设函数z=xln(xy),求对x的偏导-搜答案-搜搜考试网

由隐函数求导法可得dy/dx=-(2x-y)/(2y-x)根据复合函数的链式求导法则可得dz/dx=2x+2y*dy/dx=2x-2y(2x-y)/(2y-x)=2(y²-x²)/

第一个：z=x^xy=e^[ln(x^xy)]=e^(xylnx)令u=xy*lnx,则z=e^u∂z/∂x=(x^u)'•u'=(e^u)•(xyln

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'(

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)

1、对X求导(导数符号无,用“￡”代替)两边对x求导有：2x2z￡z/￡x=-ycos(z/x)/x^2*￡z/￡x:化简得：￡z/￡x=-2x/[2zycos(z/x)/x^2]:2、对y求导两边求

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)

x^2+y^2+z^2-3xyz=0两边对x求偏导,2x+2z*dz/dx-3yz-3xydz/dx=0从中解得：dz/dx=(3yz-2x)/(2z-3xy)(1)同理：dz/dy=(3xz-2y)

令G(X,Y,Z)=F(xy,z-2x)GZ'=F'2GX'=yF'1-2F'2∂z/∂x=-GX'/GZ'=(2F'2-yF'1)/F'2Gy'=xF'1∂z/&

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,

设u=xy,v=y/x,则z=f(u,v),所以ðz/ðx=f'1*ðu/ðx+f'2*ðv/ðx=yf'1-yf'2/x^2,注意到f'1

点击放大,右键查看图片可以进一步放大：

z=(x^2)*ln(2xy),Zx=(2x)ln(2xy)+(x^2)/2xy*(2xy)'=(2x)ln(2xy)+xZxx=2ln(2xy)+(2x)/2xy*(2xy)'+1=2ln(2xy)

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)

dz=[yIn(xy)+y]dx+[xIn(xy)+x]dy分开求导

那个符号用a表示了哈(1)az/ax=y^2+3x^2yaz/ay=2xy+x^3a^2z/ax^2=6xya^2z/(axay)=a^2z/(ayax)=2y+3x^2a^2/ay^2=2x(2)a

z=x^4+3x²y+y³∂z/∂x=4x³+6xy∂z/∂y=3x²+3y²∂²

这是隐函数.二阶导再导一次就是.方程两边对x求导,得z'=cos(xz)(xz)'+y(y不是关于x的函数吧?）=zcos(xz)+xz'cos(xz)+y所以z'=[zcos(xz)+y]/[1-x

两边对x求导先求出Z‘,然后再两边对x求导,这次得到z’和x,y,z表示的z“

二阶偏导数有四个Z''xx=(lin(x+y)+x/(x+y))'=1/(x+y)+y/(x+y)^2Z''yy=(x/(x+y))'=-x/(x+y)^2Z''yx=Z''xy=(x/(x+y))'