殳 =(u)^2ln v =dfrac (y)(x), =2x-3y,-|||-则 dfrac (partial z)(partial y)= ()-|||-A dfrac (y)({x)^2}[ 2ln (2x-3y)+dfrac (3y)(2x-3y)] -|||-B dfrac (y)({x)^2}[ 2ln (2x-3y)-dfrac (3y)(2x-3y)] -|||-C -dfrac (2{y)^2}({x)^2}[ dfrac (ln (2x-3y))(x)+dfrac (1)(2x-3y)] -|||-D) dfrac (2{y)^2}({x)^2}[ -dfrac (ln (2x-3y))(x)+dfrac (1)(2x-3y)]

sin (x+2y-3z)=x+2y-3z, 则 dfrac (partial z)(partial x)+dfrac (partial z)(partial

5.设 sin (x+2y-3z)=x+2y-3z, 证明: dfrac (partial z)(partial x)+dfrac (partial z)(pa

2.设 (xy,x+y)=(x)^2+(y)^2+xy (其中, =xy =x+y), 则 dfrac (partial f)(partial u)+dfrac

3设 +2y+z-2sqrt (xyz)=0, 求 dfrac (partial z)(partial x) 及 dfrac (partial z)(parti

dfrac ({sigma )^2z}(partial {y)^2} 和 dfrac ({partial )^2z}(partial xpartial y):-

3.设 +2y+z-2sqrt (xyz)=0, 求 dfrac (partial z)(partial x) 及 dfrac (partial z)(part

3.设 +2y+z-2sqrt (xyz)=0 ,求 dfrac (partial z)(partial x) 及 dfrac (partial z)(part

3.设 =dfrac (y)(f({x)^2-(y)^2)} ,其中f为可微函数,验证-|||-dfrac (1)(x)dfrac (partial z)(pa

1.已知 sin (3x-2y+z)=3x-2y+z, 则 dfrac ({partial )^2z}(partial xpartial y)= __

求下列函数的 dfrac ({a)^2z}(a{x)^2} ,dfrac ({partial )^2z}(partial xpartial y) ,dfrac