求由方程y=tan(x y)所确定的隐函数y=y(x)的二阶导数

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)

xy+lnx+lny=1对x求导y+xy'+1/x+y'/y=0(其中y'表示dy/dx)所以y'=(-1/x-y)/(x+1/y)=-(y+xy^2)/(x^2y+x)

方程两边求关x的导数ddx(xy)＝(y+xdydx)；     ddxex+y＝ex+y(1+dydx)；所以有  （y+xdy

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)]^

左右对x求导有y'/y=sec²(xy)(y+xy')整理有y'=y²/(cos(xy)-xy)所以dy=(y²/(cos(xy)-xy))dx

隐函数求导设z=x²y²-cos(xy)dy/dx=-(δz/δx)/(δz/δy)=-(2xy²+ysin(xy))/(2x²y+xsin(xy))=-y/x

-sin(xy)[ydx+xdy]=2xy^2*dx+x^2*2ydy-sin(xy)ydx-sin(xy)xdy=2xy^2*dx+2x^2*ydy-2x^2*ydy-sin(xy)xdy=2xy^

两边对x求导：-(y+xy')sin(xy)=2xy^2+2x^2yy'解得：y'=-[ysin(xy)+2xy^2]/[2x^2y+xsin(xy)]所以dy=-[ysin(xy)+2xy^2]/[

1、两边同时微分,y^3dx+3xy^2dy=dy,sody/dx=(y^3)/(1-3xy^2)2、dy=(e^x)/(1+e^(2x))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

z对x的偏导xy+yz+zx=1y＋yfx'+z+xfx'＝0z对y的偏导x＋z+yfy'＋xfy'＝0z对y的偏导1+fx'+yfxy"+fy'+xfxy"=01+(fx'+fy')+(x+y)fx

对y^2－2xy＝7求微分,得2ydy-2(ydx+xdy)=0,∴（y-x)dy=ydx,∴dy/dx=y/(y-x).

y+xy'+y'/y=0//对xy和lny分别求导,注意y是x的函数y'(x+1/y)=-y//移项,合并同类项y'=-y²/(xy+1)

xy+lny=1两边求导y+xy'+y'/y=0y'=-y/(x+1/y)=-y^2/(xy+1)

两边求导：y+xy'+y‘/y=0将x=0带入得到：y'=--y^2

xy+e^y=1e^y(0)=1y(0)=0xy'+y+e^yy'=00+y(0)+y'(0)=0y'(0)=0xy''+y'+y'+e^yy''+(y')^2e^y=00+2y'(0)+y''(0)

设dy/dx=y'.求导,2yy'-2y-2xy'=0dy/dx=y'=y/(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'(

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（

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)]=[