I need the steps and result of Log - Differentiation wrt "time" of the following equation: \begin{equation} \dot{D}(t) = \mu (D(t) - a - Ah(t)^{\gamma}) \end{equation}
Where: a, $\mu$, A are constants
Log - Differentiation (with respect to) time
0
$\begingroup$
ordinary-differential-equations
derivatives
1 Answers
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It's simple. The log-derivative is nothing but the derivative of the function over the function.
$$\frac{d\ln(\dot D(t))}{dt} = \frac{\mu(\dot D(t) - A \gamma h(t)^{\gamma-1}\dot h(t))}{\mu(D(t) - a - Ah(t)^{\gamma})} = \frac{\dot D(t) - A\gamma h(t)^{\gamma-1}\ \dot h(t)}{D(t) - a - Ah(t)^{\gamma}} $$
The steps are trivial, just remember that
$$\frac{d\ln(f)}{dx} = \frac{f'}{f}$$