设隐函数y=y(x)由x-y 1 2siny=0所确定,求y撇

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'^

df(x,y,z)/dx=[d(z^2)/dx]*y*e^x+y*z^2*(de^x/dx)=2zye^x(dz/dx)+y*z^2*e^x另,由x+y+z+xyz=0求dz/dx两边对x求偏导1+0

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)]^

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)]．

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

.y/x=ty=txy=xtdy/dx=t+t'xdy=(t+t'x)dxy^2(x-y)=x^2t^2(x-tx)=1x=1/[t^2(1-t)]y=1/[t(1-t)]1/y^2=t^2(1-t)

两边求微分：d(x^y+y^x)=d(f(x^2+y^2))对x^y可以这么看：先把X看成常数,对Y求微分相当于a^Y,再把Y看成常数对X求微分相当于X^a.那么就好用公式了如下：d(x^y）=X^Y

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'(

1.对x=ln(x+y）求微分,得dx=(dx+dy)/(x+y),∴dy=(x+y-1)dx,∴dy/dx=x+y-1.2.e^(xy)+y^3-5x=0,①求微分得e^(xy)*(ydx+xdy)

x+2y+z=e^(x-y-z)两边对x求偏导注意到z=z(x,y)1+z'=e^(x-y-z)*(1-z')...(1)再对x求偏导z"=e^(x-y-z)(1-z')^2-z"e^(x-y-z).

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'

y=(1-C1-C2)y1(x)+C1y2(x)+C2y3(x)即y=y1(x)+C1*[y2(x)-y1(x)]+C2*[y3(x)-y1(x)]而y1(x),y2(x),y3(x)都是线性方程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]

y=2xarctan(y/x)(y/x)'=(y'/x-y/x^2)(arctanu)'=1/(1+u^2)y'=2arctan(y/x)+2x*(y'/x-y/x^2)*[1/(1+(y/x)^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)]=[