y=2cos^2x sin2x最小正周期是
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limx→0(ln(x^2+1))/xsin2x=lim(ln(1+x^2)/2x^2)/(sin2x/2x)ln(1+x)~xln(1+x^2)~x^2lim(ln(1+x^2)/2x^2)/(si
Y′=(xsin2x)′+(cos3x)′=sin2x+2x²cos2x-3xsin3x
解1)答案e^(4xsin2x)*(4sin2x+8xcos2x)2)答案[(4lny)-(3lnx)-3-(y/x)]/[(3lnx)-(4x/y)+(2lny)+2]3)答案(3y^2-2xy)/
可以根据:asinα+bcosα=√(a^2+b^2)*sin(α+θ),其中θ由a,b的符号和tanθ=b/a确定具体到这题,就是:y=√[(√3+2)/2]^2+(1/2)^2*sin(2x+θ)
sinx+siny=1/3sinx=1/3-siny(siny)^2+(cosy)^2=1(cosy)^2=1-(siny)^2u=sinx-cos^2yu=(1/3-siny)-[1-(siny)^
特征方程是a^2-2a+5=0,解是一对共轭复数:1+2i,1-2i,因此齐次方程的通解是y=e^x(Ccos2x+Dsin2x).再考虑非齐次方程的特解.设特解为y=xe^x*(mcos2x+nsi
COS(X+Y)COS(X-Y)=(COSX*COSY-SINX*SINY)(COSX*COSY+SINX*SINY)=(COSX*COSY)^2-(SINX*SINY)^2=COS^2X(1-SIN
y1=xsin2x,y2=xcos2x,y3=(x+2)e^x=>y1-y2,y3-y2分别是其对应齐次方程的解,y1-y2=x(sin2x-cos2x),y3-y2=(x+2)e^x-xcos2x其
先求对应的齐次方程的通解y′-y/x=0,dy/y=dx/x,y=Cx.用常数变易法,把C换成u,设y=ux,则u′x+u-ux/x=2xsin2x,u′=2sin2x,u=-cos2x+C.原方程通
sin^2x+cos^2y=1/2∴sin^2x=1/2-cos^2y3sin^2x+sin^2y=3(1/2-cos^2y)+sin^2y=1.5-3cos^2y)+sin^2y又有sin^2y+c
y=cos^2x-sin^2-√3cos(3π/2+2x)+1=cos2x-√3sin2x+1=2cos(2x+π/3)+1当2kπ+π/2≤2x+π/3≤2kπ+3π/2时,函数单调递减2kπ+π/
最小正周期π最大值2最小值0
1)y=xsin2x+cos3xy′=sin2x+2xcos2x-3sin3x;(2)y=cos(-2x+π/6)y′=2sin(-2x+π/6)(3)Y=2^(2x+1)y′=2ln2*2^(2x+
y=sin^2(x)+2sin(x)cos(x)+3cos^2(x)=1+2cos^2(x)+sin2x=2+sin2x+cos2x构造向量a=(sin2x,cos2x),b=(1,1)a+b=(si
y=cosx*(-2)sin[(x+x+2/3)/2]sin[(-2/3)/2]=2cosxsin(x+1/3)sin(1/3)=2sin(1/3)cosxsin(x+1/3)=2sin(1/3)(1
∵齐次方程y''-2y'+5y=0的特征方程为r^2-2r+5=0解得特征根r1=1+2i.r2=1-2i又非齐次方程y''-2y'+5y=e^xsin2x其中f(x)=e^xsin2xα=1,l=0
y=2asinx-cos²x+a²+2=2asinx+sin²x+a²+1=(sinx+a)²+1当a>=0时最小值为f(x)=(a-1)²
1.y=sinx-cosx,求y'|x=π/6y'=cosx+sinxy'|x=π/6=(√3)/2+1/22.y^2-2xy+9=0,求dy/dx直接对方程求关于x的导数:2y*y'-2(y+xy'