y-1-x*siny=0.求dy dx

z对x的偏导=cosx+cos(x+y)=0时,cosx=-cos(x+y)=cos(pi-x-y),所以x=pi-x-y.同理z对y的偏导=0时,有y=pi-x-y.所以x=y=pi/3.此时z=3

再问：大哥，你题目看错了。。。再答：哪里有错？再问：第一条等式就错了。。是sin(x+y)=sinx+siny。后面是cos（x+y）·（1+y'）=cosx+cosy·y'？再答：OK，那我改下

两边求导：cos(x+y)*(1+y')=cosx+cosy*y'y'=(cosx-cos(x+y))/(cos(x+y)-cosy)e^x+1=e^y*y'+y'y'=(e^x+1)/(e^y+1)

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

cos(x-y)=cosxcosy+sinxsiny=[1-(sinxsiny)^2]^(1/2)+sinxsiny=0+1=1

/>sinx+siny=1/3,cosx-cosy=1/5两式分别平方得sin²x+sin²y+2sinxsiny=1/9cos²x+cos²y-2cosxco

【1】sinx+siny=1/4左右平方,【2】cosx+cosy=1/3左右各平方,【1】【2】相加可求出cos{x-y}那么sin[x-y]2+cos[x-y]2=1可求出sin[x-y]的值上下

1.由x-y=派/3得y=x-派/32.将y=x-派/3代入sinx-siny=1/2中得y=派/6x=派/23.代入cos(x+y）得cos2派/3=1/2

1+y'=cosy*y'y'=1/(cosy-1)dy/dx=1/(cosy-1)

y=(1+x平方)/5x的导数y'=(1/5x+x/5)'=-1/5x²+1/52、y=x+1/2siny、两边关于x求导得：y'=1+1/2cosy*y'整理得：y'=2/(2-cosy)

siny=√10/10由sin^y+cos^y=1得:cosy=3√10/10tany=siny/cosy=1/3在由tan的2倍角公式tan2y=2tany/(1-tan^y)=3/4tan(a+b

两边对x求导有1-y'+y'cosy=0所以y'=1/(cosy-1)

x-y+1/2siny=0F(x,y)=y-x-1/2siny=0F,Fx,Fy在定义域的任意点都是连续的,F(0,0)=0Fy(x,y)>0f'(x)=-Fx(x,y)/Fy(x,y)=1/(1-1

两边关于x求导,注意y是x的函数y'cosy=[1/(x+y)]*(1+y').①解得y'=1/(x+y)÷[cosy-1/(x+y)].②对①两边关于x求导可得y''cosy-(y')²s

x*e^y+siny=0e^y+x*e^y*y'+cosy*y'=0=>y'=-e^y/[xe^y+cosy]再问：你好！我数学太烂。。能不能补充一下完整的答案。。。再答：x*e^y+siny=0两边

y=siny+(sinx)^2-1.(sinx^2+cosx^2=1)=1/3-sinx+(sinx)^2-1=(sinx)^2-sinx-2/3=(sinx-1/2)^2-2/3-1/4=(sinx

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+

cosx-cosy=1/4cos²x-2cosxcosy+cos²y=1/16sinx-siny=1/3sin²x-2sinxsiny+sin²y=1/9两式相

两边对x求导有1-y'+y'cosy=0所以y'=1/(cosy-1)