《高等数学习题课教程》

《高等数学》(北大第二版)第02章习题课《高等数学》(北大第二版)课件《高等数学》(北大第二版)课件一、学习本章的主要要求是：学习本章的主要要求是：(1)掌握导数、微分(及高阶导数)的定义，它们的联系与区别及几何意义，会用定义求导数、微分及高阶导数. (2) 熟练地掌握计算导数与导函数、微分及高阶导数的各种方法，并善于运用相应公式、法则和方法熟练地进行计算；(3)会用微分进行近似计算并估计误差. 二、综合例题f ( x) 处连续，存在，证明f(x)在x=0处可导处可导. 例1 设f(x)在x=0处连续，且lim 在处连续存在，证明在处可导x →0f ( x ) f (0) lim x →0 x 0x 存在，故只要证f(0)=0. 分析需证证设lim f ( x) = A, 则lim f ( x) = lim x f ( x) = 0 A = 0, x →0 x →0 x →0 x x 因为f(x)在x=0处连续，所以f (0) = lim f ( x) = 0. x→0 f ( x ) f ( 0) f ( x) f ′(0) = lim = lim = A 存在，即f(x)在x=0处可导. 故x →0 x→0 x 0 x《高等数学》(北大第二版)课件例2 设f(u)的一阶导数存在，求1 r r lim [ f (t + ) f (t )] r →0 r a r a r f (t + ) f (t ) + f (t ) f (t ) a a 解原式= lim r →0 r r r [ f (t + ) f (t )] [ f (t ) f (t )] 1 1 a a 令r =h = lim + lim r r r r a →0 a →0 a a a a a1 f (t + h) f (t ) 1 f (t ) f (t h) = lim + lim h →0 a h a h →0 h 1 f (t + h) f (t ) 1 f (t h) f (t ) = lim + lim h →0 a h a h →0 hh = x1 12 = f ′(t ) + f ′(t ) = f ′(t ) a a a《高等数学》(北大第二版)课件例3 已知y = xln(x + 1 + x 2 ) 1 + x 2解′( ′ y′ = xln(x + 1 + x 2 )) 1 + x 2) (求y′.x 1+ x2 = ln(1 + 1 + x ) + x. x + 1+ x2 1+ x221+x= ln( 1 + 1 + x ) +2x 1+ x2x 1+ x2= ln( 1 + 1 + x 2 )例4 求y = 解x x x 的导数 .y= x1 1 1 + +2 4 8= x , 所以27 87 8 7 ′= x = y . 8 8 8 x1练习: y = ln1 1+ x, 求y ′.《高等数学》(北大第二版)课件例5设y =a1 x 3x log b14arctan x 2 ( a 0 , b 0 ), 求y ′.1 1 1 x ∵ ln y = ln a + ln log b x + ln arctan x2 , 解2 6 24 1 1 1 ln y = ln a + (ln ln x ln ln b ) + ln arctan x 2 , 2x 6 24 对上式两边求导,得ln a 1 x ′ = y[ y + + ] 2 4 2 2x 6 x ln x 12 (1 + x ) arctan x1 = 2a1 x 3x log b4arctan x 2x 1 ln a [ 2 + ]. 4 2 x 3 x ln x 6 (1 + x ) arctan x《高等数学》(北大第二版)课件例6 设y = y ( x) 由方程e xy + tg ( xy ) = y 确定，求y′(0)解由方程知当x = 0 时y = 1.对方程两变求导：1 e ( y + xy ′) + ( y + xy ′) = y ′2 cos ( xy ) 1 0 1 e (1 + 0 y′(0)) + (1 + 0 y ′(0)) = y ′(0) 2 cos (0)xy故y ′(0) = 2例7 已知xy = e x + y 求y′′解将方程两边对x求导，得y + xy′ = e x + y (1 + y′)(A)y + xy′ = e x + y + y′e x + y再将(B)两边对x求导,得(B)y - ex+y y′ = x + y e x(C)y′ + y′ + xy′′ = e x + y (1 + y′) + y′′e x + y + y′e x + y (1 + y′)《高等数学》(北大第二版)课件e x + y (1 + y′) 2 2 y′ y′′ = x e x+ yy - ex+y 其中y′ = x + y e x.x = ln(1 + t2 ), 例7 已知求y′, y′′, y′′′. y = t arctan t. 1 1 (t - arctant)′ 1+ t2 = t , 解y′ = = 2 2t 2 (ln(1 + t )′ 1+ t2 t ( )′ 1+ t 2 2 y′′ = = , 2 ′ (ln(1 + t )) 4t1+ t 2 ( )′ t 4 1 4t y′′′ = = 3 . (ln(1 + t 2 ))′ 8t《高等数学》(北大第二版)课件例8 设y = f 2 ( x) + f ( x 2 ), 其中f ( x)具有二阶导数, 求y′′. 解y′ = 2 f ( x) f ′( x) + f ′( x 2 )2 x. y′′ = 2[ f ′( x)]2 + 2 f ( x) f ′′( x) + 2 f ′( x 2 ) + 2 xf ′′( x 2 ) 2 x = 2[ f ′( x)]2 + 2 f ( x) f ′′( x) + 2 f ′( x 2 ) + 4 x 2 f ′′( x 2 ).