A. $\frac{|x|}{x^2 + y^2}$
B. $\frac{-|y|}{x^2 + y^2}$
C. $\frac{|y|}{x^2 + y^2}$
D. $\frac{-|x|}{x^2 + y^2}$
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) +
2.设 (xy,x+y)=(x)^2+(y)^2+xy (其中, =xy =x+y), 则 dfrac (partial f)(partial u)+dfrac
若z=f(x,y),x=u+v,y=u-v,则(partial^2z)/(partial upartial v)=(partial^2z)/(partial x
设 z = (u)/(x+y),而 u = e^xy,则 (partial z)/(partial y) = ()A. $\frac{xe^{xy}}{x+y}
设 z = e^u sin v,而 u = xy,v = x + y,则 (partial z)/(partial x) = ( )A. $z = e^{xy}
设z=sin(uv),u=x+y,v=x-y,则(partial z)/(partial y)=【】 设$z=\sin(uv)$,$u=x+y$,$v=x-y
殳 =(u)^2ln v =dfrac (y)(x), =2x-3y,-|||-则 dfrac (partial z)(partial y)= ()-|||-A
=f(x,xy), 则 dfrac (partial u)(partial y)=
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
设f(u)可导,z=xyf((y)/(x)),若x(partial z)/(partial x)+y(partial z)/(partial y)=xy(lny