z＝xf(e∧x y),求

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

(太麻烦拉,给点分啊!)设v=x*x-y*y,u=exp{xy}那么dv/dx=2x(这里应该用偏导符号,代替一下),dv/dy=2y,du/dx=y*exp{xy},du/dy=x*exp{xy}那

e^y-e^x=xy两边求导,得e^y*y'-e^x=y+xy'(e^y-x)y'=(e^x+y)所以y'=(e^x+y)/(e^y-x)x=0时,e^y-e^0=0,则e^y=1,则y=0所以y'(

Z=e^xy在x处的导函数为ye^(xy)在y处的导函数为xe^(xy)dz=ye^(xy)dx+xe^(xy)dy=2e^2dx+e^2dy

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)

根据一阶全微分形式不变得dz=d(xf(x^y,e^xy)=f(x^y,e^xy)dx+xd(f(x^y,e^xy))=f(x^y,e^xy)dx+x[f1'd(x^y)+f2'(de^xy)]=f(

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)

两端对x求偏导得：-ye^(-xy)-2(z/x)+(z/x)e^z=0,所以,z/x=ye^(-xy)/(e^z-2)两端对y求偏导得：-xe^(-xy)-2(z/y)+(z/y)e^z=0,所以,

求二元函数全微分z=f[x²-y²,e^(xy)]设z=f(u,v),u=x²-y²,v=e^(xy)则dz=(∂f/∂u)du+(&#

z=arctan(x*e^x)z'={1/[1+(x*e^x)^2]}*(x*e^x)'(x*e^x)'=x'*e^x+x*(e^x)'=e^x+x*e^x=(x+1)*e^x所以dz/dx=(x+1

(z对x的偏导)=y+F(u)+x[F'(u)(-y/x^2)](z对y的偏导)=x+F'(u)/x代入,左边=[xy+xF(u)-yF'(u)]+[xy+yF'(u)]=xy+xF(u)+xy=z+

传了张图片,不怎么清楚,凑合一下思路就是按照多元复合函数求导来一步一步求解.有问题再追问.先打这么多了. 答案是a^2z/axay=y*f ''(xy)+g'

求曲面(e^z)-z+xy=4的切平面及法线方程.设曲面方程F(x,y,z)=(e^z)-z+xy-4=0；点M(xo,yo,zo)是该曲面上的任意一点.∂F/∂x=y；

挺好的题f(xy)=xf(y)+yf(x)---(1)设y=c=常量则：f(cx)=cf(x)+f(c)x两边求导数f'(cx)*c=cf'(x)+f(c)cf'(cx)-cf'(x)=f(c)此式对

令u=xy,v=e^(x+y)Z'x=Z'u*U'x+Z'v*V'x=f'u*y+f'v*e^(x+y)Z'y=Z'u*U'y+Z'v*V'y=f'u*x+f'v*e^(x+y)

dy/dx=dy/du*du/dx+dy/dv*dv/dx=v*e^(x+y)+u*y/x=ln(xy)*e^(x+y)+e^(x+y)*y/x=e^(x+y)[ln(xy)+y/x]所以dy=e^(

求二元函数全微分z=f[x²-y²,e^(xy)]设z=f(u,v),u=x²-y²,v=e^(xy)则dz=(∂f/∂u)du+(&#

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)-2xy