已知y=e^xy*y,求dy
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xy-eˆ(2x)=siny两边对x求导,得y+x(dy/dx)-2eˆ(2x)=(cosy)*(dy/dx)(x-cosy)*(dy/dx)=2eˆ(2x)-ydy/d
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
方程两边对x求导,得:y+xy'+y'e^y=2y+2xy'y'e^y-xy'=y得y'=y/(e^y-x)因此dy=ydx/(e^y-x)
e^(x+y)-xy=1两边同时求导,e^(x+y)*(1+dy/dx)-y-xdy/dz=0(1)验证x=0,y=0在原曲线上.令x=0,y=0代入到(1)e^0*(1+dy/dz)-0-0*dy/
e^(xy)+sin(xy)=y(y+xy')e^(xy)+(y+xy')cos(xy)=y'y'=(ye^(xy)+ycos(xy))/(1-xe^(xy)-xcos(xy))
对方程取导数y+x(dy/dx)+(dy/dx)=0(dy/dx)(x+1)=-ydy/dx=(-y)/(x+1)
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
两边同时微分.e^ydy-ydx-xdy=0.变下形.答案就出来了
ydx/dy+x=(e^x)(e^y)dx/dy+(e^x)(e^y)dx/dy=[(e^x)(e^y)-x]/[y-(e^x)(e^y)]dx/dy=(xy-x)/(y-xy)dx/dy=x(y-1
f(x,y)=e^(x+y)+cos(xy)=0 //: 利用隐函数存在定理:f 'x(x,y)=e^
两边求导e^y×y'=xy'+yy'=y/(e^y-x)dy/dx=y/(e^y-x)
三种方法1式中同时对x求导-(y+xy‘)cosxy+2yy'=0解出y’2式中同时取微分d{sin(xy)+y^2-e^2}=dsin(xy)+dy^2-de^2=-cosxydxy+2ydy=-c
直接求导:e^y*y'+y+x*y'=2y*y'解得y'=dy/dx=(-y)/(e^y+x-2y).
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
y+xy'=(1+y')e^(x+y)则y'=(y-e^(x+y))/(e^(x+y)-x),dy=(y-e^(x+y)/(e^(x+y)-x)dx
对xy-e^x+e^y=0求微分得ydx+xdy-e^xdx+e^ydy=0(y-e^x)dx+(x+e^y)dy=0dy/dx=(x+e^y)/(e^x-y)
对x求导y+x*y'=e^(x+y)*(1+y')y+x*y'=e^(x+y)+e^(x+y)*y'所以dy/dx=[e^(x+y)-y]/[x-e^(x+y)]
x=0,e^y=e,y=1xy+e^y=ey+xdy/dx+e^ydy/dx=0dy/dx=-y/(x+e^y)dx/dy|x=0=-1/e
令e^(xy)=u,y=lnu/xDy/dx=[(x/u)*(du/dx)-lnu]/x²,∴(1/ux)*(du/dx)-lnu/x²+lnu/x²=u即du/u