由方程e^y=2xy确定的y是x的函数.求 dy dx-搜答案-百度作业网

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)

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

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﹚

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

(-2y^2)/(4xy+e^y)

e^y+xy=e两边求导e^y*y'+y+xy'=0∴y'(e^y+x)=-yy'=-y/(e^y+x)即dy/dx=-y/(e^y+x)当x=0时,e^y=e,y=1∴dy/dx|（x=0)=-1/

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/

再答：隐函数高阶求导。再答：

e^y+xy=1两边同时对x求导得：e^y*y'+y+xy'=0所以y'=-y/(e^y+x)即dy/dx=-y/(e^y+x)如果不懂,祝学习愉快!

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)

两边对x求导得y+xy'=(1+y')/(x+y)y(x+y)+x(x+y)y'=1+y'y'[x(x+y)-1]=1-y(x+y)y'=[1-y(x+y)]/[x(x+y)-1]dy=[1-y(x+

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

用隐函数求导一般得出的还是隐函数用WPS纯手打的,如果我理解错了你的式子,请指出,我改一下就行了,但方法是一样的再问：对不起没看到你的答案

x（y^2）-e^xy+2=0两端同时求导：(y^2+2xy'y)-e^xy(y+xy')=0集项：(2xy-xe^xy)y'=(ye^xy-y^2)则：dy/dx=y'=(ye^xy-y^2)/(2

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)

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'(