A. $f'\left(-\frac{x}{y}\right)$
B. $\frac{x}{y^2} f'\left(-\frac{x}{y}\right)$
C. $-\frac{1}{y} f'\left(-\frac{x}{y}\right)$
D. $-x f'\left(-\frac{x}{y}\right)$
设f(u)可导,z=xyf((y)/(x)),若x(partial z)/(partial x)+y(partial z)/(partial y)=xy(lny
设 =f(xy,(x)^2+(y)^2), 其中 f 可微,则 dfrac (partial z)(partial x)= __
设F(x, y, z)= 0定义z为x和y的隐函数,则(partial z)/(partial x)等于() 设$F(x, y, z)= 0$定义$z$为$x
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
143.设由方程F(x,y,z)=0所确定的函数关系中,已知(partial F)/(partial x)=ye^z-e^y,(partial F)/(part
若z=f(x,y),x=u+v,y=u-v,则(partial^2z)/(partial upartial v)=(partial^2z)/(partial x
[例3](2007,数一)设f(u,v)为二元可微函数, =f((x)^y,(y)^x), 则 dfrac (partial z)(partial x)= __
已知函数z=f(xy,sin x+sin y),f具有连续的偏导数,则(partial z)/(partial x)=()A. $xf_1' + \cos 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
sin (x+2y-3z)=x+2y-3z, 则 dfrac (partial z)(partial x)+dfrac (partial z)(partial