d/dx(log tan x) is equal to ??

We are required to find the derivative of (log tan x)

Assume tan x to be y then the question becomes to find the derivative of log y.

Hence, d/dx log (tan x) = 1/tan x . d/dx (tan x) ( using d/dx log x = 1/x)

= 1/tan x . sec² x.

= (cos x)/ (sin x). 1/ cos²x

= 1/ sin x cos x

This is the answer which if required can further be simplified as 2/ 2sin x cos x

which becomes 2/ sin 2x = 2 cosec 2x.

Thanks & Regards

Latika Leekha

askIITians Faculty

Y = log tanx
we know, if any function y = logf(x) is given then, dy/dx = 1/f(x).df(x)/dx
use this concept here ,

y = log tanx
differentiate wrt x
dy/dx = 1/tanx .d(tanx)/dx
= 1/tanx .sec²x
= sec²x/tanx
[ sec²x = 1 + tan²x use this ]
dy/dx = (1 + tan²x)/tanx
= 1/tanx + tan²x/tanx
= cotx + tanx
= sinx/cosx + cosx/sinx
= (sin²x + cos²x )/sinx.cosx
= 1/sinx.cosx
= 2/2sinx.cosx
[ 2sinx.cosx = sin2x ]
= 2/sin2x
= 2cosecx

hence, dy/dx = 2cosecx