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

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

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

九、设f(x,y)在 ^2+(y)^2leqslant 1 上二次连续可微,且满足 dfrac ({partial )^2f}(partial {x)^2}+d

11.设 =f((x)^2+(y)^2) ,其中f具有二阶导数,求 dfrac ({a)^2z}(d{x)^2} -dfrac ({partial )^2z}(

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

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

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

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

2.设z=f(xy,(y)/(x))，其中f具有二阶连续偏导数，则(partial^2z)/(partial xpartial y)=（）.A. $f_{1}^