Nối tiếp phần 1, phần 2 của tài liệu Lý thuyết và bài tập môn Toán cao cấp (Tập 2) tiếp tục trình bày các nội dung chính sau: Phép tính vi phân hàm một biến; Đạo hàm; Vi phân; Phép tính vi phân hàm nhiều biến. Mời các bạn cùng tham khảo để nắm nội dung chi tiết.
Chu.o.ng Ph´ ep t´ınh vi phˆ an h` am mˆ o.t biˆ e´n 8.1 8.2 8.3 - a.o h` D am 61 8.1.1 - a.o h` D am cˆ a´p 61 8.1.2 - a.o h` D am cˆ a´p cao 62 Vi phˆ an 75 8.2.1 Vi phˆ an cˆ a´p 75 8.2.2 Vi phˆ an cˆ a´p cao 77 `e h` y co ba’n vˆ am kha’ vi Quy C´ ac di.nh l´ ´ t˘ ac l’Hospital Cˆ ong th´ u.c Taylor 84 8.3.1 `e h` y co ba’n vˆ am kha’ vi 84 C´ ac d i.nh l´ 8.3.2 ´ ac Lˆ opitan ac da.ng vˆ o di.nh Quy t˘ Khu’ c´ (L’Hospitale) 88 8.3.3 Cˆ ong th´ u.c Taylor 96 http://tieulun.hopto.org - a.o h`am 8.1 D 8.1 8.1.1 61 - a.o h` D am - a.o h` D am cˆ a´p Gia’ su’ h`am y = f(x) x´ac di.nh δ-lˆan cˆa.n cu’a diˆe’m x0 (U (x0 ; δ) = {x ∈ R : |x − x0 | < δ) v`a ∆f (x0) = f (x0 + ∆x) − f (x0 ) l`a sˆo´ gia cu’a n´o ta.i diˆe’m x0 tu.o.ng u ´.ng v´o.i sˆo´ gia ∆x = x − x0 cu’a dˆo´i sˆo´ `on ta.i gi´o.i ha.n h˜ Theo di.nh ngh˜ıa: Nˆe´u tˆ u.u ha.n f(x0 + ∆x) − f (x0) ∆x→0 ∆x am cu’a h`am f(x) ta.i ∆x → th`ı gi´o.i ha.n d´o du.o c go.i l`a da.o h` diˆe’m x0 v`a du o c chı’ bo’ i mˆo.t c´ac k´ y hiˆe.u: lim f(x0 + ∆x) − f(x0) dy d ≡ ≡ f (x) ≡ f (x) ≡ y ∆x→0 ∆x dx dx Da.i lu.o ng lim ∆y ∆y = lim ∆x→0 ∆x ∆x→0+0 ∆x f+ (x0) = f (x0 + 0) = lim ∆x>0 v`a ∆y ∆y = lim ∆x→0 ∆x ∆x→0−0 ∆x f− (x0 ) = f (x0 − 0) = lim ∆x0 ex ax(lna)n (−1)n−1 (n − 1)! n , x > x (−1)n−1 (n − 1)! n , x > x lna nπ sin x + http://tieulun.hopto.org - a.o h`am 8.1 D 63 f(x) f (x) cos x − sin x tgx cotgx arc sin x arccosx arctgx arccotgx f (n) (x) cos x + nπ cos2 x − sin x √ , |x| < 1 − x2 −√ , |x| < 1 − x2 1 + x2 − + x2 Viˆe.c t´ınh da.o h`am du.o c du a trˆen c´ac quy t˘a´c sau dˆay d d d [u + v] = u + v 1+ dx dx dx du d (αu) = α , α ∈ R 2+ dx dx du dv d (uv) = v +u 3+ dx dx dx d u dv du 4+ = v −u , v = dx v v dx dx d df du f[u(x)] = · (da.o h`am cu’a h`am ho p) 5+ dx du dx dy ≡ yx = th`ı 6+ Nˆe´u h`am y = y(x) c´o h`am ngu.o c x = x(y) v`a dx dx ≡ xy = · dy yx http://tieulun.hopto.org Chu.o.ng Ph´ep t´ınh vi phˆan h`am mˆo.t biˆe´n 64 u.c kha’ vi 7+ Nˆe´u h`am y = y(x) du.o c cho du.´o.i da.ng ˆa’n bo’.i hˆe th´ F (x, y) = v`a Fy = th`ı F dy =− x dx Fy ´.ng cu’a h`am F (x, y) d´o Fx v`a Fy l`a da.o h`am theo biˆe´n tu.o.ng u xem biˆe´n khˆong dˆo’i 8+ Nˆe´u h`am y = y(x) du.o c cho du.´o.i da.ng tham sˆo´ x = x(t), y = y(t) (x (t) = 0) th`ı dy y (t) = · dx x (t) 9+ dn dn u dn v (αu + βv) = α + β ; dxn dxn dxn dn uv = dxn n Cnk k=0 dn−k dk u v dxn−k dxk (quy t˘´ac Leibniz) u.c d˜a cho ta c´o thˆe’ Nhˆ a.n x´et 1) Khi t´ınh da.o h`am cu’a mˆo.t biˆe’u th´ biˆe´n dˆo’i so bˆo biˆe’u th´ u.c d´o cho qu´a tr`ınh t´ınh da.o h`am do.n gia’n ho.n Ch˘a’ng ha.n nˆe´u biˆe’u th´ u.c d´o l`a logarit th`ı c´o thˆe’ su’ du.ng c´ac `oi t´ınh da.o h`am Trong