已知:e^(xy) ln(y (x 1))=0,求:y(0)-搜答案-搜狗做题网

y/x=ln(xy) 求详 dy/dx

方法一（微分法）d(y/x)=d(ln(xy))(xdy-ydx)/x²=1/xy*d(xy)即（xdy-ydx)/x²=(ydx+xdy)/xy∴dy/dx=(xy+y²

y/x=ln(xy) 求dy/dx

两边求导(y'x-y)/x^2=(y+xy')/xyxy+x^2y'=xyy'+y^2y'=(xy-y^2)/(xy+x^2)

直接在等式中零,x=0,y=y（0）,可得关于y（0）的方程解出y（0）即可.具体：e^0*y(0)+lny(0)/1=0即-y（0）=lny（0）作图y1=-x,y2=ln（x）,两者的交点的横坐标

答：xy+ln(x+e^2)+lny=0……（1）两边对x求导：y+xy'+1/(x+e^2)+y'/y=0……（2）x=0代入（1）和（2）得：0+2+lny=0y+0+1/e^2+y'/y=0解得

y'e^y+e^x-y²-2xyy'=0y'=(e^x-y²)/(2xy-e^y)即：dy/dx=(e^x-y²)/(2xy-e^y)祝你开心!希望能帮到你,如果不懂,请

设Y=y'降阶：Y'=(Y/x)ln(Y/x)这就是一个一阶齐次方程.设Y/x=u,所以Y=ux,Y'=u+x(du/dx),代回原方程,解得：lnu=C1x+1Y=xe^(C1x+1)所以y=[(C

运用函数连续性,化成一元函数求极限x→0,y→2lim[ln(x+e^xy)/x]=x→0lim[ln(x+e^(2x)]/x【0/0型】=x→0lim[ln(1+(x+e^(2x)-1)]/x=x→

ln(x+2y)=sin(xy)+1对x求导1/(x+2y)*(x+2y)'=cos(xy)*(xy)'+0(1+2y')/(x+2y)=cos(xy)*(y+x*y')x=0则ln(0+2y)=0+

[ln(xy)]'=[e^(x+y)]'(xy)'/(xy)=e^(x+y)*(x+y)'(y+xy')/(xy)=e^(x+y)*(1+y')y'=y[e^(x+y)-1]/[x(1-ye^(x+y

将x=0代入方程得：lny=1,得y=e方程两边对x求导：y+xy'+e^xlny+y'e^x/y=0代入x=0,y=e得：e+lne+y'/e=0,得y'=-e(e+1)即y'(0)=-e(e+1)

y=ln(x+e^x^2)dy/dx=y'=[1/(x+e^x^2)]*(1+2xe^x^2)

dy/dx=dy/du*du/dx+dy/dv*dv/dx=v*e^(x+y)+u*y/x=ln(xy)*e^(x+y)+e^(x+y)*y/x=e^(x+y)[ln(xy)+y/x]所以dy=e^(

y=ln(x^2+e^x) 求Y'X

如果是求导数的话,y'=(2x+e^x)/(x^2+e^x)

ln（xy）=e^x+ylnx+lny=e^x+y两边同时对x求导1/x+(1/y)(dy/dx)=e^x+dy/dxdy/dx=[(1/x)-e^x]/[1-(1/y)]=(y-xye^x)/(xy

两边求导得y'·e^y+(y+xy')/(xy)+e^(-x)=0

y`=ex^(e-1)+e^x+1/x

min是指f(x)g(x)h(x)三个函数中的最小值

sin(xy)-ln((x+1)/y)+1=0对x求导有：（y+xy')cos(xy)-y/(x+1)·[y-(x+1)y']/y^2-y/(x+1)·(x+1)(-1/y^2)y'=0x=0代入有：

已知,lny=(sinx)*e^y.求dy/dx解一：利用隐函数求导公式求解【建议你用这个】设F(x,y)=lny-(sinx)e^y=0则dy/dx=-(∂F/∂x)/(&#