函数y=y(x)由方程e^x-e^y-xy=0确定
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/27 07:21:28
函数y=arctane^x求dyy'=e^x/(1+e^2x)dy=e^xdx/(1+e^2x)函数y=y(x)由方程x-y-e^y=0确定,求y'(0)两边对x求导:1-y'-y'e^y=0y'=1
两边同时对x求导有e^x²'-e^y²'-(xy)'=02e^x²-2e^y²y'-y-xy'=02e^x²-y=2e^y²*y'+xy'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
两边同时对X求导y+xy`=e^x+y`y`=(e^x-y)/(x-1)
xy=e^(x+y)两边对x求导得y+xy'=e^(x+y)(1+y')y-e^(x+y)=[e^(x+y)-x]y'y'=[y-e^(x+y)]/[e^(x+y)-x]
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!
两边对x求导:y'e^y+(1+y')cos(x+y)=0,1)这里可得到y'=-cos(x+y)/[e^y+cos(x+y)]再对1)求导:y"e^y+(y')^2e^y+y"cos(x+y)-(1
e^y=sin(x+y)两边求导得e^y*y'=cos(x+y)(x+y)'=cos(x+y)(1+y')=cos(x+y)+y'cos(x+y)[e^y-cos(x+y)]y'=cos(x+y)y'
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.
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]
同意楼上的,两边同时微分e^xdx-e^ydy-xdy-ydx=0所以dy/dx=(e^x-y)/(e^y+x)
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'(
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'
两边对x求导xy^2+sinx=e^yy^2+2xyy'+cosx=e^y*y'y'(e^y-2xy)=y^2+cosxy'=(y^2+cosx)/(e^y-2xy)
两边对x求导:1+y'=y'e^y得dy/dx=y'=1/(e^y-1)
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)]=[