若函数f(x,y)={sin(x^2xy) xy,xy=0

函数f（x)=sin(3x+y)是偶函数,则f(-x)=sin(-3x+y)=f(x)=sin(3x+y)由正弦函数的性质sin(π-x)=sinx及周期性可得(-3x+y)+(3x+y)=π+2kπ

f(x)=0.5(1-cos2x)+0.5sin2x+cos2x=0.5+0.5(sin2x+cos2x)=0.5+0.5根号2sin(2x+π/4)f(x)的最小值是（1-根号2)/2

由条件知sin²x≤1/4,-1/2≤sinx≤1/2由图像知,-π/6+2kπ≤x≤π/6+2kπ所以f(sin²x)的定义域为[-π/6+2kπ,π/6+2kπ],k∈Z

2kπ－π/2≤2x＋π/3≤2kπ＋π/2得：kπ－5π/12≤x≤kπ＋π/12增区间是：[kπ－5π/12,kπ＋π/12],其中k∈Zx∈[－π/6,π/6],则：2x＋π/3∈[0,2π/3

y=f(x)=(sinx)平方-根号3sin(pai+x)*cosx=(1/2)(1-cos2x)+根号3sinxcosx=(√3/2)sin2x-(1/2)cos2x+1/2=sin(2x-π/6)

y'=cos(x+y)(1+y')y'=cos(x+y)/(1-cos(x+y))

设函数f（x,y)=sin(x+y),那么f(0,xy）=(sinxy)应该是sin0+sinsy=0+sinxy=sinxy再问：limsinxy\2x=()补充x→0，y→3另外一道题

∵f(x)＝2sin2x−23sinxsin(x−π2)=2sin2x+23sinxcosx=1−cos2x+3sin2x=1+2sin(2x−π6)∵0＜x＜2π3∴−π6＜2x−π6＜7π6∴−1

第一题对x求偏导,那么y就是常数因为在xy=0出不连续所以要这么求=(lim△x->0)(f(x+△x,y)-f(x,y))/△x把x=0y=1带入得(lim△x->0)sin△x²/△x&

dy/dx=cos{f[sinf(x)]}*{f[sinf(x)]}'=cos{f[sinf(x)]}*f‘[sinf(x)]*[sinf(x)]’=cos{f[sinf(x)]}*f‘[sinf(x

f(π/3)=2sin(π+y)=-2siny=√2f(3π)=2sin(9π+y)=-2siny=√2

f(x)=sin(πx/2-π/6)-2cos²πx/4+1-3=sinπx/2cosπ/6-cosπx/2sinπ/6-cosπx/2-3=√3/2sinπx/2-1/2cosπx/2-c

这道题你先看sinx必然大于等于零吧,sin((1-y)x)也必然大于等于零的吧?整个函数都是大于等于0的吧?那么你只要找到可以让函数取到零的x和y就可以得到最小值0那么试着凑一下,y=1,x=pai

它是先得出：pi/2〈=ωx

y'=f'(sin^2x)*(sin^2x)'=g(sin^2x)*2sinxcosx

f(x)=根号下2sin(2x-π/4)+1x∈(0,π/2)则-π/4

f(x)=sin2(x+y/2)由于sin2x对称轴为π/4+kπ/2;故x+y/2=π/4+kπ/2x=π/4+kπ/2-y/2;将x=x=π/8代入,得y=π/4+kπ,根据y的范围可知：y=-3

sin2x的周期为2π/2=π只需sin2x的图像向左移π/4+kπ/2（k∈Z）即关于y轴对称.向左移π/4+kπ/2后的函数为sin2(x+π/4+kπ/2)=sin(2x+π/2+kπ),k∈Z

关于y对称的f(x)=sin(x-90度);2x+z=2(x+z/2)则z/2=-90度z=-180度或者加减整数个180度----------------------------哎,各种符号不会打~

12T=2πT=π22kπ-π/2