How would you us the composite rule to differentiate
k(x)=ln(6cos((1/3)x))?
thanks for the quick reply.
The way I have done it has given a completly wrong answer. would you mind showing us where I went wrong
This is the meathod I used to get to my answer
k(x)=ln(6cos((1/3)x))
so k(x)=g(f(x)) where u = f(x) = 6cos((1/3)x) and g(u) = ln u
so f '(x)= -2 sin (x/3) and g ' (u)= ln u
so k' (x)= g'(f(x))f '(x)
k '(x)= g '(u)f '(x) where u = f(x)
k'(x)= (1/u)(-2sin (x/3))
k '(x)= (-2sin(x/3))/(6cos((1/3)x))