设z z x y 是由方程xy yz zx=1-搜答案-拍题作业网

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)

方程两边微分就行了dx*y+x*dy+e^y*dy=2xdx得dy/dx=(2x-y)/(x+e^y)

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,

xy+e^y=y+1（1）求d^2y/dx^2在x=0处的值：（1）两边分别对x求导：y+xy'+e^yy'=y'y/y'+x+e^y=1(2)(2)两边对x再求导一次：(y'y'-yy'')/y'^

两边对x求导：y'=(1+y')[sec(x+y)]^2得y'=[sec(x+y)]^2/{1-[sec(x+y)]^2}=1/{[cos(x+y)]^2-1}因此dy=dx/{[cos(x+y)]^

如图所示,最后求解是自上而下带入的

对方程两边同时求导得,﹣﹙y＋xy′﹚sin﹙xy﹚＋e^y＋﹙x＋1﹚y′e^y＝0令x=0则方程cos(xy)+(x+1)*e^y=2为1＋e^y=2,得y=0,即切点坐标为﹙0,0﹚将﹙0,0﹚

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

dy/dx=(dy/dt)/(dx/dt)=[2t/(1+t^2)]/[1-1/(1+t^2)]=2/t

e^y-xy=ee^y·dy/dx-(y+x·dy/dx)=0e^y·dy/dx-y-x·dy/dx=0(e^y-x)·dy/dx=ydy/dx=y/(e^y-x)dy/dx不能叫做dx分之dy,因为

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/

对X的偏导=yz/(e^z-xy)对Y的偏导=xz/(e^z-xy)

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

dz=-dx-dy

1、2x+2y*dy/dx-y-x*dy/dx=02x-y=(x-2y)dy/dx所以dy/dx=（2x-y）/(x-2y)2、2y*dy/dx-2ay-2ax*dy/dx=0(2y-2ax)dy/d