z=arctan x/(1+y^2),则dz=?
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z=arctan x/(1+y^2),则dz=?
![z=arctan x/(1+y^2),则dz=?](/uploads/image/z/18852060-12-0.jpg?t=z%3Darctan+x%2F%281%2By%5E2%29%2C%E5%88%99dz%3D%3F)
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
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
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