z=arctan x/(1+y^2),则dz=?
问题描述：
z=arctan x/(1+y^2),则dz=?
答
z=f(x,y)=(arctanx)/(1+y²)
fx(x,y)=1/[(1+x²)(1+y²)]
fy(x,y)=-2yarctanx/(1+y²)²
∴dz=fx(x,y)dx+fy(x,y)dy=dx/[(1+x²)(1+y²)]-(2yarctanx)dy/(1+y²)²