Giải Tích 1 - Đạo Hàm và Vi Phân - Toán Học Ứng Dụng - ĐH BK TP HCM
Trư ng Đ i h c Bách khoa H Chí Minh B mơn Tốn ng d ng - Gi i tích Chương 2: Đ o hàm vi phân • Gi ng viên Ts Đ ng Văn Vinh (9/2008) dangvvinh@hcmut.edu.vn N i dung - – Đ o hàm – Vi phân – Đ nh lý giá tr trung bình – Công th c Taylor, Maclaurint I Đ o hàm Đ nh nghĩa (đ o hàm) Hàm s y = f(x) xác đ nh lân c n c a m x0 f ' ( x0 ) = lim ∆x → f ( x0 + ∆x) − f ( x0 ) ∆x f ' ( x0 ) đư c g i đ o hàm c a f t i m x0 Ví d Tìm đ o hàm c a hàm f ( x) = cos x t i m x0 f ( x0 + ∆x) − f ( x0 ) ∆x → ∆x cos( x0 + ∆x) − cos x0 = lim ∆x →0 ∆x ∆x ∆x sin x0 + ⋅ sin = − lim ∆x → ∆x = − sin( x0 ) f ' ( x0 ) = lim Ví d 1 x sin , x ≠ ' x Tìm f (0) , bi t f ( x) = 0, x=0 f (0 + ∆x) − f (0) f (0) = lim ∆x → ∆x ' = lim ∆x →0 ( ∆x ) sin (1/ ∆x ) − ∆x = = lim ∆x ⋅ sin ∆x → ∆x (b ch n x vô bé) Đ nh nghĩa (đ o hàm ph i) Hàm s y = f(x) xác đ nh lân c n c a m x0 f ( x0 + ∆x) − f ( x0 ) ' f + ( x0 ) = lim+ ∆x →0 ∆x f '+ ( x0 ) đư c g i đ o hàm ph i c a f t i m x0 Đ nh nghĩa (đ o hàm trái) Hàm s y = f(x) xác đ nh lân c n c a m x0 f ( x0 + ∆x) − f ( x0 ) f − ( x0 ) = lim− ∆x → ∆x ' f '− ( x0 ) đư c g i đ o hàm trái c a f t i m x0 Đ nh lý Hàm s y = f(x) có đ o hàm t i m x0 , ch có đ o hàm trái đ o hàm ph i t i m x0 hai đ o hàm b ng Đ nh nghĩa (đ o hàm vô cùng) N u f ( x0 + ∆x) − f ( x0 ) lim = ∞ , ta nói hàm ∆x → ∆x có đ o hàm vơ t i m x0 Ví d Tìm f +' (0); f −' (0) , bi t e1/ x , x ≠ f ( x) = 0, x = f (0 + ∆x) − f (0) e1/ ∆x − = +∞ f +' (0) = lim+ = lim+ ∆x → ∆x → ∆x ∆x f −' (0) f (0 + ∆x) − f (0) = lim− ∆x → ∆x e1/ ∆x − = lim− ∆x → ∆x =0 Đ o hàm trái đ o hàm ph i không b ng nhau, nên đ o hàm t i x = không t n t i Ví d Tìm f ' ( x) , bi t f ( x) = x − | x | +2 x − 3x + 2, x ≥ f ( x) = x + x + 2, x < x − 3, x > ⇒ f ( x) = 2 x + 3, x < ' ' ' T i m x = 0: f + (0) = −3; f − (0) = Đ o hàm trái đ o hàm ph i không b ng nhau, suy không t n t i đ o hàm t i x = Ví d Tìm f +' (0); f −' (0) , bi t f ( x) = sin x f (0 + ∆x) − f (0) sin 2∆x f (0) = lim+ = lim+ ∆x → ∆x ∆x → ∆x ' + f −' (0) = lim− ∆x → f (0 + ∆x) − f (0) ∆x = lim− ∆x → sin 2∆x ∆x =2 = −2 Đ o hàm trái đ o hàm ph i không b ng nhau, nên đ o hàm t i x = khơng t n t i 10 Ví d Tính gi i h n ( ) x sin x = x − + ο x 3! tan x − sin x x →0 x3 I = lim ( ) x3 tan x = x + + ο x 3 x3 x3 3 tan x − sin x = x + + ο ( x ) − x − + ο ( x ) 3! x3 tan x − sin x = + ο ( x3 ) x3 + ο ( x3 ) tan x − sin x ο ( x3 ) I = lim = lim = + lim = + x →0 x →0 x x x →0 x 73 I = lim Ví d Tính gi i h n ) x3 x →0 ln(1 + x ) = x + ο ( x ) ( ln + x3 − 2sin x + x cos x ( ) x3 sin x = x − + ο x 3! ( ) x2 cos x = − + ο x3 2! x4 x3 x + ο ( x ) − x − + ο x + x 1 − + ο x 3! 