show that d/dx (ln tan 0.5x) = 1/sinx
I just want to see where im going wrong
d/dx ln f(x) = f'(x)/f(x)
so tan0.5x = sec20.5x
so d/dx ln f(x) = sec20.5x/tan 0.5x
= 1/cos20.5x/( sin 0.5x/ cos 0.5x)
multiply both sides by cos 0.5x
cos 0.5x / cos20.5x / (sin 0.5x)
cancel nominators cos 0.5x
1/cos 0.5x / sin0.5x
multiply both sides by cos 0.5x
1/(sin 0.5x)(cos0.5x)
okay so now im scratching my head thinking what next when it hits me.
sin 2x = 2sinxcosx
so 1/2 x sin2x = sinxcosx
in this case x = 0.5x
so 1/(sin 0.5x)(cos0.5x) = 1/0.5sinx = 2/sinx
So can someone check what ive done and point the error out thanks.