为自己加油

函数z=ln(1+x²-y²)的全微分dz=____．

[2/(1+x²-y²)](xdx-ydy)。解析:对于函数z=ln(1+x²-y²)，∂z/∂x=2x/(1+x²-y²)，∂z/∂y=-2y/(1+x²-y²)所以dz=(∂z/∂x)dx+(∂z/∂y)dy=[2x/(1+x²-y²)]dx+[-2y/(1+x²-y²)]dy=[2/(1+x²-y²)](xdx-ydy)