设z z x y 是由方程x=lnz y确定的隐函数,求-搜答案-智慧作业帮

方程两边同时求x对y的导：y+xdy/dx+1/x+2ydy/dx=0,dy/dx=-(y+1/x)/(x+2y),dy=-(y+1/x)dx/(x+2y)

dz=d[xyP(z)]=yP(z)dx+xP(z)dy+xyP'(z)dz所以dz=[yP(z)dx+xP(z)dy]/[1-xyP'(z)]du=df(x,z)=f'x(x,z)dx+f'z(x,

dcos(xy)=dx-sin(xy)d(xy)=dx-sin(xy)(ydx+xdy)=dx-ysin(xy)dx-xsin(xy)dy=dxdy=-[ysin(xy)+1]dx/[xsin(xy)

∂z/∂x=(∂f/∂x)+(∂f/∂y)(dy/dx)//:g(y)+y=xg'(y)y'+y'=1y'=1/[1+g'(y)

将z对x的偏导记为dz/dx,（不规范,请勿参照）(e^x)-xyz=0两边对x求导数(e^x)'-(xyz)'=0e^x-x'yz-xy(dz/dx)=0e^x-yz-xy(dz/dx)=0xy(d

两边对x求导：2cos(x^2+y)*(-sin(x^2+y))*(2x+y')=1所以y'=-1/sin(2x^2+2y)-2x再问：求f'(x)```再答：y'就是f'(x)啊。。。。。

e^z-z+xy^3=0偏z/偏x：z'e^z-z'+y^3=0y^3=z'(1-e^z)z'=y^3/(1-e^z)偏z/偏y:z'e^z-z'+3xy^2=0z'=3xy^2/(1-e^z)偏z/

不就是对x求导吗?把y看成中间变量y=y(x)说明要想导x要通过y这个中间变量两边对x求导:y^3+(3x*y^2)*dy/dx+(e^x)*siny+(e^x)*cosy*dy/dx=1/x下面你自

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)

再答：隐函数高阶求导。再答：

对X的偏导=yz/(e^z-xy)对Y的偏导=xz/(e^z-xy)

两边对X求导数就行了撒,把y看成是一个常数,Z看成对x函数就行了撒e^x-(z*y+y*x*zx)=0所以z对x的偏导数zx=(zy-e^x)/(y*x)

网上有很多高数课后习题答案,你可以下载一个参考~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,原式

两端对x求导数（把y看作x的函数）,则1-y'=e^(xy)*(1*y+x*y')y'[xe^(xy)+1]=1-ye^(xy)dy/dx=y'=[1-ye^(xy)]/[xe^(xy)+1]

dz=-dx-dy

原式两边微分2ydx+2xdy-2ydy=2dx故dy=(1-y)dx/(x-y)

1.对x=ln(x+y）求微分,得dx=(dx+dy)/(x+y),∴dy=(x+y-1)dx,∴dy/dx=x+y-1.2.e^(xy)+y^3-5x=0,①求微分得e^(xy)*(ydx+xdy)

若z=f(x,y)由方程F(x,y,z)=0确定,则将F(x,y,z)=0两边对x,y求导(x,y视为独立变量,z视为x,y的函数)这个是没有问题的,但此处x,y为两个独立的变量；题1.设y=f(x,

F(x,y)=x^2+y^2-ln(x+2y)Fx=2x-1/(x+2y)Fy=2y-2/(x+2y)F(x)=-Fx/Fy=-[2x(x+2y)-1]/[2y(x+2y)-2]