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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

sin(x²+y²)=xy²-e^x,两边对x求导 cos(x²+y²)(2x+2yy')=y²+2xyy'-e^x 2x·cos(x²+y²)+e^x-y²=y'[2xy-2ycos(x²+y²)] ∴dy/dx=y'=[2x·cos(x²+y²)+e^x-y²]/[2xy-2ycos(x²...

∂Z/∂x= y*cos(xy) -2cos(xy)*sin(xy)*y = y*cos(xy) - y*sin(2xy) ∂Z/∂y= x*cos(xy) -2cos(xy)*sin(xy)*x = x*cos(xy) - x*sin(2xy)

对x求导是ycos(xy).对y求导是xcos(xy)

r^4=ar^3(cos³θ-3cosθsin²θ)r=acos3θ(cos3θ=cosθcos2θ-sinθsin2θ=cosθ(cos²θ-sin²θ)-sinθ*2sinθcosθ=cos³θ-3cosθsin²θ)这是三叶玫瑰线的极坐标方程面积如下(图来自百度),结果前面要乘以a²

不妨设z

Zy=ycos(xy)+2cos(xy)*[-sin(xy)]*x =ycos(xy)-xsin(2xy)

题出错了 二货 第一卦限xyz全是正数 尼玛的x+y+z=0怎么让xyz全为正的时候成立啊

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