y=3^tan(x 1) x,求dy dx
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/07 10:32:59
由3x-π3≠kπ+π2,k∈Z,得x≠kπ3+5π18,k∈Z.∴函数y=tan(3x-π3)的定义域为{x|x≠kπ3+5π18,k∈Z}.值域为:(-∞,+∞).由−π2+kπ<3x−π3<π2
由已知2sin(4x1-π/3)-a=2sin(4x2-π/3)-a=0也就是sin(4x1-π/3)=sin(4x2-π/3)=a/2所以说(4x1-π/3)+(4x2-π/3)=kπ(k是整数)或
sin(x+y)=sinxcosy+cosxsiny=1/2sin(x-y)=sinxconsy-cosxsiny=1/3sinxcosy=5/12,cosxsiny=1/12tanx/tany=si
=(secX+tanX)'/(secX+tanX)=(secxtanx+sec²x)/(secX+tanX)=secx(tanx+secx)/(secX+tanX)=secx
∵y=tan(2x-π3),∴其周期T=π2.
dy/dx=sec²(x+y)*(1+dy/dx)则[1-sec²(x+y)]dy/dx=sec²(x+y)则dy/dx=sec²(x+y)/[1-sec
x1+x2=a+d,x1x2=ad-bc则:x1³+x2³=(x1+x2)[(x1+x2)²-3x1x2]=(a+d)[(a+d)²-3(ad-bc)]=(a+
y=3tan(π6-x4)=-3tan(x4-π6),∴T=π|ω|=4π,∴y=3tan(π6-x4)的周期为4π.由kπ-π2<x4-π6<kπ+π2,得4kπ-4π3<x<4kπ+8π3(k∈Z
2x-tan(x-y)=∫(0,x-y)[sec(t)]^2dt两边对x求导得:2-sec²(x-y)(1-y')=sec²(x-y)(1-y')sec²(x-y)(1-
令a=tanx则a属于Ry=f(x)=(a²-a+1)/(a²+a+1)ya²+ya+y=a²-a+1(y-1)a²+(y+1)a+(y-1)=0a是
tan(x)=4tan(y)tany=(tanx)/4arc[(tanx)/4]=y再问:需要x,y的数值关系再答:什么意思?是要求x等于几时,y等于几?再问:要求x=ky,求k值再答:可不可以用特殊
第一道题目,tanX的单调区间是-π∕2﹢nπ,π∕2+nπ.开区间,那个是不是写错了,再把-π∕2+nπ﹤2x-3x∕4﹤π∕2+nπ,前面有负号,所以这就是单调减区间第二道题目cosx的减区间是2
tan9x+45)+tan(x-45)=(1+yanx)/(1-tanx)-(1-tanx)/(1+tanx)=4tanx/(1-tan^2x)=2tan2xy=tan(x+45)+tan(x-45)
复合函数求导法则:y=u,u=v,v=f(x)=>dy/dx=dy/du*du/dv*dv/dx
sin(x+2y)=3sinx,sin[(x+y)+y]=3sin[(x+y)-y],sin(x+y)cos(y)+cos(x+y)sin(y)=3[sin(x+y)cos(y)-cos(x+y)si
y'=1/tan(x/2)*1/cos^2(x/2)*1/2=1/sinx
函数y=tan^2(x)-2tan(x),=(tanx-1)^2+1-60°
当x∈[-π/3,π/4]时-√3
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)