sln一 例 14 y=e x, 求 sin sin. 解①=( e x=e x ( sin -=e x COS D x SIn 2e x cOS结束 解 例 14 y e x1 sin =  求dx dy  解 ) 1 ( 1 ) cos 1 ( ) (sin 1 sin 1 sin 1 sin =  =   =    x x e x e e dx dy x x x 解 ) 1( 1 ) cos 1 ( ) (sin 1 sin 1 sin 1 sin =  =   =    x x e x e e dx dy x x x 解 ) 1( 1 ) cos 1 ( ) (sin 1 sin 1 sin 1 sin =  =   =    x x e x e e dx dy x x x x e x x 1 cos 1 1 sin 2 = −   