆa . p) v`a d 2 v = d(xdy + ydx)=dxdy + dxdy =2dxdy. ´ Ap du . ng (9.12) ta c´o d 2 f = ∂ 2 f ∂t 2 (dx + dy) 2 +2 ∂ 2 f ∂t∂v (dx + dy)(xdy + ydx) + ∂ 2 f ∂v 2 (xdy + ydx) 2 + ∂f ∂t · 0+ ∂f ∂v (2dxdy) =  f  tt +2yf  tv + y 2 f  vv  dx 2 +  f  tt +2xf  tv + x 2 f  vv  dy 2 +2  f  tt +(x + y)f  tv + xyf  vv + f  v  dxdy.  134 Chu . o . ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe ` ubiˆe ´ n V´ı du . 5. ´ Ap du . ng vi phˆan dˆe ’ t´ınh gˆa ` nd´ung c´ac gi´a tri . : 1) a =(1,04) 2,03 2) b = arctg  1, 97 1, 02 − 1  3) c =  (1, 04) 1,99 + ln(1, 02) 4) d = sin 1, 49 · arctg0, 07 2 2,95 . Gia ’ i. Dˆe ’ ´ap du . ng vi phˆan v`ao t´ınh gˆa ` nd´ung ta cˆa ` n thu . . chiˆe . n c´ac bu . ´o . c sau dˆay: Th´u . nhˆa ´ t l`a chı ’ r˜o biˆe ’ uth´u . c gia ’ it´ıchd ˆo ´ iv´o . i h`am m`a gi´a tri . gˆa ` n d ´ung cu ’ a n´o cˆa ` n pha ’ i t´ınh. Th´u . hai l`a cho . ndiˆe ’ mdˆa ` u M 0 sao cho gi´a tri . cu ’ a h`am v`a cu ’ a c´ac da . o h`am riˆeng cu ’ a n´o ta . idiˆe ’ mˆa ´ y c´o thˆe ’ t´ınh m`a khˆong cˆa ` nd`ung ba ’ ng. Cuˆo ´ ic`ung ta ´ap du . ng cˆong th´u . c f(x 0 +∆x, y 0 +∆y)=f(x 0 ,y 0 )+f  x (x 0 ,y 0 )∆x + f  y (x 0 ,y 0 )∆y. 1) T´ınh a =(1, 04) 2,03 . Ta x´et h`am f(x, y)=x y .Sˆo ´ a cˆa ` n t´ınh l`a gi´a tri . cu ’ a h`am khi x =1,04 v`a y =2, 03. Ta lˆa ´ y M 0 = M 0 (1, 2). Khi d´o∆x =0, 04, ∆y =0, 03. Tiˆe ´ p theo ta c´o ∂f ∂x = yx y−1 ⇒ ∂f ∂x   M 0 =2 ∂f ∂y = x y lnx ⇒ ∂f ∂y   M 0 =1·ln1 = 0. Bˆay gi`o . ´ap du . ng cˆong th´u . cv`u . anˆeuo . ’ trˆen ta c´o: a = f(1, 04; 2, 03) = (1, 04) 2,03 ≈ f(1, 2) + 2 · 0, 04 = 1 + 0, 08 = 1, 08. 2) Ta nhˆa . nx´etr˘a ` ng arctg  1, 97 1, 02 − 1  l`a gi´a tri . cu ’ a h`am f(x, y) = arctg  x y − 1  9.2. Vi phˆan cu ’ a h`am nhiˆe ` ubiˆe ´ n 135 ta . idiˆe ’ m M(1, 97; 1, 02). Ta cho . n M 0 = M 0 (2, 1) v`a c´o ∆x =1, 97 − 2=−0, 03, ∆y =1, 02 − 1=0, 02. Tiˆe ´ pd ˆe ´ n ta c´o ∂f ∂x = 1 y 1+  x y − 1  2 = y y 2 +(x − y) 2 ∂f ∂y = − x y 2 +(x −y) 2 · T`u . d´o ∂f ∂x (M 0 )=f  x (2, 1) = 1 1 2 +(2− 1) 2 =0, 5 ∂f ∂y (M 0 )=f  y (2, 1) = −1. Do d´o arctg  1, 97 1, 02 − 1  = arctg  2 1 −1  +(0, 5) · (−0, 03) + 1 · (0, 02) = π 4 −0, 015 − 0, 02=0, 785 −0, 035 =0, 75. 