nhiˆ `eu t´ınh chˆa´t cu’a logarit dˆe’ biˆe´n dˆo’i rˆ `oi ´ap tru `o ng ho p t´ınh da.o h`am ta nˆen lˆa´y logarit h`am d˜a cho rˆ du.ng cˆong th´ u.c da.o h`am loga d y (x) lny(x) = · dx y(x) 2) Nˆe´u h`am kha’ vi trˆen mˆo.t khoa’ng du.o c cho bo’.i phu.o.ng tr`ınh F (x, y) = th`ı da.o h`am y (x) c´o thˆe’ t`ım t` u phu.o.ng tr`ınh d F (x, y) = dx ´ V´I DU CAC http://tieulun.hopto.org - a.o h`am 8.1 D 65 V´ı du T´ınh da.o h`am y nˆe´u: ex ; x = π(2n + 1), n ∈ N 1) y = ln + cos x + x2 2) y = √ , x = πn, n ∈ N x4 sin7 x u.c cu’a h`am y b˘`ang c´ach Gia’i 1) Tru.´o.c hˆe´t ta do.n gia’n biˆe’u th´ du a v`ao c´ac t´ınh chˆa´t cu’a logarit Ta c´o x 1 y = lnex − ln(1 + cos x) = − ln(1 + cos x) 3 3 Do d´o y = 1 sin x 1 (cos x) − = + = 3 + cos x 3 + cosx + tg x · 2) O’ dˆay tiˆe.n lo i ho.n ca’ l`a x´et h`am z = ln|y| Ta c´o dz dy dy dy dz dz = · = ⇒ =y · dx dy dx y dx dx dx (*) Viˆe´t h`am z du.´o.i da.ng x = ln|y| = ln(1 + x2 ) − ln|x| − 7ln| sin x| 2x cos x dz = −7 · ⇒ − dx 1+x 3x sin x Thˆe´ biˆe’u th´ u.c v` u.a thu du.o c v`ao (∗) ta c´o + x2 cos x dy 2x =√ − − dx 3x sin x x4 sin7 x + x x V´ı du T´ınh da.o h`am y nˆe´u: 1) y = (2 +cos x)x, x ∈ R; 2) y = x2 , x > Gia’i 1) Theo di.nh ngh˜ıa ta c´o y = exln(2+cos x) http://tieulun.hopto.org Chu.o.ng Ph´ep t´ınh vi phˆan h`am mˆo.t biˆe´n 66 T` u d´o y = exln(2+cos x) xln(2 + cos x) = exln(2+cos x) ln(2 + cos x) − x 2) V`ı y = e2 x lnx sin x , + cos x x ∈ R nˆen v´o.i x > ta c´o x + 2x ln2 · lnx x x + ln2 · lnx = 2x x2 x V´ı du T´ınh da.o h`am cˆa´p cu’a h`am ngu.o c v´o.i h`am y = x + x5, x ∈ R Gia’i H`am d˜a cho liˆen tu.c v`a do.n diˆe.u kh˘´ap no.i, da.o h`am y = + 5x4 khˆong triˆe.t tiˆeu ta.i bˆa´t c´ u diˆe’m n`ao Do d´o y = e2 x lnx [2x lnx] = e2 xy = x lnx 1 = · yx + 5x4 Lˆa´y da.o h`am d˘a’ng th´ u.c n`ay theo y ta thu du.o c −20x3 xyy = · x = · + 5x4 x y (1 + 5x4)3 V´ı du Gia’ su’ h`am y = f(x) du.o c cho du.´o.i da.ng tham sˆo´ bo’.i c´ac cˆong th´ u.c x = x(t), y = y(t), t ∈ (a; b) v`a gia’ su’ x(t), y(t) kha’ vi cˆa´p v`a x (t) = t ∈ (a, b) T`ım yxx Gia’i Ta c´o dy dy y y = dt = t ⇒ yx = t · dx dx xt xt dt Lˆa´y da.o h`am hai vˆe´ cu’a d˘a’ng th´ u.c n`ay ta c´o yt y · tx = t xt t xt xy −y x = t tt t tt · xt yxx = · t xt http://tieulun.hopto.org - a.o h`am 8.1 D 67 V´ı du Gia’ su’ y = y(x), |x| > a l`a h`am gi´a tri du.o.ng cho du.´o.i da.ng ˆa’n bo’.i phu.o.ng tr`ınh x2 y − = a2 b T´ınh yxx Gia’i Dˆe’ t`ım y ta ´ap du.ng cˆong th´ u.c d F (x, y) = dx Trong tru.`o.ng ho p n`ay ta c´o d x2 y − − = dx a2 b Lˆa´y da.o h`am ta c´o 2x 2y − yx = 0, a2 b b2x ⇒yx = , |x| > 0, y > a y (8.1) (8.2) Lˆa´y da.o h`am (8.1) theo x ta thu du.o c 1 − yx a b v`a t` u (8.2) ta thu du o c yx : − y y =0 b2 xx b2 b2 b4 x2 = − y − x y a2 y a2 a4 y b4 x2 y b4 = − − = − , y > ay a b ay yxx = ; 2) y = x2 cos 2x x2 − ˜e n h`am d˜a cho du.