例9 求下列函数的n 阶导数y ( n ) ( n 3). x4 1 (1) y = ; (2) y = 2 . 2 1 x x ax4 1+1 1 y= = ( x 3 + x 2 + x + 1) 1 x 1 x n! (n) . 当n 3时, y = n +1 (1 x) 1 ( 2) y = 2 (练习). 2 x a解(1)《高等数学》(北大第二版)课件例10 求由方程先求微分，易得导数] 解[先求微分，易得导数将方程两边同时取微分，因为y ln x + y = arctan 所确定的隐函数的导数和微分. x2 22 2d ln x + y ==1 x +y2 2d x + y =2 21 x +y2 2d (x2 + y2 ) 2 x2 + y21 x2 + y22 xdx + 2 ydy 2 x2 + y2=而xdx + ydy , 2 2 x +yy 1 xdy ydx xdy ydx d arctan = = 2 x 1 + ( y )2 x2 x + y2 x∴xdx + ydy xdy ydx = 2 2 2 x +y x + y2∴x+ y dy = dx, x y∴dy x + y y′ = = . dx x y《高等数学》(北大第二版)课件a xb a x b 例11 设f(x) 可导, 求y = f (sin x ) + ( ) ( ) ( ) .的导数, b x a a 其中, a 0, b 0, ≠ 1, x ≠ 0. b a x b a x b 2 解记y1 = f (sin x ) , y2 = ( ) ( ) ( ) , b x a ′ 则y1 = f ′(sin 2 x ) 2 sin x cos x = sin 2 x f (sin 2 x ).2ln y 2 = x (ln a ln b ) + a (ln b ln x ) + b (ln x ln a ),a xb a x b a b a a b ′ ). ∴ y 2 = y 2 [(ln a ln b ) + ] = ( ) ( ) ( ) (ln + b x a b x x x 例12 设y = (ln x ) x x ln x , 求y ′. ln y = x ln(ln x ) + (ln x ) 2 , 解两边取对数, 两边关于x求导1 y ′ = ln(ln x ) + 1 + 2 ln x , y ln x x 1 2 ln x x ln x y ′ = (ln x ) x [ln(ln x ) + ∴ + ]. ln x x练习：设( cosx) y = (sin y ) x求y′《高等数学》(北大第二版)课件例13 解dy 已知y = a + x , a 0为常数, (a ≠ 1), 求 . dx arctan x 2 sin x 设y1 = a , y2 = x .arctan x 2 sin x)′ = ln a a (arctan x 2 )′ 1 arctan x 2 2 ′ = ln a a arctan x 2 2 x . = ln a a (x ) 4 1+ x 1+ x4 对y2 = x sin x两边取对数，得ln y2 = sin x ln x 1 sin x ′ y2 = cos x ln x + , 两边对x求导，得x y2 sin x sin x ′ y2 = x (cos x ln x + ). xarctan x 2arctan x 2′ y1 = (a《高等数学》(北大第二版)课件2 - x, 1 x +∞, 2 例13 设f(x) = x , 0 ≤ x ≤ 1, x 3 , - ∞ x 0. 解第一步，在各开区间内分别求导：1, 1 x f ′( x) = 2x, 0 x 1, 3x 2 , - ∞ x 0.求f ′ (x).第二步，在分段点用导数定义求导，分段点为x = 0,1f (0 + x) f (0) ( x) 2 0 f +′ (0) = lim+ = lim+ =0 x →0 x →0 x x 《高等数学》(北大第二版)课件f (0 + x)f (0) ( x)3 0 f ′ (0) = lim = lim = 0, ∴ f ′(0) = 0 x →0 x →0 x xf (1 + x) f (1) 2 (1 + x) 12 x = lim+ = lim+ = 1 f +′ (1) = lim+ x → 0 x → 0 x → 0 x x xf (1 + x) f (1) (1 + x) 2 12 2 x + ( x) 2 = lim = lim =3 f ′ (1) = lim x → 0 x →0 x → 0 x x x∴ f(x)在x = 1的导数不存在1, 1 x +∞, 故f ( x) = 2x, 0 ≤ x 1, 3x 2 , - ∞ x 0.在x = 1 处f(x)不可导.x ≤ c, sinx, 例14 设f(x) = c 为常数ax + b , x c.试确定a, b的值，使f ′(c) 存在.《高等数学》(北大第二版)课件解因为f ′ (c) 存在，所以f(x) 在c处连续.x →clim- f ( x) = lim- sin x = sin cx →c x →cx →clim+ f ( x) = lim+ (ax + b) = ac + bf ′ (c) = lim∴ sinc = ac + b (1)因为f(x) 在c处可导，sin x sin c f ( x ) f (c ) = lim x →c x →c x c x c x c x c x+c sin 2 sin cos 2 cos x + c = cos c. 2 2 = lim = lim x →c x c x →c 2 x c 2 f ( x ) f (c ) ax + b sin c ax + b (ac + b) = a.f +′ (c) = lim = lim = lim + + + x →c x →c x →c x c x c x c所以，cosc = a (2) 解(1), ( 2) 得，= cosc , b = sinc - ccosc. a《高等数学》(北大第二版)课件x2, x ≤ 1, 习题2-1 15. 