z=sin(xy^2) 求dz全微分
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/22 02:13:12
我来试试吧...z=e^xy*cos(x+y)Z'x=ye^xycos(x+y)-e^xysin(x+y)Z'y=xe^xycos(x+y)-e^xysin(x+y)故dZ=[ye^xycos(x+y
dz/dx=arctan(xy)+xy/[1+(xy)^2](dz/dx)|(1,1)=π/4+1/2(dz/dy)|(1,1)=x^2/[1+(xy)^2]=1/2
对方程两边求全微分得:(e^z-1)dz+y^3dx+3xy^2dy=0(方法和求导类似)移项,有dz=-(y^3dx+3xy^2dy)/(e^z-1)
对方程e^(-xy)+2z-e^z=2两边微分,有:e^(-xy)*d(-xy)+2*dz-e^z*dz=0-e^(-xy)*(x*dy+y*dx)+2*dz-e^z*dz=0移项,得:(e^z-2)
(y^2+2xy-cos(y+z))/(e^z+cos(y+z))再问:没有过程吗?再答:求导:e^z*dz-y^2-2xy+cos(y+z)(1+dz)=0把含有dz的项移到一起:(e^z+cos(
z=(x+y)^2*cos(x^2*y^2)dz/dx=2*(x+y)*cos(x^2*y^2)-2*(x+y)^2*sin(x^2*y^2)*x*y^2dz/dy=2*(x+y)*cos(x^2*y
解;z(x)=2x+2y²z(y)=4xy+12y²dz=(2x+2y²)dx+(4xy+12y²)dy
全微分啊dz=(1+xy)^x[ln(1+xy)+xy/(1+xy)]dx+(1+xy)^xx^2/(1+xy)dy
dz=2xdy+2ydx
直接凑微分即可,yz(2x+y+z)dx=d(yzx^2+xzy^2+xyz^2)(这里y,z看成常数),同理xz(x+2y+z)dy=d(yzx^2+xzy^2+xyz^2),xy(x+y+2z)d
再问:可以再帮我答题吗,我这边有很多财富值可以给你再问:
dz=Z'xdx+Z'ydy=2xcos(x^2+y^2)dx+2ycos(x^2+y^2)dy
再问:啊不好意思搞错了。。是z=e^(x^2+y^2),求dz,谢谢你帮我解答一下吧。。再答:
dz=2e^(2x+y^2)dx+2ye^(2x+y^2)dy把对x和对y的偏导分别求了出来再乘以各自的微分项即可.
u=x^2+y∂u/∂x=2x∂u/∂y=1du=(∂u/∂x)dx+(∂u/∂y)dy=2xdx+dy
z=x^2+2xy两边同时求导数,得到:dz=2xdx+2ydx+2xdy即:dz=2(x+y)dx+2xdy.
再问:就是这个吗?再答:是的。如还有不懂请追问,懂了请采纳。再问:还有这三题