4.[单选题]-|||-设 =f(x,y,z), 其中f具有一阶连续偏导数,而-|||-=(x)^2-2(y)^2, 则 dfrac (partial u)(partial x)= __-|||-_-|||-A dfrac (ef)(ex)-|||-B)-|||-dfrac (partial f)(partial x)+dfrac (partial f)(partial z)-|||-C dfrac (partial f)(partial x)+2xdfrac (partial f)(partial z)-|||-D dfrac (partial f)(partial x)+dfrac (partial f)(partial y)+2xdfrac (partial f)(partial z)

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

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}=

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

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}^

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

(3)已知函数 =f(xy,(e)^x+y) ,且f(x,y)具有二阶连续偏导数.则-|||-dfrac ({partial )^2z}(partial xpa

[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}+

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

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)

设 r = sqrt(x^2 + y^2), u = f(r), 其中 f 具有二阶连续导函数, 则 (partial^2 u)/(partial x^2) +