五、设 z=z(x,y) 是由方程 (z+dfrac (1)(x),z-dfrac (1)(y))=0 确定的隐函数,且具有连续的-|||-二阶偏导数。求证: ^2dfrac (partial z)(partial x)-(y)^2dfrac (partial z)(partial y)=1 和 ^3dfrac ({partial )^2z}(partial {x)^2}+xy(x-y)dfrac ({partial )^2z}(partial xpartial y)-(y)^3dfrac ({partial )^2z}(partial {y)^2}+2=0

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

设f具有二阶连续偏导数, =xf(x,dfrac (y)(x)), 求 dfrac ({partial )^2z}(partial xpartial y)

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

[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}+

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)=

5.设 sin (x+2y-3z)=x+2y-3z, 证明: dfrac (partial z)(partial x)+dfrac (partial z)(pa

17.设 =varphi (x+y,(x)^2), 且φ具有二阶连续偏导数,求 dfrac (partial x)(partial x) ,dfrac ({a)

sin (x+2y-3z)=x+2y-3z, 则 dfrac (partial z)(partial x)+dfrac (partial z)(partial

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}(

3设 +2y+z-2sqrt (xyz)=0, 求 dfrac (partial z)(partial x) 及 dfrac (partial z)(parti