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

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}(

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

[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}+

求下列函数的 dfrac ({a)^2z}(a{x)^2} ,dfrac ({partial )^2z}(partial xpartial y) ,dfrac

[题目]-|||-设变换 ^2)+dfrac ({partial )^2z}(partial xpartial y)-dfrac ({partial )^2

dfrac ({sigma )^2z}(partial {y)^2} 和 dfrac ({partial )^2z}(partial xpartial y):-

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

33.设函数 z=z(x,y) 由方程 +x=(e)^z-y 所确定,则 dfrac ({partial )^2z}(partial ypartial x)=

设 z=f(u,v) 具有二阶连续偏导数, =xv =dfrac (x)(y), 以u,v为新变量变换方程-|||-^2dfrac ({partial )^2z

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