先对外函数y = cos(u)求导,再乘以内函数u = arcsinx / 2的导数
y = cos(arcsinx / 2)
y' = -sin(arcsinx / 2) × 1/2*1/√(1-x²)
= [-sin(arcsinx / 2)]/[2√(1-x²)]就这样就？额，进一步化简就是= [√(1-x) - √(x+1)]/[4√(1-x^2)]sin[(arcsinx)/2]怎样化简设y = arcsinxsiny = xcosy = √(1-x^2)cosy = 1-2sin^2(y/2)√(1-x^2) = 1-2sin^2(y/2)sin^2(y/2) = [1-√(1-x^2)] / 2sin(y/2) = √[1-√(1-x^2)] / √2sin(arcsinx / 2) = √[1-√(1-x^2)] / √2，这个很难再化简了