设 y=f(x)的导函数是sinx ,则 的全体原函数为
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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
对f(x)作图,你可以很容易的发现,在0到π范围内,是sin(3x)的叠加,在π到2π范围内,sin(3x)为负,|sin(3x)|是正,它们抵消掉了,所以这个范围内f(x)是等于0的.后面的以此类推
(1)f(x)=-f(x+2)=-[-f(x+2+2)]=f(x+4)故f(x)是周期为4的周期函数.(2)当1≤x≤3时,-1≤x-2≤1,故f(x-2)=sin(x-2),又因为f(x+2)=-f
y'=f'(sin²x)*(sin²x)'+f'(cos²x)*(cos²x)'=f'(sin²x)*(2sinxcos)+f'(cos²x
见图,复合函数求导.
y=sin(x+y),y'=cos(x+y)*(1+y'),y'=cos(x+y)/(1-cos(x+y))=dy/dx
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
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
dy=dsin(x+y)dy=cos(x+y)d(x+y)dy=cos(x+y)(dx+dy)dy=cos(x+y)dx+cos(x+y)dy所以dy/dx=cos(x+y)/[1-cos(x+y)]
3*x^2*f`(x^3)
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
(1)由已知,f(x)=sin(2x+b)的周期为π.且当x=π/8时,sin(2*π/8+b)=sin(π/4+b)=±1,所以π/4+b=kπ+π/2.(k∈Z),b=kπ+π/2-π/4=(k+
f'(x)=f'(e^-(x^2))*(e^-(x^2))'=f'(e^-(x^2))*(e^-(x^2))*(-(x^2))'=f'(e^-(x^2))*(e^-(x^2))*(-2x)这个是复合函
用辅助角公式~sin的就不用管它外面就只有y再问:能不能写一下详细过程,谢谢再答:不好意思我竞赛没认真读一般一试的填空我都是用猜的我现在高三备高考而且三角的转化我不在行你可以去竞赛吧问问里面的人比较牛
将已知项展开,得3sinβ=sin2acosβ+cos2asinβ将两边同除以cosβ,得3tanβ=sin2a+cos2atanβ移项合并,将非tanβ除向另一边,得tanβ=2sin2a/3-co
上图吧再问:哦,wodongle~
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
1.f(x)=sin|x|是有界函数正确,因为|sin|x||≤1.2.设X和Y分别是同一变化过程中的两个无穷大量,则X-Y是无穷小量,不对.设X=n,Y=n-1,当n→∞时,X和Y都趋于无穷大,但X
【2kπ,2kπ+π】(k∈Z)再问:函数y=sin²x的n阶导数的表达式会么?再答:不一定对n为奇数,y(n)=(-1)的n次幂乘以2的(n-1)次幂乘以sin2xn为偶数,y(n)=(-