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Given a logistic function of the form. \begin{align*} f(t) = \frac{\alpha}{1 + \beta e^{\gamma t}} \end{align*} Harvey (1984) differentiates this and takes logs to yield: \begin{align*} \ln f' = 2 \ln f + \ln \frac{-\beta \gamma}{\alpha} + \gamma t \end{align*}

I'm obviously missing something, hoping somebody can help.

Noting, $1 - \frac{f}{\alpha} = \frac{ \beta e^{\gamma t} }{1 + \beta e^{\gamma t}}$ and $\ln \alpha = 2 \ln f + \ln \beta + \gamma t$. \begin{align*} f' = - \alpha ( 1 + \beta e^{\gamma t} )^{-2} \gamma \beta e^{\gamma t} \\ f' = - \gamma f \frac{\beta e^{\gamma t}}{1 + \beta e^{\gamma t}} = - \gamma f \left( 1 - \frac{f}{\alpha} \right) \\ f' = - \gamma f + \frac{\gamma}{\alpha} f^2 \end{align*} Now taking logs (note the mistake below as highlighted by comment with regards to the log of the sum - the derivative is correct as shown in the answer - thanks both) \begin{align*} \ln f' = - \ln \gamma - \ln f + ( \ln \gamma - \ln \alpha + 2 \ln f ) \\ \ln f' = \ln f - \ln \alpha \\ \ln f' = \ln f - (2 \ln f + \ln \beta + \gamma t) \\ \ln f' = - \ln f - \ln \beta - \gamma t \end{align*}

Ref: Harvey, A. (1984). Time series forecasting based on the logistic curve. Journal of the Opera- tional Research Society 35(7), 641–646

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    $\begingroup$ I don't follow the expression after "Now taking logs," because it looks like you are replacing $\log(a+b)$ by $\log(a)+\log(b)$ where $a=-\gamma f$ and $b=\gamma f^2/a.$ That would be invalid. $\endgroup$ – whuber Apr 19 '18 at 22:47
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Let $$ \begin{align*} f(t) = \frac{\alpha}{1 + \beta e^{\gamma t}} \end{align*}. $$ Then $$ f'(t) = -\alpha (1 + \beta e^{\gamma t})^{-2} \cdot \beta e^{\gamma t} \cdot \gamma $$ $$ = - \frac{\alpha }{1 + \beta e^{\gamma t}} \cdot \frac{1 }{1 + \beta e^{\gamma t}} \cdot \beta \gamma \cdot e^{\gamma t} $$ $$ = - f(t) \cdot \frac{1}{\alpha} \frac{\alpha }{1 + \beta e^{\gamma t}} \cdot \beta \gamma \cdot e^{\gamma t} $$ $$ = f(t)^2 \cdot \frac{-\beta \gamma}{\alpha} \cdot e^{\gamma t} $$ if we assume $\alpha \neq 0$ so $$ \log f'(t) = \log \left(-\frac{\beta \gamma}{\alpha}\right) + 2 \log f(t) + \gamma t. $$


Your derivative is correct but it looks like you just put it into a less helpful form; in particular, you turned $f'$ into a sum which won't be very amenable to taking logs. Normally we always try to find something like $f \cdot (1 - f)$ with logistic functions but in this case the main trick was multiplying by $\alpha / \alpha$.

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