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

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

由圆的方程得到圆心O(0,0),半径r=5,∵圆心O到直线l的距离d=1cos2θ+sin2θ=1<5,且r-d=5-1>1,∴圆O上到直线l的距离等于1的点的个数为4,即k=4.故答案为:4

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)

不妨设z

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

两边同时对x求导 e^x-e^yy'=cos(xy)(y+xy') y'=[e^x-ycos(xy)]/【xcos(xy)+e^y】 dy/dx=[e^x-ycos(xy)]/【xcos(xy)+e^y】

你的解题过程是正确的。

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