(3x^2-2)sin2x的一百阶导数
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/01 15:37:24
不管是(sinx)^2也好,是sin(2x)也好,极限都是∞.因为1/SIN2X当X趋于0时为∞,而SIN1/X^2为有界函数,故其差值必然为∞.再问:为啥答案上面是-1/2。。。再答:你把题目再写清
(Ⅰ)因为f(x)=sin2x-(1-cos2x)=2sin(2x+π4)-1,所以函数f(x)的最小正周期为T=2π2=π(Ⅱ)由(Ⅰ)知,当2x+π4=2kπ−π2,即x=kπ−π8(k∈Z)时,
(Ⅰ)f(x)=12cos2x+32sin2x+sin2x-cos2x=12cos2x+32sin2x-cos2x=sin(2x-π6)∴周期T=2π2=π,由2x-π6=kπ+π2(k∈Z),得x=
y=1/2sin2x+sin^2x=1/2sin2x+(1-cos2x)/2=1/2sin2x-1/2cos2x+1/2=√2/2(sin2xcosπ/4-cos2xsinπ/4)=√2/2sin(2
1,f(x)=√3sin2x+2cos²x+m=√3sin2x+1+cos2x+m=2sin(2x+π/6)+m+1∵0≤x≤π/2∴π/6≤2x+π/6≤7π/6那么f(x)max=2+m
f(x)=2cos^2x+根号3sin2x-2=cos2x+1+√3sin2x-2=2sin(2x+π/6)-1f(π/3)=2sin(5π/6)-1=1-1=0最小正周期为2π/2=π单调增区间:2
把函数f(x)=sin²x+2sin2x+3cos²x写成f(x)=Asin(wx+φ)+k(A>0,ω>0,|φ|<π/2)的形式f(x)=sin²x+2sin2x+3
(Ⅰ)∵f(x)=3sin2x-2sin2x=3sin2x+cos2x-1=2sin(2x+π6)-1故函数f(x)的最大值等于2-1=1(Ⅱ)由f(x)=0得23sinxcosx=2sin2x,于是
(1)f(x)=cos2xcosπ3−sin2xsinπ3+1−cos2x2=12−32sin2x所以T=2πω=π;当x=kπ−π4(k∈Z)时,fmax(x)=12+32.(2)由f(C2)=−1
先化简cos2x=2cos²x-1原式=cos2x+根号3sin2x+α+1=0.5sin(π/3+2x)+α+1①然后解题2kπ-0.5π
2x趋于无穷所以sin2x在[-1,1]震荡,即有界而1/2x趋于0有界乘无穷小还是无穷小所以原式=0再问:拜托写解题过程式子谢了再答:就是这样啊
原式=4(√3/2sin2x+1/2cos2x)-1=4(cosπ/6sin2x+sinπ/6cos2x)-1=4sin(2x+π/6)-1sin(2x+π/6)的最大值是1最大值是y=4-1=3当2
tan(x-π/4)=?题中没有说?=(tanx-tanπ/4)/(1+tanxtanπ/4)=(tanx-1)/(1+tanx)由此可以得到tanx则tan2x=2tanx/(1-tan^2x)(s
lim(x→0)(2sinx-sin2x)/x^3=lim(x→0)(2sinx-2sinxcosx)/x^3=lim(x→0)2sinx(1-cosx)/x^3=lim(x→0)2x*x^2/2*1
sin²2x+sin2x+cos2x=1sn2x+cos2x=1-sin²2x=cos²2xsin2x=cos²2x-cos2x----------------
(Ⅰ) f(x)=sin2x+3cosxcos(π2-x)=sin2x+3cosxsinx=1−cos2x2+3sin2x2…(5分)=3sin2x2−12cos2x+12=sin(2x-π
2sin2x+cos2x=1,用倍角公式展开,将1移到等号左边,化简得:2sinx(cosx-sinx)=0,cosx-sinx=0则tanx=1,x=45度,最后带进去算就行了
1、y=(x^2-3x+2)e^x,故y'=(x^2-3x+2)'*e^x+(x^2-3x+2)*(e^x)'而显然(x^2-3x+2)'=2x-3,(e^x)'=e^x所以y'=(2x-3)*e^x
y=lg(2sin2x+√3)-√(9-x²)令2sin2x+√3>0,9-x²≥0得sin2x>-√3/2,-3≤x≤3所以2kπ-π/3<2x<2kπ+4π/3,k∈Z,-3≤
0/0型,分子分母同时求导,可得最后结果为1