´o.i da.ng tˆo’ng c´ac phˆan th´ Gia’i 1) Biˆe’u diˆ u.c co ba’n V´ı du T´ınh y (n) nˆe´u: 1) y = 1 1 = − x2 − 4 x−2 x+2 http://tieulun.hopto.org Chu.o.ng Ph´ep t´ınh vi phˆan h`am mˆo.t biˆe´n 68 v`a d´o x −4 (n) = x−2 (n) − x+2 (n) Do x±2 (n) = (−1)(−2) · · · (−1 − n + 1)(x ± 2)−1−n = (−1)n n! (x ± 2)n+1 nˆen x −4 (n) (−1)n n! = − n+1 (x − 2) (x + 2)n+1 2) Ta ´ap du.ng cˆong th´ u.c Leibniz dˆo´i v´o.i da.o h`am cu’a t´ıch (x2 cos 2x) = Cn0x2 (cos 2x)(n) + Cn1 (x2) (cos 2x)n−1 + Cn2 (x2) (cos 2x)n−2 `eu = v`ı C´ac sˆo´ ha.ng c`on la.i dˆ x2 (k) =0 ∀ k > ´ du.ng cˆong th´ Ap u.c (cos 2x)(n) = 2n cos 2x + nπ ta thu du.o c nπ n(n − 1) cos 2x + nπ n + nx sin 2x + (x2 cos 2x)(n) = 2n x2 − V´ı du V´o.i gi´a tri n`ao cu’a a v`a b th`ı h`am ex , x 0, f(x) = x2 + ax + b, x > http://tieulun.hopto.org - a.o h`am 8.1 D 69 c´o da.o h`am trˆen to`an tru.c sˆo´ Gia’i R˜o r`ang l`a h`am f(x) c´o da.o h`am ∀ x > v`a ∀ x < Ta chı’ `an x´et diˆe’m x0 = cˆ V`ı h`am f (x) pha’i liˆen tu.c ta.i diˆe’m x0 = nˆen lim f(x) = lim f (x) = lim f (x) x→0+0 x→0−0 x→0 t´ u.c l`a lim (x2 + ax + b) = b = e0 = ⇒ b = x→0+0 Tiˆe´p d´o, f+ (0) = (x + ax + b) x =0 = a v`a f− (0) = ex x =0 = 0 `on ta.i nˆe´u a = v`a b = Nhu vˆa.y v´o.i a = 1, b = Do d´o f (0) tˆ h`am d˜a cho c´o da.o h`am ∀ x ∈ R ` TA ˆP BAI T´ınh da.o h`am y cu’a h`am y = f (x) nˆe´u: √ 10 3 − + 4) y = x3 + − + (DS √ x x x x x y = log2 x + 3log3x (DS ln24 ) xln2 · ln3 x (DS 5x ln5 + 6x ln6 − 7−x ln7) √ ) y = ln(x + + x2 + 2x + 3) (DS √ x2 + 2x + 10 ) y = tg5x (DS sin 10x √ √ ) y = ln(ln x) (DS 2xln x y = 5x + 6x + y = ln + 2x − 2x (DS ) − 4x2 http://tieulun.hopto.org `eu biˆe´n Chu.o.ng Ph´ep t´ınh vi phˆan h`am nhiˆ 144 ˜ Chı’ dˆ a n X´et h`am f = ln(x3 + y ), M0(0, 1) ii) b = 5e0,02 + (2, 03)2 ˜ Chı’ dˆ a n X´et h`am f = (DS ≈ 3, 037) 5ex + y 2, M0 (0, 2) ´.ng du.ng dˆe’ t´ınh 35 T´ınh vi phˆan cu’a h`am f(x, y) = x3 + y U xˆa´p xı’ (1, 02)3 + (1, 97)3 (DS ≈ 2, 95) Trong c´ac b`ai to´an sau dˆay (36-38) h˜ay t´ınh vi phˆan cˆa´p cu’a h`am ˆa’n z(x, y) x´ac di.nh bo’.i c´ac phu.o.ng tr`ınh tu.o.ng u ´.ng 36 z + 3x2 z = 2xy (DS dz = (2y − 6xz)dx + 2xdy ) 3(x2 + z 2) 37 cos2 x + cos2 y + cos2 z = (DS dz = − sin 2xdx + sin 2ydy ) sin 2z 38 x + y + z = e−(x+y+z) (DS dz = −dx − dy) 39 Cho w l`a h`am cu’a x v`a y x´ac di.nh bo’.i phu.o.ng tr`ınh w x = ln + w y T´ınh vi phˆan dw, d2 w (DS dw = w(ydx + wdy) , y(x + w) d2 w = − w2 (ydx − xdy)2 ) y (x + w)2 40 T´ınh dw v`a d2 w nˆe´u h`am w(x, y) du.o c x´ac di.nh bo’.i phu.o.ng tr`ınh y w − x = arctg w−x (w − x)dy , (w − x)2 + y + y 2(y + 1)(w − x)[(w − x)2 + y 2] d2 w = − dy ) [(w − x)2 + y + y]3 (DS dw = dx + http://tieulun.hopto.org `eu biˆe´n 9.3 Cu c tri cu’a h`am nhiˆ 9.3 9.3.1 145 `eu biˆ Cu c tri cu’a h` am nhiˆ e´n Cu c tri H`am f (x, y) c´o cu c da.i di.a phu.o.ng (ho˘a.c cu c tiˆe’u di.a phu.o.ng) b˘a`ng `on ta.i δ-lˆan cˆa.n cu’a diˆe’m M0 f (x0, y0 ) ta.i diˆe’m M0 (x0, y0 ) ∈ D nˆe´u tˆ cho v´o.i mo.i diˆe’m M = M0 thuˆo.c lˆan cˆa.n ˆa´y