Find the second Taylor approximation for the following
I think I have to use the chain rule, but i'm not sure how. I really need help.
Wouldn't it just be -ln(x^2/2!)?
How I got my answer was just by identifying that cos(x) = 1 - x^2/2!, then just take the ln of each nth. This left me with 0 - ln(x^2/2!).
Please keep in my that I could be wrong, fore I am learning Taylor/MacLaurin series at this moment.