Show that \(\del_{\log}(fg) = \del_{\log} f + \del_{\log} g\), i.e. \[ { (fg)' \over fg} = {f'\over f} + {g' \over g} .\]
Solution
Goal: Show that \(\del_{\log}(fg) = \del_{\log} f + \del_{\log} g\), i.e. \((fg)'/(fg) = f'/f + g'/g\), wherever \(f, g\) are nonzero holomorphic functions.
1. Differentiate \(fg\) by the product rule.
\((fg)' = f'g + fg'\).
2. Divide both sides by \(fg \neq 0\).
\(\frac{(fg)'}{fg} = \frac{f'g + fg'}{fg} = \frac{f'g}{fg} + \frac{fg'}{fg} = \frac{f'}{f} + \frac{g'}{g}\).
3. Q.E.D.
1 and 2 establish the identity, valid on any region where \(f, g\) are holomorphic and \(fg \neq 0\).