设方程x y^2-e^2y*cosx=1,确定的隐函数为y=f(x),求微分

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'^

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﹚

方程两边对x求导,得：y+xy'+y'e^y=2y+2xy'y'e^y-xy'=y得y'=y/(e^y-x)因此dy=ydx/(e^y-x)

e^(xy)(y+xdy/dx)-4x-dy/dx=0；dy/dx(xe^(xy)-1)=-ye^(xy)+4x；dy/dx=(4x-ye^(xy))/(xe^(xy)-1).

1)x=0代入方程：1-e^y=0,得y(0)=0两边对X求导：e^x-y'e^y=cos(xy)(y+xy')y'=[e^x-ycos(xy)]/[xcos(xy)+e^y]代入x=0,y(0)=0

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

把x=0代入原方程得0+e^0+y=2∴y=1方程两边对x求导得：y+xy'+e^(xy)(y+xy')+y'=0移项、整理得：[x+xe^(xy)+1]y'=y+ye^(xy)∴y'=[y+ye^(

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)

答：xy+ln(x+e^2)+lny=0……（1）两边对x求导：y+xy'+1/(x+e^2)+y'/y=0……（2）x=0代入（1）和（2）得：0+2+lny=0y+0+1/e^2+y'/y=0解得

e^(xy)+sin(xy)=y(y+xy')e^(xy)+(y+xy')cos(xy)=y'y'=(ye^(xy)+ycos(xy))/(1-xe^(xy)-xcos(xy))

=-[ysin(xy)+2e^(2x+y)]/[ysin(xy)+e^(2x+y)]*(dx)再问：麻烦给我写出解的过程。。再答：等式两边取对数，得：d[e^(2x+y)]-d[cos(xy)]=0(

两端对x求偏导得：-ye^(-xy)-2(z/x)+(z/x)e^z=0,所以,z/x=ye^(-xy)/(e^z-2)两端对y求偏导得：-xe^(-xy)-2(z/y)+(z/y)e^z=0,所以,

x+2y-z=3e^(xy-xz)两边对x求导,z看成是x的函数求偏导得,y看成常数,得1-əz/əx=3(y-z-xəz/əx)e^(xy-xz)=><

Fx=e^x-y^2Fy=cosy-2xydy/dx=-Fx/Fy=(y^2-e^x)/(cosy-2xy)

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'(

x=0时,代入方程得：1+1=y,得：y=2对x求导：(y+xy')e^xy-sin(xy)*(y+xy')=y'将x=0,y=2代入得：2=y'故dy(0)=2dx

两边对x求导数,得y'*e^y+y+xy'=0,在原方程中令x=0可得y=1,因此,将x=0,y=1代入上式可得y'+1=0,即y'(0)=-1.再问：对x求导时y可以当成一个常数吗？为什么要用公式（

/>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'

两边对x求导xy^2+sinx=e^yy^2+2xyy'+cosx=e^y*y'y'(e^y-2xy)=y^2+cosxy'=(y^2+cosx)/(e^y-2xy)