i know the derivative of ln (tan x) is sec^2 (x) divided by tan x but how do find derivative of ln tan (x/2)?

Answer 1

y = ln tan (x/2)
Let u = x/2
du / dx = (1/2)
y = ln (tan u)
dy/du = sec ² u

dy/dx = ( dy/du) (du/dx)
dy/dx = (sec²u) (1/2)
dy/dx = sec² (x/2) (1/2)

Answer 2

Certain problems require chain rule to be applief several times and therefore needs time to think about. This problem requires you to apply chain rule twice.

1/tan (x/2) * sec^2 (x/2) * 1/2

1/2 (sec^2 (x/2) / tan (x/2))

1/2 (1/(cos(x/2)sin(x/2)))

If you wish to simplify this.....

1/2 (1/(1/2[sin x + sin 0])

1/2 (2 / sin x)

1 / sin x = csc x

csc x

Answer 3

Just multiply everything by 1/2 and change your x from your previous result to x/2.

A neater way of expressing it is 1/2 csc(x/2)sec(x/2).