y+sinx+cosx的导数

堵肺奇3934函数y=sinx+cosx+2的最小值怎么算主要是过程,sinx和cosx是怎么变化成最后得数2 - 根号2的? -
浦茂往14733943886 ______[答案] y=asinx+bcosx =√(a^2 + b^2) sin(x+c) 这是基本公式 y=sinx+cosx+2=√(1+1)sin(x+c)+2=√2sin(x+π/4)+2 x+π/4=-π/2 即sin(x+π/4)=-1 y=2-√2

堵肺奇3934求y=sinx+cosx+2sinxcosx+2的最值和相应X的值 -
浦茂往14733943886 ______[答案] y=sinx+cosx+2sinxcosx+2=sinx+cosx+2sinxcosx+(sinx)^2+(cosx)^2+1=(sinx+cosx)^2+sinx+cosx+1 t=sinx+cosx=根号2sin(x+pi/4),y=t^2+t+1=(t+1/2)^2+3/4,t=-1/2时有最小值3/4,此时x=-pi/4+arcsin(-根号2/4)+2kpi或-5pi/4-arcsin(-根号2/4)+2kpi t=...

堵肺奇3934设y=sinx+cosx+tanx+cotx+secx+cscx ,则绝对值y的最小值为 答案是根号2 - 1 -
浦茂往14733943886 ______[答案] y=sinx+cosx+sinx/cosx+cosx/sinx+1/cosx+1/sinx =sinx+cosx+1/(sinxcosx)+(sinx+cosx)/(sinxcosx) =sinx+cosx+2/sin2x+2(sinx+cosx)/sin2x 设u=sinx+cosx,则u^2=1+sin2x,y=u+2/(u^2-1)+2u/(u^2-1)=u+2/(u-1)=(u-1)+2/(u-1)+1 因为u=根号2sin(x+π/4),u属于[...

堵肺奇3934怎么画y=|sinx|+|cosx|的图象 -
浦茂往14733943886 ______[答案] 有没有那么烦琐? 我来说: 首先 Y=sinx+cosx=根号2*sin(x+π /4) Y=sinx-cosx=根号2*sin(x-π /4) 这个公式的原体是 asinx+bcosx=根号下(a^2+b^2)*sin(x+arctanb/a) asinx-bcosx=根号下(a^2+b^2)*sin(x-arctanb/a) 这个楼主应该知道. 那么我们来分...

堵肺奇3934y=sinx+cosx+sinxcosx+1的最值 -
浦茂往14733943886 ______[答案] y = sinx+cosx+sinxcosx+1y' = cosx - sinx - (sinx)^2 + (cosx)^2= (cosx+1/2)^2- (sinx+1/2)^2= (cosx+sinx+1)(cosx-sinx)y' =0=> sinx = cosx or cosx+sinx+1=0=> tanx = 1 or sin(π/4 +x ) = -√2/2=> x = π/4...

堵肺奇3934y=sinx+cosx的最大值和最小值分别是多少 -
浦茂往14733943886 ______[答案] y=sinx+cosx =√2(√2/2cosx+√2/2sinx) =√2sin(x+45°) -1 ≤sin(x+45°)≤1 -√2 ≤y ≤√2 y的最大值是√2,最小值是-√2

堵肺奇3934函数y=sinx/sinx+cosx的导数y'= -
浦茂往14733943886 ______[答案] y'=[(sinx)'(sinx+cosx)-sinx(sinx+cosx)']/(sinx+cosx)^2=[cosx(sinx+cosx)-sinx(cosx-sinx)]/(sinx+cosx)^2=[cosxsinx+(cosx)^2-sinxcosx+(sinx)^2]/(sinx+cosx)^2=[(cosx)^2+(sinx)^2]/(sinx+cosx)^2=1/(sinx+cos...

堵肺奇3934sinx+cosx+|sinx - cosx|的值域 -
浦茂往14733943886 ______[答案] y=sinx+cosx+2sinxcosx=(sinx+cosx)^2+sinx+cosx-1 sinx+cosx=根号2sin(x+π/4) 令sinx+cosx=t,则t属于[-根号2,根号2] y=t^2+t-1=(t+1/2)^2-5/4 当t=-1/2时有最小值为-5/4,当t=根号2时有最大值,为1+根号2

堵肺奇3934函数y=sinX+(cosX的平方)的值域是? -
浦茂往14733943886 ______[答案] y=sinX+cos²X =1+sinx-sin²x =5/4-(1/2-sinx)² ∵-1≤sinx≤1 ∴-1≤y≤5/4 值域:[-1,5/4]

堵肺奇3934证明:y=e^ - x(sinx+cosx)满足方程 y''+y'+2e^ - xcosx=0 -
浦茂往14733943886 ______[答案] y'=-e^(-x)(sinx+cosx)+e^(-x)(cosx-sinx) y＂=e^(-x)(sinx+cosx)-e^(-x)(cosx-sinx)-e^(-x)(cosx-sinx)-e^(-x)(sinx+cosx) =-2e^(-x)(cosx-sinx) 所以y＂+y'+2e^(-x)cosx=-e^(-x)(sinx+cosx)+e^(-x)(cosx-sinx)-2e^(-x)(cosx-sinx)+2e^(-x)cosx =0