求偏导数z=arctan(x—y^2)
问题描述：
求偏导数z=arctan(x—y^2)
答
dz/dx= 1/[1+(x-y^2)] dz/dy= -2y/[1+(x-y^2)]
答
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)...