设z=ln(1+x^2-y^2),则dz=________
问题描述:
设z=ln(1+x^2-y^2),则dz=________
答
dz=Zxdx+Zydy=[2x/(1+x^2+y^2)]dx+[-2x/(1+x^2+y^2)]dy; Zx是对x的偏导数!
设z=ln(1+x^2-y^2),则dz=________
dz=Zxdx+Zydy=[2x/(1+x^2+y^2)]dx+[-2x/(1+x^2+y^2)]dy; Zx是对x的偏导数!