6 设函数z=f(x,y)在点(1,1)处可微，且 f(1,1)=1, of =2， =3, 0x1,1) ay1,1) 小-.求品0，e考翻 解：由题设p(1)=f(1,f(1,1)=f1,1)=1 po以k-1o28-l =3[f(x,f(x,x) +fx,fx,x)x,9+f5,)川x=l =3.[2+3·(2+3)]=516 (1) = f (1, f (1,1)) 1 ( ) d d 3 x = x x  = f (1,1) =1 d 1 d 3 ( ) 2 = = x x x   = 3 ( , ( , )) 1 f  x f x x ( , ( , ))( ) 2 + f  x f x x  x =1 = 3 = 51 f (1,1) =1, (x) = f (x, f ( x, x)), 2, (1,1) =   x f 求 设函数 在点 处可微 , 且 3, (1,1) =   y f 解: 由题设 2 + 3(2 + 3) (2001考研)