xy=ex y,求隐函数y的导函数
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/07 21:45:41
方程两边求关x的导数ddx(xy)=(y+xdydx); ddxex+y=ex+y(1+dydx);所以有 (y+xdy
隐函数求导,两边同时求导,此题是对X求导!两边同时求导:y+xy'=e^x-y'y'=(e^x-y)/(x+1)由XY=e^X-y解出yy=e^x/x+1,带入上式y'=(e^x-y)/(x+1)=[
方程两边对x求导:e^y×y'=y+xy'得y'=y/(e^y-x)
E(xy)=∫xy*f(xy)dxdy
首先z'(x)=x*(a-x-2*y)=0z'(y)=y(a-y-2*x)=0计算得到四组解(0,0)(a,0)(0,a)(a/3,a/3)1.(0,0)时,f''xx=0,f''xy=a,f''yy
z=xy+x/y对x的偏导数=y+1/y对y的偏导数=x-x/y^2
两边同时对X求导y+xy`=e^x+y`y`=(e^x-y)/(x-1)
两边对x求导得y+xy'=(1+y')/(x+y)y(x+y)+x(x+y)y'=1+y'y'[x(x+y)-1]=1-y(x+y)y'=[1-y(x+y)]/[x(x+y)-1]dy=[1-y(x+
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]
∵dudx=∂f∂x+∂f∂y•dydx+∂f∂z•dzdx…(1)由exy-xy=2,两边对x求导得:exy(y+xdydx)-(y+xdydx)=0解得:dydx=-yx.又由ex=∫x-z0si
y'=(y+xy')/(xy)xyy'-xy'=yy'=y/(xy-x)所以dy/dx=y'=y/(xy-x)
ezplot('exp(x*y)-sin(x+y)=0',[-3,3])
直接两边对x求导,得1/y*(-1/y2)*dy/dx=1/xy*(y+xdy/dx)下面会了吧
y+xy'+y'/y=0//对xy和lny分别求导,注意y是x的函数y'(x+1/y)=-y//移项,合并同类项y'=-y²/(xy+1)
xy=e^x-e^yd(xy)=d(e^x-e^y)xdy+ydx=e^xdx-e^ydy(x+e^y)dy=(e^x-y)dx则由dy/dx=(e^x-y)/(e^y+x)
x=yln(xy),等式两端对x求导,1=dy/dx+y[1/ln(xy)][y+x(dy/dx)]=dy/dx+y/ln(xy)+xdy/dx,整理得(dy/dx)(1+x)=1-y/ln(xy),
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'(
第一步方程两边对x求导记y+xy'-y'/y=2x第二步解出y'记y'=(2xy-y^2)/(xy-1)
答:1)y/x=ln(xy^2)两边求导:y'/x-y/x^2=[1/(xy^2)]*(y^2+2xyy')(xy'-y)/x=(y+2xy')/yy'-y/x=1+2xy'/y(1-2x/y)y'=
两边对x求导得e^y*dy/dx+y+xdy/dx=0解得dy/dx=-y/(e^y+x)再两边对x求导,左边是所求右边会出现y的一阶导数把上式带入就得到结果了