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z sin xy

∂Z/∂x= y*cos(xy) -2cos(xy)*sin(xy)*y = y*cos(xy) - y*sin(2xy) ∂Z/∂y= x*cos(xy) -2cos(xy)*sin(xy)*x = x*cos(xy) - x*sin(2xy)

解答如下: x,y具有轮换对称性,所以写出一个的,另一个直接替换就可以了。

dz=cos(xy)xdy+cos(xy)ydx 由z/x=ycos(xy),z/y=xcos(xy)→dz=ycos(xy)dx + xcos(xy)dy. cos(x-y)dx-cos(x-y)dy

Zy=ycos(xy)+2cos(xy)*[-sin(xy)]*x =ycos(xy)-xsin(2xy)

您好,步骤如图所示: 很高兴能回答您的提问,您不用添加任何财富,只要及时采纳就是对我们最好的回报。若提问人还有任何不懂的地方可随时追问,我会尽量解答,祝您学业进步,谢谢。☆⌒_⌒☆ 如果问题解决后,请点击下面的“选为满意答案”

因为z=sin(xy2)所以?z?x=y2cos(xy2),所以?2z?x?y=2ycos(xy2)?2xy3sin(xy2).

字数补丁

由z=sin(xy)+cos2(xy),得? z? y=xcos(xy)-2xcos(xy)sin(xy)

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...

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