f(x,y)=x^2 sin(xy) 2y

sin^2(x-y)+sin^2(y-z)+sin^2(z-x)=[1-cos2(x-y)+1-cos2(y-z)+1-cos2(z-x)]/2=3/2-[(cos2xcos2y+sin2xsin2y

记g(x)=f(x^2+sin^2x)+f(arctanx)=yg'(x)=f'(x^2+sin^2x)(2x+sin2x)+f'(arctanx)/(x2+1)dy/dx|x=0,即g'(0)代入得

y'=f'(sin²x)*(sin²x)'+f'(cos²x)*(cos²x)'=f'(sin²x)*(2sinxcos)+f'(cos²x

这是个函数句柄@(x,y)表示未知数是x和ypunct-Functionhandlecreation@@在匿名函数中表示函数句柄例如ln(x),在matlab中是没有定义的,正确表示是log(x);但

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

d/dx(f(sin^2(x))+sin(f(x)^2)) = sin(2 x) f'(sin^2(x))+2 f(x) f'

y'=f'(sinx^2)*cosx^2*2x-f'(cosx^2)*sinx^2*2x

dy/dx=y'=f'[2x/(x-1)]*[[2x/(x-1)]'=sin[2x/(x-1)]²*-2/(x-1)²=-2sin[2x/(x-1)]²/(x-1)

设x0所以f(-x)=sin2(-x)+cos(-x)=-sin2x+cosx因为f(x)为奇函数,所以f(-x)=-f(x)得f(x)=-f(-x)=sin2x-cosx(x

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&

f(x)=2sin(2x+π/6)=2sin[2(x+π/12)]x+π/12看成是y=f(x+fai)中的x所以y=f(x+fai)=2sin[2(x+π/12+fai)]关于原点对称,即奇函数有f

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

f(x)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx

f(x)=sin^2x-2sinxcosx+3cos^2x=1/2（1-cos2x）-sin2x+3/2（1+cos2x）=2+cos2x-sin2x=2+√2cos（2x+π/4）y=cos2x先沿

令t=sin^2x,则sinx=√t和-√t.若sinx=√t,即x=arcsin√t所以f(t)=arcsin√t/√t.若sinx=-√t,x=-arcsin√t.f(t)=arcsin√t/√t

sinx+siny+sinz-sin(x+y+z)=4sin[(x+y)/2]sin[(x+z)/2]sin[(y+z)/2]sinx+siny+sinz-sin(x+y+z)=2sin[(x+y)/

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

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

y'=f'(sin(2x))*(sin(2x))'+(sin(f(2x)))'*f'(2x)=f'(sin(2x))*2*cos(2x)+cos(f(2

再问：谢谢了写的清楚明白