设y=y(x)是由方程xy3=y-1所确定的隐函数-搜答案-大师作文网

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)]^

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

lny+x/y=0等式两边求导：y'*1/y+1/y+x*y'(-1/y²)=0(1/y-x/y²)y'=-1/y∴y'=(-1/y)/(1/y-x/y²)=-y/(y-

由隐函数微分法可得：－sin(x＋y)(1＋y′)＋y′＝0－sin(x＋y)＋[1－sin(x＋y)]y′＝0∴y′＝sin(x＋y)／[1－sin(x＋y)]．

设y=y(x)由方程ysinx=cos(x-y)所确定,则y'(0)=x=0时cos(-y)=cosy=0,故y=π/2+2kπ,k∈ZF(x,y)=ysinx-cos(x-y)=0dy/dx=-(&

这是一个复合函数求导,y=y(x)所以求e^y的导数首先对整体求导,再对y求导即为e^y*y'xy的导数为y+x*y'(根据求导规则)所以两边求导可得e^y*y'-y-x*y'=0

再答：隐函数高阶求导。再答：

对两边求导：[-sin（x+y）]（1+dy/dx）+dy/dx=0-sin（x+y）-[sin（x+y）]dy/dx+dy/dx=0dy/dx=[sin(x+y)]/[1-sin(x+y)]

如图

cos(x+y)+y=1两边同时对x求导-(1+y~)sin(x+y)+y~=0可得：=(1+y~)sin(x+y)=sin(x+y)/(1-sin(x+y))

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

分别对y求导,求左边为1+【e^(x+y）×（dx/dy+1）】右边为2×dx/dy推的dx/dy：自己算下,没得草稿纸.

dz=-dx-dy

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'(

ln(x+y)=x·lny(1+y‘)/(x+y)=lny+x/y·y‘y+y·y‘=y(x+y)lny+x(x+y)·y‘y‘=【y(x+x)lny-y】/【y-x(x+y)】再问：лл��

/>e^y+xy+e^x=0两边同时对x求导得：e^y·y'+y+xy'+e^x=0得y'=-(y+e^x)/(x+e^y)y''=-[(y'+e^x)(x+e^y)-(y+e^x)(1+e^y·y'

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]

化为：e^(ylnx)-e^y=sin(xy)两边对x求导：e^(ylnx)(y'lnx+y/x)-y'e^y=cos(xy)(y+xy')y'[lnxe^(ylnx)-e^y-xcos(xy)]=[