设函数z z x y 由方程y*z x^2 e^z=0确定全微分dz=

d(y^2)/dx=d(y^2)/dy*dy/dx=2y*dy/dx这个复合函数求导法则正如ovtr0001仁兄所说那样,你可以翻翻课本这个……还要详细点呀?你有书么?你看书那里不懂可以提出来,我可能

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

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-

两边对x求导2x+2y*dy/dx=0dy/dx=-x/y有不明白的追问再问：刚学不太明白，2x+2y*dy/dx=0里的dy/dx哪来的，是y'吗？再答：是的复合函数求导注意这里y是x的函数不妨换个

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)]．

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

x+2y+xy-z-exp(z)=0.(1)对(1)两边同时对x求偏导1+y-Zx-(e^z)*Zx=0.(2)Zx=(1+y)/(e^z+1)故Zx(1,0)=1/(e^0+1)=1/2对(1)两边

直接求导,用xy表示导数【欢迎追问,

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.

两端对x求导数（把y看作x的函数）,则1-y'=e^(xy)*(1*y+x*y')y'[xe^(xy)+1]=1-ye^(xy)dy/dx=y'=[1-ye^(xy)]/[xe^(xy)+1]

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)

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=1,y=0代入方程：z=1+ln1-e^z,得：z=0.两边对x求偏导：∂z/∂x=1/(x+y)-e^z∂z/∂x,得：∂z/W

z=ln(x+y)az/ax=1/(x+y)所以az/ax|(1,1)=1/(1+1)=1/2

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����

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'

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)]=[