• 2021-04-14
    z=cos(xy+y^2)()的全微分为()(2.0分)A.()dz=-sin(xy+y^2)(dx+dy)()B.()dz=-sin(xy+y^2)[ydx+(x+2y)dy]()C.()dz=-sin(xy+y^2)(y^2dx+xdy)()D.()dz=-sin(xy+y^2)[xydx+(x+y^2)dy]