ta c´o f (M) < f(M0 ) (tu.o.ng u ´.ng : f (M) > f (M0 )) Go.i chung cu c da.i, cu c tiˆe’u cu’a h`am sˆo´ l`a cu c tri cu’a h`am sˆo´ `eu kiˆe.n cˆ `an dˆe’ tˆ `on ta.i cu c tri di.a phu.o.ng: Nˆe´u ta.i diˆe’m M0 h`am Diˆ f(x, y) c´o cu c tri di.a phu.o.ng th`ı ta.i diˆe’m d´o ca’ hai da.o h`am riˆeng cˆa´p `on ta.i) dˆ `eu b˘`ang ho˘a.c ´ıt nhˆa´t mˆo.t hai da.o h`am (nˆe´u ch´ ung tˆ `on ta.i (d´o l`a nh˜ riˆeng khˆong tˆ u.ng diˆe’m t´ o.i ha.n ho˘a.c diˆe’m d` u.ng cu’a `eu l`a diˆe’m cu c tri h`am f (x, y)) Khˆong pha’i mo.i diˆe’m d` u.ng dˆ `eu kiˆe.n du’: gia’ su’ Diˆ fxx (M0 ) =, fxy (M0 ) = B, fyy (M0 ) = C Khi d´o: i) Nˆe´u ∆(M0) = A B > v`a A > th`ı ta.i diˆe’m M0 h`am f c´o B C cu c tiˆe’u di.a phu.o.ng ii) Nˆe´u ∆(M0 ) = A B > v`a A < th`ı ta.i diˆe’m M0 h`am f c´o B C cu c da.i di.a phu.o.ng A B < th`ı M0 l`a diˆe’m yˆen ngu a cu’a f , t´ u.c B C l`a ta.i M0 h`am f khˆong c´o cu c tri iii) Nˆe´u ∆(M0 ) = A B = th`ı M0 l`a diˆe’m nghi vˆa´n (h`am f c´o B C thˆe’ c´o v`a c˜ ung c´o thˆe’ khˆong c´o cu c tri ta.i d´o) iv) Nˆe´u ∆(M0) = http://tieulun.hopto.org `eu biˆe´n Chu.o.ng Ph´ep t´ınh vi phˆan h`am nhiˆ 146 9.3.2 `eu kiˆ o diˆ Cu c tri c´ e.n `eu kiˆe.n cu’a h`am f (x, y) Trong tru.`o.ng ho p do.n gia’n nhˆa´t, cu c tri c´o diˆ `eu kiˆe.n c´ac biˆe´n l`a cu c da.i ho˘a.c cu c tiˆe’u cu’a h`am d´o da.t du.o c v´o.i diˆ x v`a y tho’a m˜an phu.o.ng tr`ınh ϕ(x, y) = (phu.o.ng tr`ınh r` ang buˆ o.c) `eu kiˆe.n v´o.i diˆ `eu kiˆe.n r`ang buˆo.c ϕ(x, y) ta lˆa.p Dˆe’ t`ım cu c tri c´o diˆ h` am Lagrange (h` am bˆ o’ tro ) F (x, y) = f(x, y)λϕ(x, y) d´o λ l`a h˘`ang sˆo´ nhˆan chu.a du.o c x´ac di.nh v`a di t`ım cu c tri thˆong thu.`o.ng cu’a h`am bˆo’ tro n`ay Dˆay l`a phu.o.ng ph´ ap th` u.a sˆ o´ bˆ a´t di.nh Lagrange `eu kiˆe.n cˆ `an dˆe’ tˆ `on ta.i cu c tri c´o diˆ `eu kiˆe.n l`a gia’i T`ım diˆ hˆe phu.o.ng tr`ınh ∂F ∂f ∂ϕ = +λ =0 ∂x ∂x ∂x ∂F ∂f ∂ϕ (9.15) = +λ =0 ∂y ∂y ∂y ϕ(x, y) = T` u hˆe n`ay ta c´o thˆe’ x´ac di.nh x, y v`a λ `e tˆ `on ta.i v`a d˘a.c t´ınh cu’a cu c tri di.a phu.o.ng du.o c minh di.nh Vˆa´n dˆ trˆen co so’ x´et dˆa´u cu’a vi phˆan cˆa´p hai cu’a h`am bˆo’ tro ∂ 2F ∂ 2F ∂ 2F dxdy + dx + dy d F = ∂x2 ∂x∂y ∂y 2 `eu du.o c t´ınh dˆo´i v´o.i c´ac gi´a tri x, y, λ thu du.o c gia’i hˆe (9.15) v´o.i diˆ kiˆe.n l`a ∂ϕ ∂ϕ dx + dy = (dx2 + dy = 0) ∂x ∂y Cu thˆe’ l`a: http://tieulun.hopto.org `eu biˆe´n 9.3 Cu c tri cu’a h`am nhiˆ 147 `eu kiˆe.n i) Nˆe´u d2 F < h`am f(x, y) c´o cu c da.i c´o diˆ `eu kiˆe.n ii) Nˆe´u d F > h`am f(x, y) c´o cu c tiˆe’u c´o diˆ `an pha’i kha’o s´at iii) Nˆe´u d2 F = th`ı cˆ Nhˆ a.n x´et `eu ho.n du.o c tiˆe´n h`anh i) Viˆe.c t`ım cu c tri cu’a h`am ba biˆe´n ho˘a.c nhiˆ tu.o.ng tu nhu o’ `eu kiˆe.n cu’a h`am ba biˆe´n ho˘a.c ii) Tu.o.ng tu c´o thˆe’ t`ım cu c tri c´o diˆ `eu ho n v´o i mˆo.t ho˘a.c nhiˆ `eu phu o ng tr`ınh r`ang buˆo.c (sˆ nhiˆ o´ phu.o.ng `an lˆa.p h`am bˆo’ tro v´o.i tr`ınh r` ang buˆ o.c pha’i b´e ho.n sˆ o´ biˆe´n) Khi d´o cˆ sˆo´ th` u.a sˆo´ chu.a x´ac di.nh b˘a`ng sˆo´ phu.o.ng tr`ınh r`ang buˆo.c iii) Ngo`ai phu.o.ng ph´ap th` u.a sˆo´ bˆa´t di.nh Lagrange, ngu.