Can anyone differentiate this : y=Ln(tg(x/2))
tg is tangent function ?
Yes
do you know chain rule ?
I know how to do this type of problems but I can't get the right answer and I don't know where I go wrong
so sho me your attempt, maybe i will find the error in your work...
you know derivates of ln x, tg x and x/2 , right ?
(1/(2cos^2 x/2))/tg x/2
did u mean, \(\huge \dfrac{\dfrac{1}{2\cos^2(x/2)}}{\tan(x/2)}\) if yes, thats correct, just try to simplify a bit.
Yes that's what I meant. And when I simplify it I get 1/(2sin x/2 * cos x/2)
I don't know what to do from there on. None of the answers match.
which is still correct now use sin 2x = 2 sin x cos x so, 2sin x/2 * cos x/2 will be just sin x and 1/sin x = cosec x is cosec x in choices ?
and did u get what i did ?
Oh yes, you are right! I forgot to use that equation at the end. The answer is 1/sinx. Thanks!! :)
welcome ^_^
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