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}+dfrac ({partial )^2f}(partial {y)^2}=(e)^-((x^2+{y)^2)},-|||-试求 iint (xdfrac (partial f)(partial x)+ydfrac (partial f)(partial y))dxdy.

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

13.17 设函数 =f(ln sqrt ({x)^2+(y)^2}) 有二阶连续偏导数,满足 dfrac ({partial )^2u}(partial {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

34.设函数f(x,y)有连续二阶偏导数.满足 dfrac ({partial )^2f}(partial xpartial y)=0, 且在极坐标系下可表成

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}(

[题目]设函数f w)具有二阶连续导数, =f((e)^xcos y)-|||-满足 dfrac ({partial )^2z}(partial {x)^2}+

11.设 z=f(x,y) 二次可微,且 =(e)^ucos v =(e)^usin v, 试证:-|||-dfrac ({partial )^2z}(part

设=f((x)^3+(y)^2), 其中f具有二阶连续偏导数,则 dfrac ({partial )^2z}(partial {y)^2}=