Use Newton's Method to approximate 96^(1/96) correct to 8 decimal places.
Newton's Method equation: Second Approx = (first approx) - (f(first approx))/(f'(first approx))
The Attempt at a Solution
So I know that 96^(1/96) must = some number x, so I set it up as 96^(1/96)-x=0, but what that does is cause f'(x)=1. This must not be right because when I approximate using this, I get a negative value which is significantly farther away than my first approx. Please help!