`o.i ta c`on `eu kiˆe.n d` ung phu.o.ng ph´ ap khu’ biˆe´n sˆ o´ dˆe’ t`ım cu c tri c´o diˆ 9.3.3 Gi´ a tri l´ o.n nhˆ a´t v` a b´ e nhˆ a´t cu’a h` am `en d´ong bi ch˘a.n da.t gi´a tri l´o.n nhˆa´t (nho’ nhˆa´t) H`am kha’ vi miˆ `en ho˘a.c ta.i diˆe’m d` u.ng ho˘a.c ta.i diˆe’m biˆen cu’a miˆ ´ V´I DU CAC V´ı du T`ım cu c tri di.a phu.o.ng cu’a h`am f (x, y) = x4 + y − 2x2 + 4xy − 2y `en x´ac di.nh cu’a h`am l`a to`an m˘a.t ph˘a’ng R2 Gia’i i) Miˆ ii) T´ınh c´ac da.o h`am riˆeng fx v`a fy v`a t`ım c´ac diˆe’m t´o.i ha.n Ta c´o fx = 4x3 − 4x + 4y, fy = 4y + 4x − 4y Do d´o 4x3 − 4x + 4y = 4y + 4x − 4y = http://tieulun.hopto.org `eu biˆe´n Chu.o.ng Ph´ep t´ınh vi phˆan h`am nhiˆ 148 v`a t` u d´o x1 = y1 = √ x2 = − √ y2 = √ x3 = √ y3 = − `on ta.i v´o.i mo.i diˆe’m Nhu vˆa.y ta c´o ba diˆe’m t´o.i ha.n V`ı fx , fy tˆ M (x, y) ∈ R2 nˆen h`am khˆong c`on diˆe’m t´o.i ha.n n`ao kh´ac iii) Ta t´ınh c´ac da.o h`am riˆeng cˆa´p hai v`a gi´a tri cu’a ch´ ung ta.i c´ac diˆe’m t´o i ha.n fxx (x, y) = 12x2 = 4, fxy = 4, fyy = 12y − Ta.i diˆe’m O(0, 0): A = −4, B = 4, C = −4 √ √ Ta.i diˆe’m M1(− 2, + 2): A = 20, B = 4, C = 20 √ √ Ta.i diˆe’m M2(+ 2, − 2): A = 20, B = 4, C = 20 iv) Ta.i diˆe’m O(0, 0)ta c´o −4 A B = 16 − 16 = = −4 B C Dˆa´u hiˆe.u du’ khˆong cho ta cˆau tra’ l`o.i Ta nhˆa.n x´et r˘a`ng lˆan `on ta.i nh˜ cˆa.n bˆa´t k` y cu’a diˆe’m O tˆ u.ng diˆe’m m`a f (x, y) > v`a nh˜ u.ng diˆe’m m`a f (x, y) < Ch˘a’ng ha.n do.c theo trung c Ox (y = 0) ta c´o f (x, y) y=0 = f (x, 0) = x4 − 2x2 = −x2(2 − x2 ) < `an (0, 0), v`a do.c theo du.`o.ng th˘a’ng y = x ta.i nh˜ u.ng diˆe’m du’ gˆ f (x, y) y=x = f (x, x) = 2x4 > u.ng diˆe’m kh´ac cu’a mˆo.t lˆan cˆa.n n`ao d´o cu’a Nhu vˆa.y, ta.i nh˜ `an ∆f (x, y) khˆong c´o c` diˆe’m O(0, 0) sˆo´ gia to`an phˆ ung mˆo.t dˆa´u v`a d´o ta.i O(0, 0) h`am khˆong c´o cu c tri di.a phu o ng √ √ Ta.i diˆe’m M1(− 2, 2) ta c´o 20 A B = 400 − 16 > = 20 B C http://tieulun.hopto.org `eu biˆe´n 9.3 Cu c tri cu’a h`am nhiˆ 149 √ √ v`a A > nˆen ta.i M1 (− 2, 2) h`am c´o cu c tiˆe’u di.a phu.o.ng v`a fmin = −8 √ √ Ta.i diˆe’m M2 ( 2, − 2) ta c´o AC − B > v`a A > nˆen ta.i d´o h`am c´o cu c tiˆe’u di.a phu.o.ng v`a fmin = −8 V´ı du Kha’o s´at v`a t`ım cu c tri cu’a h`am f(x, y) = x2 + xy + y − 2x − 3y Gia’i i) Hiˆe’n nhiˆen Df ≡ R ii) T`ım diˆe’m d` u.ng Ta c´o fx = 2x + y − fy = x + 2y − ⇒ 2x + y − = 0, x + 2y − = 4 Hˆe thu du.o c c´o nghiˆe.m l`a x0 = , y0 = Do d´o , l`a diˆe’m 3 3 u.ng d´o h`am f khˆong c´o diˆe’m d` u.ng n`ao kh´ac v`ı d` u.ng v`a ngo`ai diˆe’m d` `on tˆa.i ∀(x, y) fx v`a fy tˆ iii) Kha’o s´at cu c tri Ta c´o A = fx2 = 2, B fxy = 1, C = fy2 = Do d´o ∆(M0) = = > v`a A = > nˆen h`am f c´o cu c tiˆe’u ta.i diˆe’m M0 ( , 3 `eu