3) Ta thˆa ´ yr˘a ` ng c =  (1, 04) 1,99 + ln(1, 02) l`a gi´a tri . cu ’ a h`am u = f(x, y, z)= √ x y +lnz ta . idiˆe ’ m M(1, 04; 1, 99; 1, 02). Ta cho . n M 0 = M 0 (1, 2, 1). Khi d´o ∆x =1, 04 − 1=0, 04 ∆y =1, 99 − 2=−0, 01 ∆z =1, 02 − 1=0, 02. 136 Chu . o . ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe ` ubiˆe ´ n Bˆay gi`o . ta t´ınh gi´a tri . c´ac d a . o h`am riˆeng ta . idiˆe ’ m M 0 .Tac´o ∂f ∂x = yx y−1 2 √ x y +lnz ⇒ ∂f ∂x (M 0 )= 2 · 1 2 √ 1 + ln1 =1, ∂f ∂y = x y lnx 2 √ x y +lnz ⇒ ∂f ∂y (M 0 )=0, ∂f ∂z = 1 2z √ x y +lnz ⇒ ∂f ∂z (M 0 )= 1 2 · T`u . d´o suy ra  (1, 04) 1,99 + ln(1, 02) ≈ √ 1+ln1+1· (0, 04) + 0 · (−0, 01) +(1/2) · 0, 02=1,05. 4) Ta thˆa ´ y d l`a gi´a tri . cu ’ a h`am f(x, y, z)=2 x sin y arctgx ta . idiˆe ’ m M(−2, 95; 1, 49; 0, 07) Ta lˆa ´ y M 0 = M 0  − 3, π 2 , 0  . Khi d´o ∆x = −2, 95 − (−3)=0, 05 ∆y =1,49 − 1, 57 = −0, 08 ∆z =0, 07. Tiˆe ´ p theo ta c´o f(M 0 )=2 −3 sin(π/2) arctg0 = 0, f  x (M 0 )=2 x ln2 · sin y arctgz   M 0 =0, f  y (M 0 )=2 x cos y arctgz   M 0 =0, f  z (M 0 )= 2 x sin y 1+z 2   M 0 =2 −3 . T`u . d´o ta thu du . o . . c sin 1, 49 arctg0, 07 2 2,95 ≈ 2 −3 · 0, 07 ≈ 0, 01.  V´ı du . 6. Khai triˆe ’ n h`am f(x, y)=x y theo cˆong th´u . c Taylor ta . i lˆan cˆa . nd iˆe ’ m(1, 1) v´o . i n =3. [...]... w w 3w 3w2 (**) ’ ’ ´ ` Dˆ c´ biˆu th´.c d2 w qua x, y, w, dx v` dy ta cˆn thˆ dw t` (*) v`o e o e u a a e u a (**) ’ ’ e V´ du 9 C´c h`m ˆn u(x, y) v` v(x, y) du.o.c x´c dinh bo.i hˆ ı a a a a a xy + uv = 1, xv − yu = 3 ´ ` 9. 2 Vi phˆn cua h`m nhiˆu biˆn a ’ a e e 1 39 ´ T´ du(1, −1), d2 u(1, −1); dv(1, −1), d2 v(1, −1) nˆu u(1, −1) = 1, ınh e v(1, −1) = 2 ` ´ ’ a o Giai Lˆy vi phˆn hˆ d˜ cho... dx − 3 cos ydy) ´ ` Chu.o.ng 9 Ph´p t´ vi phˆn h`m nhiˆu biˆn e ınh a a e e 140 4 w = ln(x2 + y) (DS 2xdx dy + 2 ) 2+y x x +y y x−1 y x y dx + ln − dy) x x x y 2ydx 2dy 6 w = ln tg (DS − + ) 2y 2y x x2 sin x sin x x ’ ’ a a a T´ dw(M0 ) cua c´c h`m tai diˆm M0 d˜ cho (7-14) ınh e 5 w = y x x (DS y x x y 7 w = e− x , M0 (1, 0) (DS dw(1, 0) = −dy) √ 8 w = y 3 x, M0 (1, 1) 9 f (x, y) = 1 (DS dw(1, 1)... 