设f(x) = ax + b , x 1. 为了使函数f(x)在x=1处连续且可导，a,b应取什么值？解要使f(x)在x=1处连续，因为x →1lim f ( x) = lim x 2 = 1, x →1x →1x →1lim (ax + b) = a + b, +应有lim f ( x) = lim f ( x) = f (1) +x →1即a+b=1(1)要使f(x)在x=1处可导，因为(1 + x) 2 12 2 x + ( x) 2 f (1 +x) f (1) = lim = 2, f ′ (1) =lim = lim x →1 x →1 x →1 x x x代a + b =1a (1 + x) +b 12 f (1 + x) f (1) a x f +′ (1) = lim = lim = lim = a, + + + x →1 x →1 x →1 x x x应有a=2,代入(1)式得b=-1.《高等数学》(北大第二版)课件6. 假定f ′( x0 )存在，指出下式A表示什么？f ( x) = A, 其中f (0) = 0, 且f ′(0)存在；x →0 x f ( x0 + h) f ( x0 h) (3) lim = A. h→0 h 解(2) ∵ lim f ( x) = lim f ( x) f (0) = f ( x0 ), x →0 x →0 x 0 x (2) lim∴ A = f ( x0 ).(3) ∵ limh →0f ( x0 + h) f ( x0 ) + f ( x0 ) f ( x0 h) f ( x0 + h) f ( x0 h) = lim h →0 h h f ( x0 + h) f ( x0 ) f ( x0 ) f ( x0 h) + lim h →0 h h = limh →0f ( x0 h) f ( x0 ) 令h = x = f ′( x0 ) + lim ======== f ′( x0 ) + f ′( x0 ) = 2 f ′( x0 ), h →0 h∴ A = 2 f ′( x0 ).《高等数学》(北大第二版)课件9 .如果f ( x)为偶函数，且f ′(0)存在，证明f ′(0) = 0.证f ( x) f ( x0 ) f ( x) f (0) f ( x) f (0) ′( x0 ) = lim (f ) f ′(0) = lim = lim x → x0 x →0 x →0 x x0 x 0 x 0f ( x) f (0) (令x = y ) f ( y ) f (0) = f ′(0) = lim ========== lim x →0 x 0 y →0 y 0∴ 2 f ′(0) = 0,f′(0) = 0.1 例16 设f (t ) = lim t (1 + ) 2tx ,求f ′(t ). x →∞ x 1 x 2t 1 2tx 解lim t (1 + ) = lim t[(1 + ) ] = t e 2t x →∞ x →∞ x xf ′(t )= (t e 2t )′ = (2t + 1)e 2t .《高等数学》(北大第二版)课件1 2 x sin , x ≠ 0; 例15 求f(x) = x 0, x=0一阶导数和二阶导数.1 1 解当x ≠ 0时, f ′( x) =2 x sin cos , x x 1 2 1 1 1 f ′′( x ) = 2 sin cos 2 sin . x x x x x当x=0时,用导数定义先求一阶导数,再来看二阶导数.f (0 + x) f (0) = lim f ( x ) f ′(0) = lim x → 0 x → 0 x x= lim由于x 2 sinx → 01 x = lim x sin 1 = 0; x → 0 x x1 lim f ′( x) = lim(2 x sin 1 cos 1 ) = lim cos x →0x →0不存在(极限故处不连续(是振荡间断点是振荡间断点),所以不可导,即不存在极限),故f ′(x ) 在x=0 处不连续是振荡间断点所以f ′(x ) 在x=0不可导即极限不可导 f ′′(0) 不存在不存在.xxx→0x《高等数学》(北大第二版)课件1 g(x)cos , x ≠ 0, 例16 设f(x) = x 0, x = 0.且g(0) = g′(0) = 0 试问：(1) lim f ( x);x →0(2) f(x) 在x = 0处是否连续？(3) f(x) 在x = 0处是否可导？若可导，f ′(0) = ?解(1 lim f ( x) = lim g ( x) cos ) 1 =0 x →0 x →0 x 1 ( ∵ lim g(x) = g(0) = 0; cos 为有界函数) x →0 __ →0(2) ∵ lim f ( x) = 0 = f (0)∵ f(x)在x = 0 处连续.1 1 g ( x ) cos 0 g ( x) cos x x =0 lim (3) f ′(0) = lim x →0 x →0 x 0 x1 g ( x ) g ( 0) g ( x) ( ∵ g′ (0) = lim = lim = 0, cos 有界) x →0 x →0 x 0 x x。