e^(x+y)·(1+dy/dx)+dy/dx=0
dy/dx=-e^(x+y)/[1+e^(x+y)]
已知隐函数e^(x+y)+y=0,求d
e^(x+y)+y=0 e^(x+y)·(1+dy\/dx)+dy\/dx=0 dy\/dx=-e^(x+y)\/[1+e^(x+y)]
求由方程e^(x+y)-xy=0所确定的隐函数y=f(x)的微分dy
由已知得：e^(x+y)=xy.d e^(x+y)=dxy.e^(x+y)*d(x+y)=(ydx+xdy).e^(x+y)*(dx+dy)=ydx+xdy.e^(x+y)dx+e^(x+y) dy=ydx+xdy.[(e^(x+y)]dy-xdy=[y-e^(x+y)]dx.dy={[y-e^(x+y)]\/[e^(x+y)-x]}dx.
隐函数求导怎么求呀,例e^y+xy
f(x，y)=0才是隐函数 如果e^y+xy=0的话 对x求导得到 e^y *y' +y+xy'=0 可以得到y'=-y\/(e^y+x)
求由方程e∧xy+x+y=0所确定的隐函数y=f(x)的导数dy\/dx,要有过程
回答：我身上没带笔~你直接两边对X求导~
隐函数求导
因为y是x的函数，对ey的求导,相当于复合函数的求导，先对ey求导，然后对y求导。e^y*y'+y+xy'=0 y'(e^y+x)=-y y'=-y\/(e^y+x)
已知隐函数XY=e(X+Y)次方,求dy
解法一：∵xy=e^(x+y) ==>d(xy)=d(e^(x+y)) (两端取微分)==>xdy+ydx=e^(x+y)(dx+dy)==>xdy+ydx=e^(x+y)dx+e^(x+y)dy ==>xdy-e^(x+y)dy=e^(x+y)dx-ydx ==>(x-e^(x+y))dy=(e^(x+y)-y)dx ∴dy=[(e^(x+y)-y)\/(x-e^(x+y))]dx；...
设隐函数y=y(x)由方程tan(x+y)-y=0所确定,求d²y\/dx²?
隐函数求导法则
设e^(x+y)+cos(xy)=0确定y是x的函数求dy
f(x,y)=e^(x+y)+cos(xy)=0 \/\/: 利用隐函数存在定理：f 'x(x,y)=e^(x+y) - ysin(xy)f 'y(x,y)=e^(x+y) - xsin(xy)dy\/dx = - f 'x\/f 'y = -[e^(x+y) - ysin(xy)]\/[e^(x+y) - xsin(xy)]dy = -[e^(x+y) - ysin(xy)]\/[e^(x+y) - ...
隐函数求导
因为函数y=y(x)是由e^(x+y)-xy=1确定的，对方程两边求导得 e^(x+y)*(1+y')-(y+xy')=0 e^(x+y)+y'e^(x+y)-y-xy'=0 〔e^(x+y)-x〕y'=y-e^(x+y)所以，y'=〔y-e^(x+y)〕\/〔e^(x+y)-x〕。
设y=y(x)是由方程e^y=x²+y所确定的隐函数,求dy\/dx?
方法如下，请作参考：若有帮助，请采纳。
