此题太easy!
(1)利用导数的定义:
[cos(x)]'=lim{[cos(x+h)-cos(x)]/h}
注意:极限过程是h→0
(2)利用三角公式中的和差化积公式:
[cos(x)]'=lim{[cos(x+h)-cos(x)]/h}
=lim{(1/h)*[-2sin(x+h/2)*sin(h/2)]}
=lim{-sin(x+h/2)*[sin(h/2)/(h/2)]}
(3)在高数极限一章我们已经熟知的重要极限:
lim[sin(x)/x]=1(极限过程是x→0)
(4)[cos(x)]'=-sin(x),得证.