
Natural log problem
I have to find the first derivative of y= ln((1cos(x))/(1+cos(x)))
the derivative of the ln is 1+cos(x)/1cos(x)
and the derivative of the inner function is:
((1+cosx) (sinx)  (1cos(x) (sinx)

(1+cos(x))^2
i then multiplied and got
(1+cos(x))(sinx + sinxcosx+sinxsinxcosx)

(1+cos(x))^2 * 1cos(x)
to clean this up:
2sinx * (1+cos(x))
 = y'
(1+cos(x))^2 * 1cos(x)
After canceling out stuff:
2sinx
 = y'
1cosx^2
but the answer says:
 2sinx
 = y'
cosx^21
Is this merely a matter of just multiplying both of these by 1??

Yep just multiply top and bottom by 1. You've done all the hard work! Well done.

quick question, was there a certain explanation for multiplying both num and denominator by 1... just asking for future reference.

You don't have to do that last step. Your answer is just as correct. So if you stopped where you did, you're still correct. It's sort of like if you said a+b and they said b+a, then they are both the same.

oh ok... thank you very much!!