先求z_x = cos(xy) y (2x^2)y + sin(xy) (4x) y 再求z_y = cos(xy) y (2x^2)y + sin(xy) (2x^2)dz = z_x dx + z_y dy 将上面求得式子带入即可。
由z/x=ycos(xy),z/y=xcos(xy)→dz=ycos(xy)dx + xcos(xy)dy. 解析看不懂?免费查看同类题视频解析查看解答 相似问题 z=sin(xy)的全微分 若z = sin ( xy ) 则它的全微分dz = z=xe^(-xy)+sin(x+y)的全微分怎么求 想要特别特别详细的步骤 ...
偏微分方程,第二讲(Partial differential equations, 2nd Class, MTH210@XJTLU, 2022) 503 -- 3:27 App 51单片机独立按键控制LED左右位移 1720 -- 2:45 App 方程豹豹5,帝瓦雷音响实测参考。测试曲目别安乐队《遥望》。如果豹友们觉得听感ok想抄作业的话,我上一条视频有具体的音响设置哦～ 1317 2 3:37 ...
y=sin(xe^y)两边取微分 dy =dsin(xe^y)链式法则 =sin(xe^y) d(xe^y)乘积规则 =sin(xe^y) ( e^ydx + x de^y)=sin(xe^y) ( e^ydx + xe^y dy)得出结果 y=sin(xe^y) ， dy=sin(xe^y) ( e^ydx + xe^y dy)😄: y=sin(xe^y) ， dy=sin(xe^y) ( e^...
解：由z/x=ycos(xy)，z/y=xcos(xy)→dz=ycos(xy)dx + xcos(xy)dy。麻烦采纳，谢谢!
A:xcos (xy) B:(xdx+ydy) cos (xy) C:ycos (xy) D:(ydx+xdy) cos (xy) 相关知识点: 试题来源: 解析 dz=cos(xy)ydx+cos(xy)xdy=(ydx+xdy)cos(xy),选D 结果一 题目 若z = sin ( xy ) 则它的全微分dz =A:xcos (xy) B:(xdx+ydy) cos (xy) C:ycos (xy) D:(...
可以运用导数的链式法则和微分运算法则来求解。解：dy=d(sin(x+y))=cos(x+y)·d(x+y)=cos(x+y)·(dx+dy)=cos(x+y)dx+cos(x+y)dy dy-cos(x+y)dy=cos(x+y)dx dy=cos(x+y)/(1-cos(x+y))dx 所以，y=sin(x+y)的微分为 dy=cos(x+y)/(1-cos(x+y))dx ...
dz=cos(xy)ydx+cos(xy)xdy=(ydx+xdy)cos(xy)，选D
若z = sin ( xy ) 则它的全微分dz =A:xcos (xy) B:(xdx+ydy) cos (xy) C:ycos (xy) D:(ydx+xdy) cos (xy)
积分得：y'=-cosx+c1 再积分得：y=-sinx+c1x+c2 此即为通解.