f(x)=2-2sin²(x/2)
f'(x)=2'-[2sin²(x/2)]'
=0-2×2sin(x/2) [sin(x/2)]'
=-4sin(x/2) ×cos(x/2) ×(x/2)'
=-2sin(x/2)cos(x/2)
=-sinx
一样的结果
f(x)=2-2sin²(x/2)
f'(x)=2'-[2sin²(x/2)]'
=0-2×2sin(x/2) [sin(x/2)]'
=-4sin(x/2) ×cos(x/2) ×(x/2)'
=-2sin(x/2)cos(x/2)
=-sinx
一样的结果