If the given function is meant to be y = log2 {x2/(x − 1)}, then first determine
d{x2/(x-1)}/dx which goes to
[(x − 1)(2x) − (x2)(1)] / (x − 1)2 and then to [2x2 − 2x − x2] / (x − 1)2.
Take this last to [x2 − 2x] / (x − 1)2.
Next write {[x2 − 2x] / (x − 1)2} / {x2/(x − 1)}
which simplifies to
[x2 − 2x] / x2(x − 1).
Divide [x2 − 2x] / x2(x − 1) top-and-bottom by x2
to obtain [1 − 2/x] / (x − 1).
The rule of differentiation here is d[loga u]/dx = (1/ln a) [du/dx]/u.
The final answer here would then be (1/ln 2)[1 − 2/x] / (x − 1).
The answer provided in the problem statement is (1/ln 2)[2/x − 1/(x − 1)]
which goes to
(1/ln 2)[(2x − 2 − x)/x(x − 1)] or (1/ln 2)[(x − 2)/x(x − 1)] or (1/ln 2)[1 − 2/x] / (x − 1)
in agreement.
