你拆解成复合函数,就错不了了。
令y=u²,u=arcos t,t=x/2
则dy/dx =dy/du × du/dt × dt/dx
=2u · -1/√(1-t²) · ½
=-arccos(x/2) / √(1-x²/4)
=- 2 arccos(x/2) /√(4-x²)
y=[arccos(x/2)]^2
y'
=2[arccos(x/2)].[arccos(x/2)]'
=2[arccos(x/2)].[-1/√(1- (x/2)^2)] . (x/2)'
=2[arccos(x/2)].[-2/√(4- x)^2)] . (1/2)
=-2[arccos(x/2)]/√(4- x)^2)
y'=2arccos(x/2) [arccos(x/2)]'
=2arccos(x/2) [-1/√(1-x²/4) ] (x/2)'
=arccos(x/2) [-1/√(1-x²/4) ]
=-2arccos(x/2) /√(4-x²)
x/2也要求一下导。