Z=e^xy+(x^2+y^2)arctanx偏导数?

2025-01-03 06:55:56
推荐回答(2个)
回答1:

Z'x=e^(xy) · y + 2x arctanx +(x²+y²) · 1/(1+x²)
=ye^(xy) +2xarctanx +(x²+y²)/(1+x²)
Z'y=e^(xy) · x +2y arctanx =xe^(xy) +2yarctanx

回答2:

Z=e^(xy)+(x^2+y^2)arctanx,
∂z/∂x=ye^(xy)+2xarctanx+(x^2+y^2)/(1+x^2),
∂z/∂y=xe^(xy)+2yarctanx.