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求y lnsin2x的导数

y=ln(sin2x)y'=(1/sin2x)*(sin2x)'y'=(1/sin2x)*cos2x*(2x)'y'=(cos2x/sin2x)*2y'=2/tan2xy'=2cot2x

拆分为lnsin2x-lnx 视sin2x=u,2x=v 根据复合函数求导法则 求导得[(2cos2x)/(sin2x)]-(1/x)

cos(x/2) / 2sin(x/2)

(lnsin2x)'=(1/sin2x)*cos2x*2=2/tan2x

y=lnsin(x/2) y'={1/[sin(x/2)]}*cos(x/2)*(1/2) =cos(x/2)/[2sin(x/2] =cot(x/2)/2

(ln(x)^2)'=1/x^2*(x2)' =1/x^2*2x =2/x

y=ln(sin2x/x)=lnsin2x-lnxy'=1/sin2x*cos2x*2-1/x=2cot2x-1/x

y'=(lnsin(2x-1))'=(1/sin(2x-1))*(cos(2x-1))*2=2cot(2x-1)

y=ln(cosx)y'=2ln(cosx)-sinx/cosx =-2tanxln(cosx)

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