kiˆe.n l`a V´ı du T`ım cu c tri cu’a h`am f (x, y) = − 4x − 3y v´o.i diˆ x v`a y liˆen hˆe v´o.i bo’.i phu.o.ng tr`ınh x2 + y = Gia’i Ta lˆa.p h`am Lagrange F (x, y) = − 4x − 3y + λ(x2 + y − 1) Ta c´o ∂F = −4 + 2λx, ∂x ∂F = −3 + 2λy ∂y http://tieulun.hopto.org `eu biˆe´n Chu.o.ng Ph´ep t´ınh vi phˆan h`am nhiˆ 150 v`a ta gia’i hˆe phu.o.ng tr`ınh −4 + 2λx = −3 + 2λx = x2 + y = Gia’i ta c´o λ1 = , x1 = , y1 = 5 λ2 = − , x = − , y2 = − 5 V`ı ∂ 2F = 2λ, ∂x2 ∂ 2F = 0, ∂x∂y ∂ 2F = 2λ ∂y nˆen d2 F = 2λ(dx2 + dy 2) , h`am Nˆe´u λ = , x = , y = th`ı d2 F > nˆen ta.i diˆe’m 5 5 `eu kiˆe.n c´o cu c tiˆe’u c´o diˆ Nˆe´u λ = − , x = − , y = − th`ı d2 F < v`a d´o h`am c´o cu c 5 `eu kiˆe.n ta.i diˆe’m − , − da.i c´o diˆ 5 Nhu vˆa.y 16 + = 11, 5 16 − = fmin = − 5 `eu kiˆe.n cu’a h`am V´ı du T`ım cu c tri c´o diˆ 2 1) f(x, y) = x + y + xy − 5x − 4y + 10, x + y = 2) u = f (x, y, z) = x + y + z fmax = + z − x = 1, y − xz = http://tieulun.hopto.org `eu biˆe´n 9.3 Cu c tri cu’a h`am nhiˆ 151 Gia’i 1) T` u phu.o.ng tr`ınh r`ang buˆo.c x + y = ta c´o y = − x v`a f (x, y) = x2 + (4 − x)2 + x(4 − x) − 5x − 4(4 − x) + 10 = x2 − 5x + 10, ta thu du.o c h`am mˆo.t biˆe´n sˆo´ g(x) = x2 − 5x + 10 `eu kiˆe.n cu’a ung ch´ınh l`a cu c tri c´o diˆ v`a cu c tri di.a phu.o.ng cu’a g(x) c˜ ´ du.ng phu.o.ng ph´ap kha’o s´at h`am sˆo´ mˆo.t biˆe´n sˆo´ dˆo´i h`am f (x, y) Ap v´o.i g(x) ta t`ım du.o c g(x) c´o cu c tiˆe’u di.a phu.o.ng gmin = g Nhu.ng cu c tiˆe’u d´o 15 = · h`am f(x, y) `eu diˆ kiˆe.n (y = − x ⇒ y = − = ) v`a 2 c´o fmin = f ta.i d˜a diˆe’m cho c´o , 2 15 , = · 2 2) T` u c´ac phu.o.ng tr`ınh r`ang buˆo.c ta c´o z =1+x y = x2 + x + v`a thˆe´ v`ao h`am d˜a cho ta du.o c h`am mˆo.t biˆe´n sˆo´ u = f (x, y(x), z(x)) = g(x) = 2x2 + 4x + ˜e d`ang thˆa´y r˘a`ng h`am g(x) c´o cu c tiˆe’u ta.i x = −1 (khi d´o y = 1, Dˆ `eu kiˆe.n ta.i diˆe’m z = 0) v`a d´o h`am f(x, y, z) c´o cu c tiˆe’u c´o diˆ (−1, 1, 0) v`a fmin = f(−1, 1, 0) = http://tieulun.hopto.org `eu biˆe´n Chu.o.ng Ph´ep t´ınh vi phˆan h`am nhiˆ 152 u.a sˆo´ bˆa´t di.nh Lagrange t`ım cu c tri V´ı du B˘a`ng phu.o.ng ph´ap th` `eu kiˆe.n cu’a h`am c´o diˆ u = x + y + z2 `eu kiˆe.n v´o.i diˆ z−x = y − xz = (9.16) (xem v´ı du 4, ii)) Gia’i Ta lˆa.p h`am Lagrange F (x, y, z) = x + y + z + λ1 (z − x − 1) + λ2 (y − zx − 1) v`a x´et hˆe phu.o.ng tr`ınh ∂F = − λ1 − λ2 z = ∂x ∂F = + λ2 = ∂y ∂F = 2z + λ1 − λ2 x = ∂z ϕ1 = z − x − = ϕ2 = y − xz − = Hˆe n`ay c´o nghiˆe.m nhˆa´t x = −1, y = 1, z = 0, λ1 = v`a λ2 = −1 ngh˜ıa l`a M0 (−1, 1, 0) l`a diˆe’m nhˆa´t c´o thˆe’ c´o cu c tri cu’a `eu kiˆe.n r`ang buˆo.c ϕ1 v`a ϕ2 h`am v´o.i c´ac diˆ T` u c´ac hˆe th´ u.c z−x = y − xz = ta thˆa´y r˘a`ng (9.16) x´ac di.nh c˘a.p h`am ˆa’n y(x) v`a z(x) (trong tru.