Show that \(\logd(fg) = \logd f + \logd g\), i.e. \[ { (fg)' \over fg} = {f'\over f} + {g' \over g} .\]
Solution
\[ {(fg)' \over fg} = { f'g + fg' \over fg} = {f'g \over fg} + {fg' \over fg} = {f'\over f} + {g' \over g} .\]
The logarithmic derivative of a product: $\frac{(fg)'}{fg}=\frac{f'}{f}+\frac{g'}{g}$
E-PWVJS
- Area
- Complex Analysis
- Topics
- Argument Principle, Poles, Zeros
- Status
- draft