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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

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

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

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...

dcos(xy)+dlny-dx=d(1) -sin(xy)d(xy)+(1/y)dy-1=0 -sin(xy)(xdy+ydx)+(1/y)dy-1=0 -xsin(xy)dy-ysin(xy)dx+(1/y)dy-1=0 所以dy/dx=[ysin(xy)+1]/[1/y-xsin(xy)] =[y²sin(xy)+y]/[1-xysin(xy)]

y=cos(xy)-x 求导?还是求什么?

两边对x求导:-y^2+cos(y+z)Z'x=0, 得:Z'x=y^2/cos(y+z) 两边对y求导:-2xy+cos(y+z)(1+Z'y)=0, 得:Z'y=2xy/cos(y+z)-1

u=xysin(x+y) du=[ysin(x+y)+xycos(x+y)]dx+[xsin(x+y)+xycos(x+y)]dy

∂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)

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