求偏导数z=arctan(x?y^2)

z=arctan(x-y²) ∂z/∂x={1/[1+(x-y²)²]}×(x-y²)'=1/[1+(x-y²)²]，那么∂z=∂x/[1+(x-y²)²] ∂z/∂y={1/[1+(x-y²)²]}×(x-y²)'=-2y/[1+(x-y