例3 设 y=x~ (αER),求y(n)解 '=αxα-1j" =(αxa-")'= α(α-1)xα-2y" = (α(α- 1)xα-2) = α(α-1)(α- 2)xα-3y(n) =α(α-1)..(α-n+ 1)xα-n(n≥1)若α为自然数n,则y(") =(x")(") = n!,j(n+1) = (n!)' = 0.>注意：求n阶导数时.求出1-3或4阶后.不要急于合并.分析结果的规律性,写出n阶导数.（数学归纳法证明)
例3 ( ), . (n) 设 y = x R 求y 解 −1 y = x ( ) 1 = − y x 2 ( 1) − = − x 3 ( 1)( 2) − ( ( 1) ) = − − x 2 = − − y x ( 1) ( 1) ( 1) ( ) = − − + − y n x n n n 若 为自然数n,则 ( ) ( ) ( ) n n n y = x = n!, ( !) ( 1) = + y n n = 0. ➢ 注意:求n阶导数时,求出1-3或4阶后,不要急于 合并,分析结果的规律性,写出n阶导数.(数学归 纳法证明)
例4 设 y=In(1+x),求y(n)2!1解LnV(1 + x)31+x(1 + x)3!D(4)(1+x)(n -1)!=(-1)"-1(h(n≥1, 0!=1)(1 +x)"思考: y=ln(1-x), y(m) =-(n-1!(1-x)
例4 ln(1 ), . (n) 设 y = + x 求y 解 x y + = 1 1 2 (1 ) 1 x y + = − 3 (1 ) 2! x y + = 4 (4) (1 ) 3! x y + = − ( 1, 0! 1) (1 ) ( 1)! ( 1) ( ) 1 = + − = − − n x n y n n n 思考: y x = − ln(1 ) , ( ) ( 1)! (1 ) n n n y x − = − −
例5 设 y=sin x,求y(n)π-2解 j'=cosx = sin(x+元-2元元-2Ty" = cos(x += sin(x += sin(x + 2 .十-一P22元一=i(+.y" = cos(x + 2S2元j(") = sin(x + n2元(cosx)(n) = cos(x + n 同理可得-2
例5 sin , . (n) 设 y = x 求y 解 y = cos x ) 2 sin( = x + ) 2 cos( y = x + ) 2 2 sin( + = x + ) 2 sin( 2 = x + ) 2 cos( 2 y = x + ) 2 sin( 3 = x + ) 2 sin( ( ) y = x + n n ) 2 (cos ) cos( ( ) x = x + n 同理可得 n
常用高阶导数公式(e*)(n) =ex(1)(a*)(") =a*.In" a (a>0)元-2元-2(2) (sin kx)(") = k" sin(kx + n(3) (cos kx)(n) = k" cos(kx + n(4) (x~)(") =α(α-1)...(α- n+1)xα-nn!(5) (ln x)() =(-1)"-1 (n - 1)!()(") =(-1)"t+1hx
常用高阶导数公式 n n x n x − (4) ( ) = ( − 1) ( − + 1) ( ) n n n x n x ( 1)! (5) (ln ) ( 1) ( ) 1 − = − − ) 2 (2) (sin ) sin( ( ) kx = k kx + n n n ) 2 (3) (cos ) cos( ( ) kx = k kx + n n n (1) ( ) ln ( 0) ( ) a = a a a x n x n x n x e = e ( ) ( ) 1 ( ) ! ) ( 1) 1 ( + = − n n n x n x