Originally Posted by

**Soroban** Hello, c_323_h!

I tried something on #2

. . and found it can be simplified *beyond all recognition . . .*

Since $\displaystyle \ln(2u) \:=\:\ln(2) + \ln(u)$, we have:

. . $\displaystyle f(u)\;=\;\frac{\ln u }{1 + \ln2 + \ln u}$

Differentiate:

. . $\displaystyle f'(u) \;= \;\frac{[1 + \ln 2 + \ln u]\cdot\frac{1}{u} - \ln u\cdot\frac{1}{u}}{(1 + \ln2 + \ln u)^2}$

Multiply top and bottom by $\displaystyle u:$

. . $\displaystyle f'(u) \;= \;\frac{1 + \ln 2 + \ln u - \ln u}{u(1 + \ln 2 + \ln u)^2}$

Therefore: .$\displaystyle f'(u) \;=\;\frac{1 + \ln 2}{u(1 + \ln2u)^2}$

If *that's* what they did, no wonder your answer is different!