NAN DA XUE JING PIN KE CHENG 例3.设y=x",n为正整数,求y)和ym+) 解 n(n-1)xy2,y3)=n(n-1)(n-2) -1)3·21xn=nl! 而y+=(m)=0 易见,若f(x),g(x)均存在n阶导数,则 ((x)+g(x))(n=f(n(x)+g n(x) 类似,设f(x)=ax+a1xn1a2x-2+.+an1xn+an, 为n次多项式,则f(n(x)=aol,而(+x)=0 OD 高等數粤
例3. 解: y' = nxn–1 , , , . ( ) ( +1) = n n n 设y x n为正整数 求y 和y y'' = n(n–1)x n–2 , y (3)= n(n–1)(n–2) x n–3 , …, y (n)= n(n–1)… 3 ·2 ·1x n–n = n! 而 y (n+1)= (n!)' = 0 易见, 若f (x), g(x)均存在n阶导数, 则 ( ( ) ( )) ( ) ( ) ( ) ( ) ( ) f x g x f x g x n n n = 类似, 设f (x)=a0x n +a1x n–1 +a2x n–2 +…+an–1 x n +an , 为n次多项式, 则f (n) (x)=a0n!, 而f (n+1)(x)= 0
NAN DA XUE JING PIN KE CHENG 例4.设(1)y=e,(2)y=a、(a>0,a≠1),求ym) 解:(1) 故 特别,取=1,得(e n(n)= ox 取=-1,得(e))=(-1)y) (2)由于ax=emn,由(1)得 e na (na) 高等歐學
例4. (1) , (2) ,( 0, 1), . x x (n) 设 y = e y = a a a 求y 解: (1) y' = ex , y'' = ex 2 , y (3)=ex 3 ,…, 故 y (n) = ex n . 特别, 取 = 1, 得(e x ) (n) = e x (a x ) (n) = ( e xlna ) (n) 取 = –1, 得(e –x ) (n) =(–1)(n) e x . (2) 由于a x = e xlna , 由(1)得 = a x (lna) n = e xlna (lna) n
NAN DA XUE JING PIN KE CHENG 例5.求y sinx的n 阶导数y 解:我们知道y=cosx,y sinr. 13 -cOSX (4)= sinx, 但的通项公式难写,并且不好记 由于c0sx=s(x+).从而 y=(sinx)=cosx=sin(x+o) sin (x Z)=cos(x+7)=sin(x+2 OD 高等數粤
例5. 求y = sinx的n阶导数y (n) . 解: 我们知道 y' =cosx, y'' = –sinx, y (3)= –cosx, y (4)= sinx,… 但y (n)的通项公式难写, 并且不好记. ). 2 cos sin( 由于 x = x + 从而 y = (sinx) ). 2 sin( =cosx = x + = + ) 2 sin( y x ) 2 cos( = x + ). 2 sin( 2 = x +
NAN DA XUE JING PIN KE CHENG sin (x +2 cos(x+2 sin(x+3. 故y=(sinx)=sin(x+n7) 类似,(c0sx)=c0(x+n) 高等歐學
= + ) 2 sin( 2 (3) y x ) 2 cos( 2 = x + ). 2 sin( 3 = x + ( ) ( ) (sin ) n n 故 y = x ). 2 sin( = x + n ). 2 , (cos ) cos( ( ) x = x + n 类似 n
NAN DA XUE JING PIN KE CHENG 例6.设y=sin2x,求ym) 解:y=(sin2x)′=2 esInxcox=sin2x y=(sin2x)'=sin( 2x+). 2 (3) Sin(2x+2·)·2 丌 =sn2x+(m-1) OD 高等數粤
例6. 设y = sin2x, 求 y (n) . 解: y' = (sin2x)' y'' =(sin2x)' = 2sinxcox = sin2x. ) 2 2 = sin(2 + x ) 2 , 2 sin(2 2 (3) 2 = + y x …… 2 . 2 sin 2 ( 1) ( ) −1 = + − n n y x n