y∧2-2xy-12=0在点(2,6)处的切线方程为
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/23 13:17:36
设曲线方程为y=a+kx,其中k是斜率把点(0,1)带入得1=a再把斜率2xy和a=1同时带入方程y=a+kx得y=1+2xy*x整理得:y=1/1-2x^2
极限存在的条件是(x,y)以任何方式靠近(0,0)极限都相等所以证明极限不存在就是找两种不同的方式,使得极限不相等证明如下:取x=y,f(x,y)=x^2/2x=x/2显然极限=0/2=0又取x=-y
已知点P(x,y),且xy=0则P点在坐标轴上xy=0∴x=0或y=0
1)隐函数求导y'=(2x)/(x^2-2y+y^2),y在(1,0)上的导数是22)两边关于x求导得y'=(3y^2)/(3xy-1)再求导并把y‘代入得y''=(27(-y^3+2xy^4))/(
f(x,y)={xy/[2-√(4+xy)]=-2-√(4+xy),xy≠0;{4,xy=0,在点(0,0),(1,0)处不连续,在(1,2)处连续.再问:能简述下原因么?再答:f(0+,0+)=-4
y-x-2xy=0y-x=2xyx-y=-2xy(3x+xy-3y)/(y-xy-x)=[3(x-y)+xy]/[(y-x)-xy]=(-6xy+xy)/(2xy-xy)=-5xy/xy=-5
两边对x求导:[e^(2x - y)](2 - y') - [cos(xy)]*(y + xy')&nb
y=3xy³-2x=x(3y³-2)x=(3y³-2)/yx'=2(3y²+1)/y²当y=1时,x'=8x-1=8(y-1)y=x/8+7/8
若xy=0则x=0或y=01.x=0时,p点在y轴上2.y=0时,p点在x轴上
(4xy+12y)+[7x-(3xy+4y-x)]=4xy+12y+7x-3xy-4y+x=xy+8x+8y=xy+8(x+y)=(-2)+8*3=-2+24=22
1.分析:(1)实质是求x,y异号的象限.一种:x>0,且y
u=y+xy-2-zau/ax=yau/ay=1+xau/az=-1n=(y,1+x,1)=(1,2,-1)
y-x-2xy=0所以x-y=-2xyy-x=2xy所以原式=[3(x-y)+xy]\[(y-x)-xy]=[3×(-2xy)+xy]\(2xy-xy)=-5xy\xy=-5
f'x=(y·(x+y^2)-xy)/(x+y^2)²=y³/(x+y^2)²,则f'x(1,1)=1/4fy=(x·(x+y^2)-(xy)·2y)/(x+y^2)
分解因式(将x看成常数,解出y)得(x+y)(x-2y)=0所以是两条直线y=-x和y=1/2x,还要除去y=0(即x轴)
x>0;y>0积3>=(6xy)^0.5=>9>=6xy=>1.5>=xy=>max(xy)=1.5...ans此时2x=3y=3;x=1.5,y=1
(-1,2),(-2,1)啊,这不是很简单么再问:噢噢,但是谢谢啊再答:没关系,帮到你就好
x+y=5xy(2x-3xy+2y)/(x+2xy+y)=[2(x+y)-3xy]/[(x+y)+2xy]=(2×5xy-3xy)/(5xy+2xy)=7xy/7xy=1再问:若x+1/x=3,求(x
x^2-7xy+12y=0,且x.y不等于0(x-3y)(x-4y)=0x=3y或x=4y(x^2-2xy+y^2)/(2xy)=(x-y)^2/(2xy)x=3y时原式=4y^2/(6y^2)=2/
x²-7xy+12y²=0(x-3y)(x-4y)=0x1=3yx2=4yx=3y时原式=9y²-3y²+y²/6y²=7/6x=4y时原式