z=arctan(x+y)/(1-xy)求dz还有个题 f(x,y)=√（x^2+y^2）则dz 这是要求什么啊 不太明白 谢

z=arctan【(x+y)/(1-xy)】的偏导数 讨论函数z=arctan(x+y)/(1-xy)的连续性方程是z=arctan[(x+y)/(1-xy)] 高数 求全微分求函数z=arctan(x+y)/(1-xy)的全微分 求二阶偏导数Zxx,Zyy和Zxy,z=arctan[(x+y)/(1-xy)], 函数求下列二阶偏导数的Z=arctan((x+y)/(1-xy)) 高数 求二阶偏导数(f具有连续偏导数1)z=arctan(x+y)/(1-xy) z=arctan x/(1+y^2),则dz=? z=arctan(x/y),求dz 设z=arctan(xy),y=e^x ,求dz/dx ． z=arctan(xy),而y=e^x,求dz/dy 求Z=arctan(xy),而y=e^x的全导数 设z=arctan(xy)+2x^2+y^2,求dz arctan(x-y)^z求偏导 求函数偏导：z=arctan(x-y)^z z=x*arctan(xy),求(dz/dx)|(1,1),(dz/dy)|(1,1) z=arctan(x+y)/(x-y)的全微分 z=arctan(y/x) x,y的偏导数 全导数:设z=arctan(xy),而y=e*x次方,求dz/dx设z=arctan(xy),而y=e*x次方,求dz/dx,