z=tan[(x+y)/(1-xy)],求dz

dz=∂z/∂x *dx +∂z/∂y*dy 显然 ∂z/∂x =1/cos²[(x+y)/(1-xy)] * ∂[(x+y)/(1-xy)]/∂x =1/cos²[(x+y)/(1-xy)] * [∂(x+y)/∂x *(1-xy) -∂(1-xy)/W