设y=y(x)由方程cos(xy) 3x y=1所确定

cos(xy)=x+y两边微分,得dx+dy-sin(xy)*(x*dy+y*dx)=0dx(1-ysin(xy))+dy(1-xsin(xy))=0dy/dx=(ysin(xy)-1)/(1-xsi

令F（x,y)=cos(xy)-x-yF'(x,y)x=-ysin(xy)-1对x求偏导F'(x,y)y=-xsin(xy)-1对y求偏导切线方程为：（x-0)/F'(x,y)=(y-1)/F'(x,

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﹚

dcos(xy)=dx-sin(xy)d(xy)=dx-sin(xy)(ydx+xdy)=dx-ysin(xy)dx-xsin(xy)dy=dxdy=-[ysin(xy)+1]dx/[xsin(xy)

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)]．

设y=y(x)由方程ysinx=cos(x-y)所确定,则y'(0)=x=0时cos(-y)=cosy=0,故y=π/2+2kπ,k∈ZF(x,y)=ysinx-cos(x-y)=0dy/dx=-(&

两边对x求导：2cos(x^2+y)*(-sin(x^2+y))*(2x+y')=1所以y'=-1/sin(2x^2+2y)-2x再问：求f'(x)```再答：y'就是f'(x)啊。。。。。

两边求导,-sin(x+y)(1+y`)+e^yy`=1,dy=1+sin(y+x)/e^y-sin(x+y)dx再问：亲，这是正确的么？我是帮人问的==对的就给分了啊！

对两边求导：[-sin（x+y）]（1+dy/dx）+dy/dx=0-sin（x+y）-[sin（x+y）]dy/dx+dy/dx=0dy/dx=[sin(x+y)]/[1-sin(x+y)]

=-[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(

dy/dt=cost-cost+tsint=tsintdx/dt=-sintdy/dx=(dy/dt)/(dx/dt)=-t再问：为什么-tcost会分解成-cost+tsint~~~+_+知道了==

cos(x+y)+y=1两边同时对x求导-(1+y~)sin(x+y)+y~=0可得：=(1+y~)sin(x+y)=sin(x+y)/(1-sin(x+y))

B对方程x+cos(x+y)=0两边取微分,得dx-sin(x+y)d(x+y)=0即dx-sin(x+y)dx+sin(x+y)dy=0,整理得[1-sin(x+y)]dx=-sin(x+y0dy从

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.

这个是对隐函数的求导.隐函数求导时,遇到因变量时,除和自变量一样外,还要再乘以因变量的一阶导数.因此y=y（x）由方程cos（x）+y=1确定时,两端对x求导就得-sinx+y'=0y'=sinx如果

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

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����

在方程ex+y+cos（xy）=0左右两边同时对x求导，得：ex+y（1+y′）-sin（xy）•（y+xy′）=0，化简求得：y′＝dydx＝ysin(xy)−ex+yex+y−xsin(xy)．