du(x,y)=2xycos(x^2y)dx+x^2cos(x^y)dy,求u(x,y)
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/22 06:18:41
du(x,y)=2xycos(x^2y)dx+x^2cos(x^y)dy,求u(x,y)du(x,y)=2xycos(x^2y)dx+x^2cos(x^y)dy,求u(x,y)du(x,y)=2xyc
du(x,y)=2xycos(x^2y)dx+x^2cos(x^y)dy,求u(x,y)
du(x,y)=2xycos(x^2y)dx+x^2cos(x^y)dy,求u(x,y)
du(x,y)=2xycos(x^2y)dx+x^2cos(x^y)dy,求u(x,y)
因为du/dx=2xycos(x^2y)
所以u(x,y)=∫2xycos(x^2y)dx
=∫cos(x^2y)d(x^2y)
=sin(x^2y)+A(y) 其中A(y)是关于y的任意函数
因为du/dy=x^2cos(x^2y)
所以x^2cos(x^2y)+A'(y)=x^2cos(x^2y)
所以A'(y)=0
A(y)=C
所以u(x,y)=sin(x^2y)+C 其中C是任意常数
du(x,y)=2xycos(x^2y)dx+x^2cos(x^y)dy,求u(x,y)
cos(y^2-x^2)求导u(x,y)=e^2xycos(y^2-x^2)
函数y=|sinx|/sinx+cosx/|cosx|+|tanx|/tanx+cotx/|cotx|的值域速求!同时求x^2sin(1350°)+y^2tan405°-(x-y)^2cot765°-2xycos(-1080°)的值!
怎样推导出同频率互相垂直简谐振动合成的轨道方程x^2/a1^2+y^2/a2^2-2xycos(p1-p2)=sin^2(p1-p2)
设u={cos(x^2)}/y +(xy)^(y/x) 求du
已知du=2xydx+x²dy,求原函数u(x,y)
设u=f(x,y)=∫(0到xy)e^(-t^2)dt 求du答案是du=e^(-x^2*y^2)(ydx+xdy)
x=y^2+y,u=(x^2+2)^3/2,求dy/du 求详细过程
函数u(x,y,z)=4xy^3+5x^2y^6+xz的全微分du
设x=y^2+y,u=(x^2+x)^(3/2),求dy/du.(能不能结果中既有X又有Y?)
求函数u=f(2x^2-y^2,xy)的全微分du.
设u=ln√(x^2+y^2+z^2) 求du
设函数u=In(x^2+y^2+z^2),求du.
求函数u=ln(2x+3y+4z^2)的全微分du
设u=z/√(x^2+y^2 ),求全微分du(3,4,5)
设u=(e^xy)(cos(x+y^2)),求du
设y=(x^2),求dy/dx令y=sinu,u=x^2,则dy/du乘以du/dx=sinu/u乘以x^2/x为什么等于cosu乘以2x
x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2