Question

一道数学对数和导数的证明题.f(x)＝In（1＋cos x／1－cos x）证明,f'(x)＝-2cosec x

Answer 1

fun stuff...

First let ((1+cosx)/(1-cosx)) = u

f(x) = ln(u)

f'(x) = (1/u)(du/dx) = (1-cosx)/(1+cosx) * (d((1+cosx)/(1-cosx))/dx)

=(1-cosx)/(1+cosx)*((1-cosx)d(1+cosx) - (1+cosx)d(1-cosx))/((1+cosx)(1-cosx)^2)

=((1-cosx)(-sinx) + (1+cosx)(-sinx))/((1+cosx)(1-cosx))

=((cosx-1)sinx - (1+cosx)sinx)/((1+cosx)(1-cosx)

= (cosx*sinx-sinx-sinx-cosx*sinx)/((1+cosx)(1-cosx))

=(-2sinx)/(1-cos^2(x))

=-2sinx/sin^2(x)

=-2/sinx = -2cosec(x)

It's kind of messy,I can send you a word version if you want.