西安毛子科技大学高阶导数XIDIANUNIVERSITY例3设=xe2x，求(20)(uw)(n) =ZChu(n-k),(k)解 设 u=e2,=x2则u(k) = 2ké2x(k =1,2,...,20)v'= 2xv(k) =0(k=3,4, ..,20)v"=2代入莱布尼兹公式，得y(20) = C%ou(20)v +C2ou(19)v' +C2ou(18) ,"20.19(20) = 220 .e2* . x2 +20 . 219 .e2* .2x +e2x.22= 220 e2*(x? + 20x + 95)高阶导数 例3 求 代入莱布尼兹公式, 得 ( ) ( ) ( ) 0 ( ) n n k n k k n k uv C u v − = =  ( ) 2 2 ( 1,2, ,20) k k x u e k = = v x  = 2 v  = 2 ( ) 0 ( =3,4, ,20) k v k = (20) 0 20 (20) y = C u v 解 设 u e = 2x , v x = 2 则 (20) 20 2 2 2 x y = e  x 1 (1 20 9) +C v u  2 (1 20 8) +C u v  19 2 20 2 2 x +  e  x 18 2 2 20 9 2 2 1 x  e  +  2 2 , x y x e = (20) y . 20 2 2 2 ( 20 95). x = + + e x x 设