求xy=e^x=y所确定的隐函数的导数-搜答案-智慧作业帮

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'^

如图所示,最后求解是自上而下带入的

隐函数求导,两边同时求导,此题是对X求导!两边同时求导：y+xy'=e^x-y'y'=(e^x-y)/(x+1)由XY=e^X-y解出yy=e^x/x+1,带入上式y'=(e^x-y)/(x+1)=[

方程两边对x求导：e^y×y'＝y＋xy'得y'＝y/(e^y－x)

原方程是xy=1-e^y?如果是的话将等式两边对X求导数得y+xy'=e^y*y'则y‘=y/(e^y-x)y'(0)=y/e^y

两边同时对X求导y+xy`=e^x+y`y`=(e^x-y)/(x-1)

这个题目要用到微分的形式不变性e^y*dy+d(xy)=0e^y*dy+xdy+ydx=0-ydx=(x+e^y)dydy=-y*dx/(x+e^y)

先对X求导y+xy'-e^x+e^yy'=0y'=(e^x-y)/(x+e^y)再问：主要是e^y我不懂，答案是对的，老师。还有y'=0是为什么？

[ln(xy)]'=[e^(x+y)]'(xy)'/(xy)=e^(x+y)*(x+y)'(y+xy')/(xy)=e^(x+y)*(1+y')y'=y[e^(x+y)-1]/[x(1-ye^(x+y

两端对x求导得e^x+e^y*y'=y+xy'y'=(e^x-y)/(x-e^y)dy=(e^x-y)/(x-e^y)dx

两边求导e^y×y'=xy'+yy'=y/(e^y-x)dy/dx=y/(e^y-x)

恩就是用楼上的方法做的再问：可答案是y((x-1)^2+(y-1)^2)/x^2*(y-1)^3再答：我也没具体算拉y+xy'=e^(x+y)+e^(x+y)y'=xy+xyy'--->y'=(e^(

首先把x=0代入隐函数得到：e^y=e∴y=f(0)=1e^y+xy=e两边对x求导：【注意y是关于x的函数】(e^y)y'+y+xy'=0把x=0,y=1代入：(e^1)y'+1=0∴f'(0)=y

两边分别求x的导数得：e^x+(y+xy')=0,即y'=-(e^x+y)/x,即：dy/dx=-(e^x+y)/x

y+x*y'=e^(x+y)*(1+y')∴dy/dx=[e^(x+y)-y]/[x-e^(x+y)].

ln（xy）=e^x+ylnx+lny=e^x+y两边同时对x求导1/x+(1/y)(dy/dx)=e^x+dy/dxdy/dx=[(1/x)-e^x]/[1-(1/y)]=(y-xye^x)/(xy

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'(

先对X求导y+xy'-e^x+e^yy'=0y'=(e^x-y)/(x+e^y)

e^(x+y)=xy两边对x求导：e^(x+y)*(1+y')=y+xy'解得：y'=[y-e^(x+y)]/(e^(x+y)-x]=(y-xy)/(xy-x)

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)]=[