2x^2 y^2-cosy=0所确定的隐函数
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有cos(x+y)=cosxcosy-sinxsiny又将题目所给式子左右平方,在两个式子左右对应相加为:cos^2x+cos^2y+2cosxcsoy+sin^2x+sin^2y-2sinxsiny
极值就是求导fx=cosx-sin(x-y)=0fy=-siny+sin(x-y)=0x+y=pi/2f(x,y)=1+0+0=1极小值这是f(x,y)
解cosx+cosy-cos(x+y)=3/22cos[(x+y)/2]*cos[(x-y)/2]-2{cos[(x+y)/2]}^2=1/2{cos[(x-y)/2]-2cos[(x+y)/2]}^
有cos(x+y)=cosxcosy-sinxsiny又将题目所给式子左右平方,在两个式子左右对应相加为:cos^2x+cos^2y+2cosxcsoy+sin^2x+sin^2y-2sinxsiny
3:通分,将3个分式的分母化成相同的,即sin(x-y)sinz/(sinxsinysinz)+sin(y-z)sinx/(sinxsinysinz)+sin(z-x)siny/(sinxsinysi
∵(1+y²)dx+(xy-√(1+y²)cosy)dy=0==>√(1+y²)dx+(xy/√(1+y²)-cosy)dy=0(等式两端同除√(1+y
思路:利用极值和导数的关系(极值点,导数为0)函数关于x,y求偏导数,令其为0,解出x,y的值,和相应的函数值,那就是极值
cos(2x-2y)=cos2(x-y)=2cos^2(x-y)-1cos(x-y)=cosxcosy+sinxsiny(sinx+siny)^2=sin^2x+sin^2y+2sinxsiny=4/
把cosy看作新的因变量,令z=cosy,原方程化为dz/dx+2/x×z=-3,一个线性方程,套用通解公式,z=1/x^2×(-x^3+C).原方程的通解是cosy=1/x^2×(-x^3+C),即
cos(2x+y)=cos[(x+y)+x]=cos(x+y)cosx-sin(x+y)sinxcosy=cos[(x+y)-x]=cos(x+y)cosx+sin(x+y)sinx3cos(2x+y
消元sinx=1-siny----(1cosx=-cosy-----(2(1)式^2+(2)式^2--------1=(1-siny)^2+cosy^21=1+sin^2y+cos^2y-2siny1
∫∫(D)cosy²dxdy=∫(0,2)cosy²dy∫(1,y+1)dx(∫(a,b)表示从a到b积分)=∫(0,2)cosy²*[(y+1)-1]dy=∫(0,2)
x=(x+y)/2+(x-y)/2y=(x+y)/2-(x-y)/2所以左边=cos[(x+y)/2+(x-y)/2]-cos[(x+y)/2-(x-y)/2]={cos[(x+y)/2]cos[(x
题目有没有问题呢?怎么觉得应该是求tan(x+y)tanx呢?∵2cos(2x+y)=cosy∴2cos(x+x+y)=cos(x+y-x)∴2cosxcos(x+y)-2sinxsin(x+y)=c
原条件等价于cos(x-y)=1/3根据倍角公式cos2(x-y)=2cos^2(x-y)-1=2/9-1=-7/9
sinx+siny=2sin(x/2+y/2)·cos(x/2-y/2)=-1/3----①cosx+cosy=2cos(x/2+y/2)·cos(x/2-y/2)=1/2----②①/②=tan(x
两式分别求平方得:sinx2-2sinxsiny+siny2=1/4,cosx2-2cosxcosy+cosy2=1/4相加得:2-2(sinxsiny+cosxcosy)=1/2,sinxsiny+
题目应该是“证明cosx(cosx-cosy)+sinx(sinx-siny)=2sin²(x-y)/2”Pr:左边展开得cos²x-cosxcosy+sin²x-sin
y`=x^2/cosysinydy=x^2dx积分得通-cosy=x^3/3+C或:y=arccos(-x^3/3-C)再问:sinydy=x^2dx,拜托--你都cosy,变成siny了还dy啊再答
cos(2x+y)=3cosycos(x+y+x)=3cos(x+y-x)cos(x+y)cosx-sin(x+y)sinx=3[cos(x+y)cosx+sin(x+y)sinx]2cos(x+y)