nynw.net
当前位置:首页 >> lim (x→0,y→0)(xy%sin(xy))/(xy%xyCos(xy))=1/3,... >>

lim (x→0,y→0)(xy%sin(xy))/(xy%xyCos(xy))=1/3,...

令t=xy, x->0,y->0即得t->0 Taylor展开:sin(t)=t-t^3/6+o(t^3),cos(t)=1-t^2/2+o(t^2) 所以原式=lim(t->0)(t-sin(t))/(t-tcos(t))=lim(t^3/6+o(t^3))/(t^3/2+o(t^3))【o(t^3)是无穷锌 =1/3+o(1)=1/3 编辑的格式不规范,但应该不影响理解

0/0型等价交换公式,xy-sinxy等价于(xy)³/6, 1-cosxy等价于(xy)²/2,所以原式就等于见图吧 等价公式很重要呀,要记得

设u=xy,则du=ydx+xdy,于是xdy=du-ydx,原方程变为 (sinu+ucosu)dx+cosu(xdu-udx)=0, 等式两边除以cosu,得(tanu+u)dx+xdu-udx=0, 1/tanudu=-1/xdx, 两边积分得ln|sinu|=-ln|x|+c 取指数得到xsinu=c1(c1为常数) 即xsin(xy)=c1

df=ydx+xdy+cos(x+y)(dx+2dy) =[y+cos(x+y)]dx+[x+2cos(x+y)]dy 所以: df(1,2)=(2+cos5)dx+(1+2cos5)dy. 即方向导数为:((2+cos5),(1+2cos5))

不妨设z

对x的偏导数其实就将y看成是常数,当成一阶导数来求就可以了: ∂z/∂x=e^sin(xy)*cos(xy)*y 对y的偏导数同理: ∂z/∂y=e^sin(xy)*cos(xy)*x

如果相等 sin2xy=2sinxycosxy=2sinxy cosxy=1 所以当xy=2πk,k为整数时才相等

dz=cos(xy)xdy+cos(xy)ydx 由z/x=ycos(xy),z/y=xcos(xy)→dz=ycos(xy)dx + xcos(xy)dy. cos(x-y)dx-cos(x-y)dy

这涉及到大学里偏导数的概念,对x求导,可以把y看做常数,这样就很好理解了吧

解:∵xy’+x+sin(x+y)=0 ==>x(y'+1)=-sin(x+y) ==>x(dy/dx+dx/dx)=-sin(x+y) ==>xd(x+y)/dx=-sin(x+y) ==>d(x+y)/sin(x+y)=-dx/x ==>sin(x+y)d(x+y)/sin²(x+y)=-dx/x ==>d[cos(x+y)]/[1-cos²(x+y)]=dx/x ==>{1/[1+cos(x+y)]+1/[1-cos(x...

网站首页 | 网站地图
All rights reserved Powered by www.nynw.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com