先對(duì)外函數(shù)y = cos(u)求導(dǎo),再乘以內(nèi)函數(shù)u = arcsinx / 2的導(dǎo)數(shù)
y = cos(arcsinx / 2)
y' = -sin(arcsinx / 2) × 1/2*1/√(1-x²)
= [-sin(arcsinx / 2)]/[2√(1-x²)]就這樣就？額，進(jìn)一步化簡(jiǎn)就是= [√(1-x) - √(x+1)]/[4√(1-x^2)]sin[(arcsinx)/2]怎樣化簡(jiǎn)設(shè)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，這個(gè)很難再化簡(jiǎn)了