[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)= __

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

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 z = f(-(x)/(y))，且 f(x) 可导，则 (partial z)/(partial x) = ( )A. $f'\left(-\frac{x}

7.设 =dfrac (y)(f({x)^2-(y)^2)} 其中f为可导函数,验证: dfrac (1)(x)dfrac (partial z)(partia

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

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

设F(x, y, z)= 0定义z为x和y的隐函数，则(partial z)/(partial x)等于（） 设$F(x, y, z)= 0$定义$z$为$x

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

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