A. $xf_1' + \cos x \cdot f_2'$
B. $yf_1' + \cos x \cdot f_2'$
C. $xf_1' + \cos y \cdot f_2'$
D. $yf_1' + \cos y \cdot f_2'$
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
(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
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}^
设 z = e^u sin v,而 u = xy,v = x + y,则 (partial z)/(partial x) = ( )A. $z = e^{xy}
设f(u)可导,z=xyf((y)/(x)),若x(partial z)/(partial x)+y(partial z)/(partial y)=xy(lny
sin (x+2y-3z)=x+2y-3z, 则 dfrac (partial z)(partial x)+dfrac (partial z)(partial
设 z = f(-(x)/(y)),且 f(x) 可导,则 (partial z)/(partial x) = ( )A. $f'\left(-\frac{x}
设z=sin(uv),u=x+y,v=x-y,则(partial z)/(partial y)=【】 设$z=\sin(uv)$,$u=x+y$,$v=x-y
3.设 =sin (xy)+varphi (x,dfrac (x)(y)), 其中φ(u,v)具有二阶连续偏导数,求 dfrac ({partial )^2z}
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