=f(x,xy), 则 dfrac (partial u)(partial y)=

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

设 =f(xy,(x)^2+(y)^2), 其中 f 可微,则 dfrac (partial z)(partial x)= __

设f(u)可导，z=xyf((y)/(x))，若x(partial z)/(partial x)+y(partial z)/(partial y)=xy(lny

设 =f(x+y,xy), f具有一阶连续偏导数,求 dfrac (partial z)(partial x), dfrac (partial z)(parti

设 z = (u)/(x+y)，而 u = e^xy，则 (partial z)/(partial y) = ()A. $\frac{xe^{xy}}{x+y}

设 z = e^u sin v，而 u = xy，v = x + y，则 (partial z)/(partial x) = ( )A. $z = e^{xy}

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

殳 =(u)^2ln v =dfrac (y)(x), =2x-3y,-|||-则 dfrac (partial z)(partial y)= ()-|||-A

[例3](2007,数一)设f(u,v)为二元可微函数, =f((x)^y,(y)^x), 则 dfrac (partial z)(partial x)= __

若z=f(x,y),x=u+v,y=u-v,则(partial^2z)/(partial upartial v)=(partial^2z)/(partial x