Question

z=sin（xy）+cos²（xy）的偏导数，详细过程，谢谢…

Answer 1

解：
∂z/∂x
= [cos(xy)]·(xy)' + [2cos(xy)]·[2cos(xy)]‘
=ycos(xy)-4[cos(xy)]·[sin(xy)]·(xy)'
=ycos(xy)-4ysin(xy)cos(xy)
=ycos(xy) · [1-4sin(xy)]

∂z/∂y
= [cos(xy)]·(xy)' + [2cos(xy)]·[2cos(xy)]‘
=xcos(xy)-4[cos(xy)]·[sin(xy)]·(xy)'
=xcos(xy)-4xsin(xy)cos(xy)
=xcos(xy) · [1-4sin(xy)]

Answer 2

解：
∂z/∂x
= [cos(xy)]·(xy)' + [2cos(xy)]·[cos(xy)]‘
=ycos(xy)-2[cos(xy)]·[sin(xy)]·(xy)'
=ycos(xy)-2ysin(xy)cos(xy)
=ycos(xy) · [1-2sin(xy)]
∂z/∂y
= [cos(xy)]·(xy)' + [2cos(xy)]·[cos(xy)]‘
=xcos(xy)-2[cos(xy)]·[sin(xy)]·(xy)'
=xcos(xy)-2xsin(xy)cos(xy)
=xcos(xy) · [1-2sin(xy)]