2! I = lim x →0 x3 ( 3 ) x3 + ο ( x3 ) = lim 3 x →0 x ( ) ( ) ο ( x3 ) = + lim = +0 x →0 x 74 + tan x − e x + x Ví d Tính gi i h n I = lim x →0 arcsin x − sin x ( ) x3 sin x = x − + ο x 3! x3 tan x = x + + ο x 3 ( ) ( ) x3 arcsin x = x + + ο x x x3 x e = + x + + + ο ( x3 ) 2! 3! x x3 x x3 3 1+ x − + + ο ( x ) − 1 + x + + + ο ( x ) + x 2! 3! I = lim x →0 x3 x3 x + +ο x − x − +ο x 3! x3 / + ο ( x3 ) I = lim = x →0 x / + ο ( x ) ( ) ( ) 75 ( ) x3 ln x + + x − x + Ví d Tính gi i h n I = lim x →0 x − x ( ) sinh x x3 x = = x − + ο x4 cosh x ( ( ln x + + x ( )) ' ⇒ ln x + + x I = lim x →0 ( ) x2 = = − + ο x3 + x2 ) ( ) x3 = x − + ο x4 x − x3 / + ο ( x ) − x + x3 / ( x − x − x3 / + ο ( x ) ) = 18 76 Ví d Tính g n A = cos ( 0.2 ) v i đ xác 10−7 Ph n dư khai tri n Maclaurint c a hàm y = cos x f (2 n + 2) (ξ ) n+ Rn ( x) = x ,0 < ξ < x ( 2n + )! 2n+2 cos ( x + (n + 1)π ) n + ≤ ⋅ ( 0.2 ) Rn ( x) = x ( n + )! ( n + )! Tìm n đ 1 −7 ⇒n=2 Rn ≤ ⋅ n+ < 10 = (2n + 2)! 10000000 x2 x4 (0.2) (0.2) cos x ≈ − + ⇒ cos(0.2) ≈ − + =0.9800666667 2! 4! 2! 4! 77 I Tìm đ o hàm c p n ( ln 1) (x − 1)2 x −1 3+ x 2) x ln 3− x ( n−1 ) 2 x −1 ((ln 2)( x − 1) + n) (n − 2)!((3n − x)(3 − x)− n + (−1) n (3n + x)(3 + x )− n ) 3) x ln x − x + 2 4) ( x + x)cos x 2 5) ( − x ) e 2−3 x ) (−1) n (n − 2)!(( x − n)( x − 1)− n + ( x − 2n)( x − 2)− n ) nπ 2n−3 ((4 x + x − n + n) cos x + ( nπ + 2n ( x + 1) sin x + ) (−3) n−2 36 x − 12(9 + 2n) x + 81 + 32n + 4n e2−3 x 78 I Tìm khai tri n Maclaurint đ n c p n x + 3e x 1) ,n = 2x e − 3x + − 3x 2) ln ,n = 3 + 2x ln(2 / 3) − ( ) 3) ln x + 3x + , n = 13 65 793 x − x2 − x + ο ( x3 ) 72 648 17 ln + x − x + x3 − x + ο ( x ) 8 64 4) (1- x)ln(1 + x) - (1 + x)ln(1- x), n = x−4 5) x − 5x + 5 x − x + ο ( x3 ) x + x3 + x5 + ο ( x5 ) 10 13 53 187 + x+ x + x + ο ( x3 ) 18 108 648 79 x2 + 5x − 6) ,n = x + x−2 7) x cosh x, n = x4 + 8) , n = x +1 ( 5 17 − x − x − x3 + ο ( x3 ) 16 27 x + x3 + x + ο ( x5 ) − x2 + x4 + ο ( x4 ) ) 9) ln x + x + , n = x − x3 + x5 + ο ( x5 ) 40 10) sinh x ⋅ cosh x, n = x+ 11) x ⋅ cosh x, n = 13 121 x + x + ο ( x5 ) 120 x + x3 + x5 + ο ( x5 ) 80 12) 13) ( ) x − x sinh x, n = x2 + + - x2 ,n = 14) ,n = x + x +1 16) ,n = 1− x + x − x 17) 1+ 1+ x 10 x + ο ( x4 ) 2 + x + x + ο ( x8 ) 128 8192 − x + x3 − x + x − x + x + ο ( x ) 1 11 + x + x − x3 − x + ο ( x ) 24 15) e x cos x , n = − + x2 