`o.ng ˜e d`ang r´ ho p n`ay y(x) v`a z(x) dˆ ut t` u (9.16)) Gia’ su’ thˆe´ nghiˆe.m http://tieulun.hopto.org `eu biˆe´n 9.3 Cu c tri cu’a h`am nhiˆ 153 `ong nhˆa´t y(x) v`a z(x) v`ao hˆe (9.16) v`a b˘`ang c´ach lˆa´y vi phˆan c´ac dˆ th´ u c thu du o c ta c´o dz − dx = ⇒ dy − xdz − zdx = dz = dx dy = (x + z)dx (9.17) Bˆay gi`o t´ınh vi phˆan cˆa´p hai cu’a h`am Lagrange d2 F = 2(dz)2 − 2λ2 dxdz (9.18) Thay gi´a tri λ2 = −1 v`a (9.17) v`ao (9.18) ta thu du.o c da.ng to`an phu.o.ng x´ac di.nh du.o.ng l`a d2 F = 4dx2 `eu kiˆe.n ta.i diˆe’m T` u d´o suy h`am d˜a cho c´o cu c tiˆe’u c´o diˆ M0(−1, 1, 0) v`a fmin = V´ı du T`ım gi´a tri l´o.n nhˆa´t v`a nho’ nhˆa´t cu’a h`am f (x, y) = x2 + y − xy + x + y `en miˆ D = {x 0, y 0, x + y −3} `en D d˜a cho l`a tam gi´ac OAB v´o.i dı’nh ta.i A(−3, 0), Gia’i Miˆ B(0, −3) v`a O(0, 0) i) T`ım c´ac diˆe’m d` u.ng: fx = 2x − y + = fy = 2y − x + = u.ng l`a M(−1, −1) T` u d´o x = −1, y = −1 Vˆa.y diˆe’m d` Ta.i diˆe’m M ta c´o: f(M) = f(−1, −1) = −1 http://tieulun.hopto.org `eu biˆe´n Chu.o.ng Ph´ep t´ınh vi phˆan h`am nhiˆ 154 ii) Ta c´o A = fxx (−1, −1) = B = fxy (−1, −1) = −1 C = fyy (−1, −1) = u.c AC − B > Vˆa.y AC − B = − = > 0, nˆen h`am c´o biˆe.t th´ v`a A = > Do d´o ta.i diˆe’m M n´o c´o cu c tiˆe’u di.a phu.o.ng v`a fmin = −1 `en D iii) Kha’o s´at h`am trˆen biˆen cu’a miˆ +) Khi x = ta c´o f = y + y Dˆo´i v´o.i h`am mˆo.t biˆe´n f = y + y, −3 y ta c´o (fln ) x=0 (fnn ) x=0 = ta.i diˆe’m (0, −3) −1 ta.i diˆe’m 0, − = +) Khi y = ta c´o h`am mˆo.t biˆe´n f = x2 + x, −3 tu.o.ng tu : (fln ) y=0 (fnn ) y=0 x v`a = ta.i diˆe’m (0, −3) −1 ta.i diˆe’m − , = +) Khi x + y = −3 ⇒ y = −3 − x ta c´o f(x) = 3x2 + 9x + v`a (fnn ) x+y=−3 (fln ) x+y=−3 −3 3 ta.i diˆe’m − , − 2 = ta.i diˆe’m (0, −3) v`a (−3, 0) = iv) So s´anh c´ac gi´a tri thu du.o c dˆo´i v´o.i f ta kˆe´t luˆa.n fln = ta.i (0, −3) v`a (−3, 0) v`a gi´a tri fnn = −1 ta.i diˆe’m d` u.ng (−1, −1) ` TA ˆ P BAI http://tieulun.hopto.org `eu biˆe´n 9.3 Cu c tri cu’a h`am nhiˆ 155 H˜ay t`ım cu c tri cu’a c´ac h`am sau dˆay f = + 6x − x2 − xy − y (DS fmax = 13 ta.i diˆe’m (4, −2)) f = (x − 1)2 + 2y (DS fmin = ta.i diˆe’m (1, 0)) f = x2 + xy + y − 2x − y (DS fmin = −1 ta.i diˆe’m (1, 0)) f = x3y (6 − x − y) (x > 0, y > 0) (DS fmax = 108 ta.i diˆe’m (3, 2)) f = 2x4 + y − x2 − 2y (DS fmax = ta.i diˆe’m (0, 0), fmin = − ta.i c´ac diˆe’m M1 fmin = − ta.i c´ac diˆe’m M3 +xy+y ) f = (5x + 7y − 25)e−(x −1 , −1 v`a M2 ,1 2 −1 , −1 v`a M4 , −1 ) 2 (DS fmax = 3−13 ta.i diˆe’m M1 (1, 3), −1 −3 , ) fmin = −26e−1/52 ta.i diˆe’m M2 26 26 50 20 + , x > 0, y > (DS fmin = 30 ta.i diˆe’m (5, 2)) x y 2 f = x + xy + y − 6x − 9y (DS fmin = −21 ta.i diˆe’m (1, 4)) √ f = x y − x2 − y + 6x + (DS fmax = 15 ta.i diˆe’m (4, 4)) √ 10 f = (x2 + y) ey (DS fmin = − ta.i (0, −2)) e 11 f = + (x − 1) (y + 1) (DS fmin = ta.i diˆe’m (1, −1)) f = xy + ˜ n Ta.i diˆe’m M0 (1, −1) ta c´o ∆(M0) = Cˆ `an kha’o s´at dˆa´u Chı’ dˆ a cu’a f (M) − f(M0 ) = f(1 + ∆x, −1 + ∆y) − f (1, −1) 12 f = − (x − 2)4/5 − y 4/5 (DS fmax = ta.i diˆe’m (2, 0)) ˜ Chı’ dˆ a n Ta.i diˆe’m (2, 0) h`am khˆong kha’ vi Kha’o