回答:
#(DY)/ DX =(cosxy-xysinxy)/(E ^ Y +的x ^ 2(sinxy))#
说明:
#1 = E ^的y xcos(XY)#
#rArr(D1)/ DX = d / DX(E ^的y xcos(XY))#
#rArr0 =(解^ Y)/ DX-(d(xcos(XY)))/ DX#
#rArr0 =(DY / DX)E 1Ý - (((DX)/ DX)cosxy + X(dcosxy)/ DX)#
#rArr0 =(DY / DX)E 1 Y-(cosxy + X(DXY)/ DX(-sinxy))#
#rArr0 =(DY / DX)E 1 Y-(cosxy +×((Y + X(DY)/ DX)( - sinxy)))#
#rArr0 =(DY / DX)E 1 Y-(cosxy + X(-ysinxy-X(DY)/ DX(sinxy)))#
#rArr0 =(DY / DX)E 1 Y-(cosxy-xysinxy-X ^ 2(DY)/ DX(sinxy))#
#rArr0 =(DY / DX)E 1的y cosxy + xysinxy + X ^ 2(DY)/ DX(sinxy)#
#rArr0 =(DY / DX)在线^ Y + X ^ 2(DY)/ DX(sinxy)-cosxy + xysinxy#
#rArr0 =(DY / DX)(E ^ Y +的x ^ 2(sinxy)) - cosxy + xysinxy#
#rArrcosxy-xysinxy =(DY / DX)(E ^ Y +的x ^ 2(sinxy))#
#rArr(DY)/ DX =(cosxy-xysinxy)/(E ^ Y +的x ^ 2(sinxy))#