8x2 + ,n = + x + x + x + ο ( x5 ) 1 − x + x − x6 + ο ( x6 ) 64 81 I Tìm khai tri n Taylor t i x0 đ n c p n 1) (x − 1)e x , x = −1, n = ( ) −e −2 ( x + 1) + ( x + 1) + ( x + 1) + ο (( x + 1)3 ) 2 3 2) ln ( x + 1) , x = 1/ 2, n = 1 1 1 1 1 ln + x − − x − + x − + ο x − 2 2 3 2 2 2x +1 3) ln x, x = 1, n = x -1 3+ x2 + 3x 4) , x = 1, n = x +1 5) e x + x −1 , x = −1, n = 1 ( x − 1) − ( x − 1)3 + ( x − 1)4 + ο ( x − 1)4 12 10 2+ 1 ( x − 1) − ( x − 1)2 + ( x − 1)3 + ο ( x − 1)3 18 4 e −2 1 + ( x + 1) + ( x + 1) + ο ( x + 1) 82 6) ln(2 + x − x ), x = 1, n = 2x 7) , x = 2, n = 1− x 8) 2x - x ln − ( x − 1) + ( x − 1)2 − ( x − 1)3 + ο (( x − 1)3 ) 24 10 28 82 3 − + ( x − 2) − ( x − 2) + ( x − 2) + ο ( x − 2) 27 81 , x = 1, n = 1+ ( x − 1)2 + ( x − 1)4 + ο ( x − 1)4 9) 10) x− x2 , x = 1/ 2, n = x−2 x2 − 4x + , x = 2, n = 1 1 − ⋅ ln ⋅ x − + ο x − 2 2 ( x − ) − ( x − )3 + ( x − )5 ο ( x − 1)5 83 I Tính gi i h n x2 cos x − + 1) lim x →0 x4 24 arctan x − arcsin x 2) lim x →0 tan x − sin x −1 + x cos x − + x 3) lim x →0 ln(1 + x) − x −1 4) (1 + x ) lim −1 earctan x + ln(1 − x) − x →0 5) lim x →0 x x2 − + x3 84 I Tính gi i h n esin x + ln(1 − x) − 6) lim x →0 arcsin x − sin x −1 xe tan x − sin x − x 7) lim x →0 x + x3 − tan x x 2e x − ln(1 + x ) − arcsin x3 8) lim x →0 x sin x − x −6 + x3 − cos x 9) lim x →0 tan x − x e x /(1− x ) − sinh x − cos x 10) lim x →0 1+ x + 1− x − −72 85 cosh x − (1 + x) −1/ − x 11) lim x →0 x / + ln(1 + tan x ) − arcsin x esin x − + x − arcsin x 12) lim x →0 sinh( x − x ) − ln + x sin arctan x − tan x 13) lim sinh x x →0 e − (1 + x)1/ − x 14) lim x →0 arcsin x − xe x x − x − tan x tan x − ln( x + + x ) 15) lim x →0 sin x − x cos x 28 86 e x − + x + x2 16) lim x →0 x + tan x − sin x ex − x 1+ x −1 17) lim x →0 sin x cosh x − sinh x e x + ln(1 − sin x) − 1 18) lim x →0 − x4 − sin + x3 − sin1 19) lim x →0 − x ln cos x − cos1 ecos x − e − x 20) lim x →0 (1/ x ) arcsin x − 2cosh x 5e 87 ... (? ?1) 99 99! 1 (0) = − 10 0 = 10 0 2i (i ) (−i ) (? ?1) 100 10 0! 1 10 0! ? ?1 (10 1) y (0) = − 10 1 = − = 10 0! 10 1 2i 2i i i (i ) (−i ) 36 II Vi phân Đ nh nghĩa (kh vi) ... x+a y (n) (n) = (? ?1) n! ( x + a ) n +1 n (? ?1) n n! 1 = ⋅ − n +1 4i ( x − 2i ) ( x + 2i ) n +1 ⇒ y (10 0) 10 0! (? ?1) 10 010 0! 1 = = ⋅ − 10 1 10 1 ⋅ 210 0 4i (−2i ) (2i ) 30... n ? ?1 π cos(2 x + n ) 31 Ví d Tính y (10 0) (1) , bi t y = (3x + 1) ln x f ( x) = x + 1; g ( x) = ln x y = f ⋅g ( fg ) (10 0) = C100 f (0) ⋅ g (10 0) + C100 f (1) ⋅ g (99) + C100 f (2) ⋅ g (98) + 10 0