s´at dˆa´u cu’a f(M ) − f(M0 ), M0 = (2, 0) http://tieulun.hopto.org 156 `eu biˆe´n Chu.o.ng Ph´ep t´ınh vi phˆan h`am nhiˆ `eu kiˆe.n cu’a c´ac h`am sau dˆay T`ım cu c tri c´o diˆ `eu kiˆe.n x + y = 13 f = xy v´o.i diˆ 1 , ) (DS fmax = ta.i diˆe’m 2 `eu kiˆe.n x2 + y = 14 f = x + 2y v´o.i diˆ (DS fmax = ta.i diˆe’m (1, 2)) x y `eu kiˆe.n + = 15 f = x2 + y v´o.i diˆ 36 18 12 ta.i diˆe’m , ) (DS fmin = 13 13 13 `eu kiˆe.n x2 + y + z = 16 f = x − 2y + 2z v´o.i diˆ (DS fmin = −9 ta.i diˆe’m (−1, 2, −2); fmax = ta.i (1, −2, 2).) `eu kiˆe.n 2x + 3y = 17 f = xy v´o.i diˆ 25 5 ta.i diˆe’m , ) (DS fmax = 24 x y `eu kiˆe.n r`ang buˆo.c + = 18 1) f = x2 + y v´o.i diˆ 36 48 144 ta.i , ) (DS fmin = 25 25 25 `eu kiˆe.n x + y = 2) f = exy v´o.i diˆ 1 , ) (DS fmax = e1/4 ta.i diˆe’m 2 ˜ Chı’ dˆ a n C´o thˆe’ su’ du.ng phu.o.ng ph´ap khu’ biˆe´n `eu kiˆe.n x − y + z = 19 f = x2 + y + 2z v´o.i diˆ (DS fmin = 0, ta.i diˆe’m (0, 4; −0, 4; 0, 2)) `eu kiˆe.n x + y − z = 20 f = x3 + y − z + v´o.i diˆ 10 (DS fmin = ta.i diˆe’m (0, 0, 0) v`a fmax = ta.i diˆe’m − , , ) 27 3 http://tieulun.hopto.org `eu biˆe´n 9.3 Cu c tri cu’a h`am nhiˆ 157 `eu kiˆe.n x + y + z = 5, xy + yz + zx = 21 f = xyz v´o.i c´ac diˆ 4 7 4 ta.i , , ; , , ; , , (DS fmax = 27 3 3 3 3 fmin = ta.i (2, 2, 1); (2, 1, 2); (1, 2, 2)) T`ım gi´a tri l´o.n nhˆa´t v`a nho’ nhˆa´t cu’a c´ac h`am sˆo´ sau 22 f = x2y(2 − x − y), D l`a tam gi´ac du.o c gi´o.i ha.n bo’.i c´ac doa.n th˘a’ng x = 0, y = 0, x + y = (DS fln = ta.i diˆe’m (1, 2); fnn = −128 ta.i diˆe’m (4, 2)) 23 f = x + y, D = {x2 + y 1} √ √ √ 2 , ; (DS fln = ta.i diˆe’m biˆen √2 √ √ 2 ,− ) fnn = − ta.i diˆe’m biˆen − 2 24 T` u mo.i tam gi´ac c´o chu vi b˘`ang 2p, h˜ay t`ım tam gi´ac c´o diˆe.n t´ıch l´o.n nhˆa´t ˜ Chı’ dˆ a n D˘a.t a = x, b = y ⇒ c = 2p − x − y v`a ´ap du.ng cˆong th´ u.c Heron S= p(p − x)(p − y)(x + y − p) `eu) (DS Tam gi´ac dˆ 25 X´ac di.nh gi´a tri l´o.n nhˆa´t v`a nho’ nhˆa´t cu’a h`am f = x2 − y 2, D = {x2 + y 1} (DS fln = ta.i (1, 0) v`a (−1, 0); fnn = −1 ta.i (0, 1) v`a (0, −1)) 26 X´ac di.nh gi´a tri l´o.n nhˆa´t v`a nho’ nhˆa´t cu’a h`am f = x3 − y − 3xy, D = {0 x 2, −1 y 2} http://tieulun.hopto.org 158 `eu biˆe´n Chu.o.ng Ph´ep t´ınh vi phˆan h`am nhiˆ (DS fln = 13 ta.i diˆe’m (2, −1); fnn = −1 ta.i diˆe’m (1, 1) v`a (0, −1)) http://tieulun.hopto.org ... (2 cos x2 − 4x2 sin x2 )dx2 b) Nˆe´u x l`a biˆe´n trung gian th`ı n´oi chung d2 x = v`a d´o ta c´o d2 f = d(2x cos x2dx) = (2x cos x2)d2 x + [d(2x cos x2)]dx = 2x cos x2 d2 x + (2 cos x2 − 4x2... (DS −2ex (cos x + sin x)) 66 y = x2 sin x (DS −2ex (cos x + sin x)) 67 y = x32x (DS 2x (x3ln3 + 9x2 ln2 x + 18xln2 + 6)) 68 y = x2 sin 2x 69 y = (f (x2 ) (DS −4(2x2 cos 2x + 6x sin 2x − cos 2x))... cos x2 dx + 2x(− sin x2 )2xdx dx = (2 cos x2 − 4x2 sin x2)dx2 Phu.o.ng ph´ ap II T´ınh da.o h`am cˆa´p hai fxx ta c´o fx = 2x cos x2 , fxx = cos x2 − 4x2 sin x2 v`a theo (8.6) ta thu du.o c d2