1、3.3高階導(dǎo)數(shù)問(wèn)題:變速直線運(yùn)動(dòng)的加速度.設(shè)s = s(t),則瞬時(shí)速度為v(t) = s(t)Q加速度a是速度v對(duì)時(shí)間t的變化率a(t) = v(t) = s(t).定義設(shè)f(x)可導(dǎo),則稱(f (x)為函數(shù)f (x)在點(diǎn)x處的二階導(dǎo)數(shù).d 2 yd 2f ( x)記作f( x), y,dx 2.或dx 2d 3yf( x),y,.二階導(dǎo)數(shù)的導(dǎo)數(shù)稱為三階導(dǎo)數(shù),dx 3d 4y三階導(dǎo)數(shù)的導(dǎo)數(shù)稱為四階導(dǎo)數(shù),f ( 4 ) ( x),y( 4 ) ,.dx4一般地, 函數(shù)f ( x)的n - 1階導(dǎo)數(shù)的導(dǎo)數(shù)稱為函數(shù)f ( x)的n階導(dǎo)數(shù), 記作d nyd nf ( x)f ( n ) ( x),y

2、( n ) ,.或dx ndxn注意: yn表示y的n次方, y( n)表示y的n階導(dǎo)數(shù)二階和二階以上的導(dǎo)數(shù)統(tǒng)稱為高階導(dǎo)數(shù).相應(yīng)地, f ( x)稱為零階導(dǎo)數(shù); f ( x)稱為一階導(dǎo)數(shù).高階導(dǎo)數(shù)求法舉例1.直接法:由高階導(dǎo)數(shù)的定義逐步求高階導(dǎo)數(shù).例1 設(shè)y = arctanx, 求f (0),f(0).注意:求n階導(dǎo)數(shù)時(shí),求出1-3或4階后,不要急于合并,分析結(jié)果的規(guī)律性,寫出n階導(dǎo)數(shù).(數(shù)學(xué)歸納法證明)例2 求下列函數(shù)的n階導(dǎo)數(shù)(1)設(shè) y = xa (a R), 求y(n) .y = ax a-1y = (ax a-1 )= a(a - 1) x a- 2y = (a(a - 1) x

3、a- 2 ) = a(a - 1)(a - 2) x a- 3LL解= a(a - 1)L(a - n + 1)x a- n(n 1)y( n)= a(a - 1)L(a - n + 1)x a- n(n 1)y( n)若a 為自然數(shù)n,則= ( xn )( n ) = n!,= (n!)y( n+1)= 0.y( n )k(k - 1)(k - 2)L(k - n + 1)xn-kn!k n= y(k ) = (xn )(k )0若a 為- 1,則(-1)n n!(-1)n n!xn+1 11(n)=(x + 1)n+1( n)= ()x( n)=()x + 1y(2) : 設(shè)y = ln(

4、1 + x), 求y(n) .1y =解1 + x11 + xy(n) = (y)(n-1)= ()(n-1)(-1)n n!1(n)=(x + 1)n+1()x + 1Q(n - 1)!= (-1)n-1(n 1,0!= 1)y( n )(1 + x)n(3)設(shè)y = sinx, 求y(n) .= sin( x + p)y = cos x解2= sin( x + 2 p)y = -sinxy = -cos x2= sin( x + 3 p)2= sin(x + 4 p)y(4)= sinx2LL= sin( x + n p)y( n)2(cos x)( n) = cos( x + n p)同

5、理可得2(4)y = ax(ax )(n)= ax(lna)n特別地: (ex )(n) = ex由復(fù)合函數(shù)求導(dǎo)法則有:(1)(eax+ b )( n)= aneax+b= an sin(ax + b + n p )(2) (sin( ax + b)( n)2= an cos(ax + b + n p )(3) (cos(ax + b)( n)2n!(ax + b)n+1 1(4)()(n)= (-1)nanax + b2.間接法:利用已知函數(shù)的高階導(dǎo)數(shù)公式來(lái)求未知函數(shù)的高階導(dǎo)數(shù).例3：求函數(shù)的n階導(dǎo)數(shù)1(1) y =- 5 x + 6x2(2) y = sin2x3. 高階導(dǎo)數(shù)的運(yùn)算法則:設(shè)

6、函數(shù)u和v具有n階導(dǎo)數(shù), 則(1) (Cu)(n)= Cu(n)(2) (au bv)(n)= au(n) bv(n)( n-1)( n-2)(3) (u v)( n)=v + Cu+ Cu0( n)12Cuvvnnn+L+ Ck u( n-k )v( k )+L+ Cnuv( n)nnnk =0( n-k )=k( k )Cuvn-萊布尼茲公式例4：求y = x 3eax的n階導(dǎo)數(shù)。a 0n( n-k )=( x3eax )( n)k3( k )axC( x)(e)nk =0= C 0 x3 (eax )( n) + C 1 ( x3 )(eax )( n-1)nn+ C 2 ( x3 )(eax )( n-2) + C 3 ( x3 )(eax )( n-3)n= x3aneax + 3nx2an-1eaxn+ 3n(n - 1)xan-2eax + n(n - 1)(n - 2)an-3eax練習(xí)：1.求下列函數(shù)的n階導(dǎo)數(shù)1 - x(1).y = ln1 + x(