0, 0) a (DS df df M M0 = (ft cos x − fv sin x)dx + ft cos ydy + fv sin zdz, = ft (0, 0)dx + fv (0, 0)dy, t = sin x + sin y, v = cos x − cos z) ’ ´ e T´ vi phˆn dw v` d2 w tai diˆm M(x, y) ( 19- 22) nˆu: ınh a a e 19 w = f (lnz), z = x2 + y 2 (DS d2 w = (x2 2 (2x2 ftt − x2 ft + y 2ft )dx2 + y 2 )2 + (4xyftt − 4xyft )dxdy + (x2 ft − yft + 2yft2 )dy 2 ) ` ´ 20 w = f (α, β, γ), α = ax, β = by, γ = cz; a,... ` Giai Ta xem phu.o.ng tr` d˜ cho nhu mˆt dˆng nhˆt v` lˆy vi o ´ ´ phˆn cua vˆ tr´i v` vˆ phai: a ’ e a a e ’ 3w2 dw + 6xydx + 3x2 dy + wdx + xdw + 2y · w2 dy + 2y 2 wdw − 2dx + dy = 0 ´ ` Chu.o.ng 9 Ph´p t´ vi phˆn h`m nhiˆu biˆn e ınh a a e e 138 v` t` d´ r´t ra dw Ta c´ a u o u o (6xy + w − 2)dx + (3x2 + 2yw2 + 1)dy + (3w2 + x + 2y 2w)dw = 0 v` do d´ a o dw = 2 − 6xy − w 3x2 + 2yw2 + 1 dx − dy... ra (15-18) ım a ’ a a 15 f (x, y) = f (x − y, x + y), M(x, y), M0 (1, −1) (DS df df M M0 = (ft + fv )dx + (fv − ft )dy, = ft (2, 0) + fv (2, 0) dx + fv (2, 0) − ft (2, 0) dy, t = x − y, v = x + y) ´ ` 9. 2 Vi phˆn cua h`m nhiˆu biˆn a ’ a e e 16 f (x, y) = f xy, (DS df df 141 x , M(x, y), M0 (0, 1) y 1 x = yft + fv dx + xft − 2 fv dy, y y M = ft (0, 0) + fv (0, 0) dx, t = xy, v = M0 x ) y 17 f (x, y,...´ ` 9. 2 Vi phˆn cua h`m nhiˆu biˆn a ’ a e e 137 ’ o a o u o Giai Trong tru.`.ng ho.p n`y cˆng th´.c Taylor c´ dang sau dˆy a df (1, 1) d2 f (1, 1) d2 f (1, 1) + + + R3 (*) f (x, y) = f (1, 1) + 1! 2! 3!... 2 + c2 fγ 2 dz 2 + 2(fαβ abdxdy + fβγ bcdydz + fαγ acdxdz)) 21 w = f (x + y, x − y) (DS x + y = u, x − y = v; d2 w = (fu2 + 2fuv + fv2 )dx2 + (fu2 − 2fv2 )dxdy + (fu2 − 2fuv + fv2 )dy 2 ) ´ ` Chu.o.ng 9 Ph´p t´ vi phˆn h`m nhiˆu biˆn e ınh a a e e 142 x2 x (DS dw = f + f dx − 2 f dy, y y x 22 w = xf y d2 w = 2 x 4x 2x2 f + 2 f )dx2 − f + 3 f y y y2 y dxdy − 2x2 x3 f − 4f y3 y dy 2 ) ’ ´ ’ a a a a . z)= √ x y +lnz ta . idiˆe ’ m M(1, 04; 1, 99 ; 1, 02). Ta cho . n M 0 = M 0 (1, 2, 1). Khi d´o ∆x =1, 04 − 1=0, 04 ∆y =1, 99 − 2=−0, 01 ∆z =1, 02 − 1=0, 02. 136 Chu . o . ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe ` ubiˆe ´ n Bˆay. tri . : 1) a =(1,04) 2,03 2) b = arctg  1, 97 1, 02 − 1  3) c =  (1, 04) 1 ,99 + ln(1, 02) 4) d = sin 1, 49 · arctg0, 07 2 2 ,95 . Gia ’ i. Dˆe ’ ´ap du . ng vi phˆan v`ao t´ınh gˆa ` nd´ung ta cˆa ` n. = ∂ 2 f ∂t 2 dt 2 +2 ∂ 2 f ∂t∂v dtdy + ∂ 2 f ∂v 2 dv 2 + ∂f ∂t d 2 t + ∂f ∂v d 2 v. (9. 12) 9. 2. Vi phˆan cu ’ a h`am nhiˆe ` ubiˆe ´ n 1 29 9.2.5 Cˆong th´u . c Taylor Nˆe ´ u h`am f(x, y)l`an +1 lˆa ` n kha ’ vi