证明cosX导数为-sinX

3个回答

  • 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.