设函数z=sin(xy)dz
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dz=[yIn(xy)+y]dx+[xIn(xy)+x]dy分开求导
dz=2xdy+2ydx
dz=(y+y/(X^2))dx+(x-1/x)dy,
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:
∂z/∂x=cos(x-y)∂z/∂y=-cos(x-y)dz=∂z/∂x*dx+∂z/∂y*dy=cos(x-y)dx-cos(x-y)dy
先对x求偏导数得z'(x)cosz=yz+z'(x)y所以z'(x)=yz/(cosz-y)同理对y求偏导数得z'(y)=xz/(cosz-x)所以dz=yz/(cosz-y)dx+xz/(cosz-x)dy
两边同时微分zdx+xdz+zdy+ydz+xdy+ydx=0(x+y)dz+(y+z)dx+(z+x)dy=0dz=-[(y+z)dx+(z+x)dy]/(x+y)
因为x、y都为自变量,不是宗量,故此题没有全微分,应只有偏微分.详解如下:对方程两边微分:左边:de^z=e^z*dz右边d[xyz+cos(xy)]=xydz+yzdx+xzdy-(sinxy)*(ydx+xdy)则有 e^z*dz=x
z=x^2+2xy两边同时求导数,得到:dz=2xdx+2ydx+2xdy即:dz=2(x+y)dx+2xdy.
e^(-xy)-x^2*y+e^z=z,令F(x,y,z)=e^(-xy)-x^2*y+e^z-z=0分别对F取x,y,z的偏导数,可得əF/əx=e^(-xy)*(-y)-2xyəF/əy=e^(-
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)
对左右两边求导:(1+ez)dz=ydx+xdy.dz=1/(1+ez).(ydx+xdy).
z=sin(x²y²)+3x-5y²+1所以δz/δx=cos(x²y²)*2xy²+3δz/δy=cos(x²y²)*2x²y-10y所以dz=(δ
u=x^2+y∂u/∂x=2x∂u/∂y=1du=(∂u/∂x)dx+(∂u/∂y)dy=2xdx+dyv=2xy∂v/∂
x+2y-z=3e^(xy-xz)两边对x求导,z看成是x的函数求偏导得,y看成常数,得1-əz/əx=3(y-z-xəz/əx)e^(xy-xz)=>əz/əx=[1-3(y-z
zx=1/(1+(x/y)²)*1/y=y/(x²+y²)zy=1/(1+(x/y)²)*(-x/y²)=-x/(x²+y²)所以dz=y/(x²+y²
e^x(1/y1)x^(y1)再问:亲,经多方证实你的答案是错误的,不过你是唯一回答我的人,我还是采纳了
设F(x,y,z)=sinz-xyz则F′(X)=-yzF′(y)=-xzF′(z)=cosz-xyz对x的谝导数等于-yz/(cosz-xy)z对y的谝导数等于-xz/(cosz-xy)dz=[-yz/(cosz-xy)]dx+[-xz/