Could someone help me find the derivative of ln(2cosx) ? So far, I have
(1/2cosx)(-2sinx), so (-2sinx/2cosx). Is this right???
I would appreciate all the help I can get!
Yes, you have solved the problem correctly. Alternatively, you could break up the expression to $\displaystyle \ln{2} + \ln{(\cos x)}$ and then differentiate. Either way, the 2 isn't going to change the result of $\displaystyle -\tan {x}$.