例6设y=x2e2,求y12 庄解设n=e2,”=x,则由莱布尼兹公式知 y2)-=(e2)20)·x2+20(e2)y.(x2 20(20-1)2x e (x2)”+0 2r 20g2x·x2+20.2l2x2x 20·19 十 218g2x.2 2! =2e(x2+20x+95) 上页例 6 , . 2 2 (20) y x e y 设 = x 求 解 设u = e 2 x , v = x2 ,则由莱布尼兹公式知 ( ) ( ) 0 2! 20(20 1) ( ) 20( ) ( ) 2 (18) 2 (20) 2 (20) 2 2 (19) 2   + − + =  +   e x y e x e x x x x 2 2 2! 20 19 2 20 2 2 18 2 20 2 2 19 2  + =  +   x x x e e x e x 2 ( 20 95) 20 2 2 = e x + x + x