把f(x)的图像向左移π/6个单位即可.f'(x)=[sin(π/3-x)]' * (π/3-x)' = -cos(π/3-x)=-sin[π/2-(π/3-x)]=-sin(x-π/6)f(x)=sin(π/3-x)=-sin(x-π/3)f(x)的图像是把y=sinx的图像向右移了π/3个单位,f'(x)的图像是...答案是向左平移π/2个单位把f(x)的图像向左移π/6个单位即可。f'(x)=[sin(π/3-x)]' * (π/3-x)' = -cos(π/3-x)=-sin[π/2-(π/3-x)]=-sin(x+π/6)f(x)=sin(π/3-x)=-sin(x-π/3)f(x)的图像是把y=sinx的图像向右移了π/3个单位，f'(x)的图像是把y=sinx的图像向左移π/6个单位，因此，要得到f'(x)的图像，只需要把f(x)的图像向左移π/3+π/6=π/2个单位即可。 不好意思，f'(x)=-sin(x+π/6)所以，答案是向左平移π/2个单位