设 =dfrac (y)(f({x)^2-(y)^2)} ,其中f(u)为可导函数,验证-|||-.dfrac (1)(x)dfrac (dz)(partial x)+dfrac (1)(y)dfrac (partial z)(partial y)=dfrac (z)({y)^2} .

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

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=dfrac (y)(f({x)^2-(y)^2)},其中f为可导函数.验证=dfrac (y)(f({x)^2-(y)^2)}.设,其中f为可导函数.验证.

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

1.设f,φ是C^(2)类类函数, =yf(dfrac (x)(y))+xvarphi (dfrac (y)(x)), 求:(1) a2/ey; (2) dfr

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

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

设(X,Y)的分布函数为(x,y)=dfrac (1)({pi )^2}(dfrac (pi )(2)+arctan dfrac (x)(2))(dfrac (

(B) dfrac (x)(y)((y+1))^2-|||-(C) ^2((x+dfrac {1)(x))}^2. (D) dfrac (y)(x)((y+1)

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