设y=y（x）由方程e^（xy）+Sin（xy）=y确定，求y'（0）

设y=y(x)由方程e^(xy)+Sin(xy)=y确定,求y'(0)
先代入x＝0求得y.后两边求导,化简最后得y'(0)＝2

设y=y(x)由方程e^xy+cos(xy)=y确定,求dy(0).
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

设y=y(x)由方程e^xy+sin(xy)=y确定,求dy\/dx.
e^(xy) + sin(xy) = y (y+xy')e^(xy) + (y+xy')cos(xy) = y'y'=(ye^(xy)+ycos(xy))\/(1-xe^(xy)-xcos(xy))

设y=y(x)由方程e^xy+sin(xy)=y确定,求dy\/dx.
e^(xy) + sin(xy) = y (y+xy')e^(xy) + (y+xy')cos(xy) = y'y'=(ye^(xy)+ycos(xy))\/(1-xe^(xy)-xcos(xy))

y=y(x)由方程 [e^(x+y)]+sin(xy)=1确定,求y'(x)及y'(0)
e^(x+y) + sin(xy) = 1 e^(x+y)*(1+y')+cos(xy)(y+xy')=0 y'*[e*(x+y)+xcos(xy)]=-[ycos(xy)+e^(x+y)]y'=-[ycos(xy)+e^(x+y)]\/[e*(x+y)+xcos(xy)]x=0，求出 y=0，代入上式，得到y'（x=0）=-1....

设y=y(x)是由方程e^x+y=sin(xy)确定的隐函数,求y‘
e^x+y=sin(xy)两边同时对x进行求导，得：e^x+y'=cos(xy)*(y+xy')∴[xcos(xy)-1]y'=e^x-ycos(xy)∴y'=[e^x-ycos(xy)]\/[xcos(xy)-1]望采纳

y=y(x)由方程 [e^(x+y)]+sin(x+y)=1确定,求y'(x)及y'(0)
等式两边对x求导 e^(x+y)*(1+y')+cos(x+y)*(1+y')=0 (1+y')*[e^(x+y)+cos(x+y)]=0 y'=-1 所以y'(x)恒等于-1，y'(0)=-1

y=y(x)由方程 [e^(x+y)]+sin(xy)=1确定,求y'(x)及y'(0)
e^(x+y) + sin(xy) = 1， x = 0 时， y = 0 两边对 x 求导， 得 （1+y')e^(x+y) + (y+xy')cos(xy) = 0 y'(x) = -[e^(x+y)+ycos(xy)]\/[e^(x+y)+xcos(xy)]y'(0) = -1

高数题一枚,函数y=y(x)是由方程e∧(x+y)+cos(xy)=0确定,则dy\/dx=?
隐函数 直接求导数.e^(x+y)(1+y')-sin(xy)(y+xy')=0 解出y'即为dy\/dx=[e^(x+y)-y*sin(xy)]\/[e^(x+y)-x*sin(xy)]

函数y=y(x)由方程e^x -e^y = sin(xy)确定,求y(0) 要步骤
把x=0代入,所以1-e^y=0,所以e^y=1,所以y=0.