Chromosome painting reveals that these X and Y chromosomes contain pairing (XY shared) and differential (X- or Y-specific) segments. Y differential regions must contain male-determining genes, and X differential regions should be dosage-compensated in the female. Two models for the evolution of ...
N. Tzanakis, The Diophantine equation x3 -3xy2 -y3 = 1 and related equations, J. Number Theory 18 (1984), no. 2, 192-205.The Diophantine equation x 3 −3xy 2 −y 3 = 1 and related equations - Tzanakis - 1984 () Citation Context ...e four solutions of the Diophantine ...
运江回【】求曲线x3+y3-3xy=0在点(,)处的切线方程和法线方程. 求曲线x 3 +y 3 -3xy=0在点( , )处的切线方程和法线方程.运江回
During the evolution of vertebrates, the XY and ZW chromosomes developed independently through translocations and fusion/fission rearrangements, generating multiple sex chromosomes1,2,3. In the specific case of reciprocal translocations, sex chromosomes and autosome pairs suffer breaks, and the resulting s...
首先，我们设函数为f(x,y)=x³+y³-3xy，为了找到其极值点，我们需要计算一阶偏导数并令其等于零。计算一阶偏导数：fx(x,y)=3x²-3y=0 fy(x,y)=3y²-3x=0 解上述方程组，得到两个驻点：x=0,y=0；和x=1,y=1。接下来，我们需要计算二阶偏导数，并将驻点代入...
X3 + Y3曲线隐函数表示 - 对4xy = 0(2,2)点要求提供切线方程。 翻译结果2复制译文编辑译文朗读译文返回顶部 由隐式函数曲线 x 3 + y3-4 xy 上问方程的切线点 = 0 (2,2)。 翻译结果3复制译文编辑译文朗读译文返回顶部 由隐式函数曲线 x 3 + y3-4 xy 上问方程的切线点 = 0 (2,2)。
y0)fyy(x0,y0)-fxy(x0,y0)^2>0 在该稳定点出我们算出的f(x0,y0)就是极大值。（3）若fxx(x0,y0)fyy(x0,y0)-fxy(x0,y0)^2<0，在该稳定点取不到极值。（4）若fxx(x0,y0)fyy(x0,y0)-fxy(x0,y0)^2=0，无法确定是否取到极值。【这些内容书上都有的，就是比较散】
x3+y3-3xy=0+经不经过第三象限解题x³+y³-3xy=0两边同时对x进行求导得3x²+3y²y’-3y-3xy’=0y’=（y-x²）/（y²-x）令y’=0既（y-x²）/（y²-x）=0解得x=0，x=1所以函数f（x，y）=x³+y³-3xy在（-∞，0）（1，+∞）上单调递增在（0，1）上...
求方程y3+x3-3xy=0的隐函数的导数yˆ 相关知识点: 试题来源: 解析 隐函数求导,将y看成x的函数,将等式两边分别对x求导,3(y2)y'+3(x2)-3y-3xy'=0,提公因子y'=(3y-3(x2))/(3y2-3x),化简得: y'=(y-x2)/(y2-x) 反馈 收藏 ...
y³+x³-3xy=0,differentiate both sides w.r.t.x 3y²*y'+3x²-3(y+xy')=0 3y²*y'-3y-3xy'=-3x²y'(3y²-3x)=3y-3x²y'=(3y-3x²)/(3y²-3x)=(y-x²)/(y²-x)...