Can anyone please help with taking the derivative of the following equation with respect to time?

$ln(\frac{\gamma_t+(n+\delta)}{sA}) = (\beta - 1)ln{k_t}+(\alpha+\beta-1)ln{N_t}$

This is not a homework question, just reading a text on economic growth and not sure how the author arrived at the final expression.


  • $\begingroup$ Since the right hand side contains symbols that do not appear in the left-hand side, you must provide more context about what each symbol represents and how they are interrelated, so this is not just about taking a time derivative. $\endgroup$ Feb 23 '16 at 13:04
  • $\begingroup$ @Alecos thanks and never mind, I figured it out. $t$ - indexed elements are variables and the rest are constants. $\endgroup$
    – london
    Feb 23 '16 at 13:27

I'll give a try, perhaps it is just this first derivative giving you trouble:

$$ \frac{d}{dt} ln \left( \frac{ \gamma_t + (n+\delta)}{sA} \right) = \frac{sA}{\gamma_t + (n+\delta)} \frac{d}{dt} \left( \frac{ \gamma_t + (n+\delta)}{sA} \right) $$

$$ = \frac{sA}{\gamma_t + (n+\delta)} \frac{\dot{\gamma}_t + 0}{sA} = \frac{\dot{\gamma}_t}{\gamma_t + (n+g)} $$

And the rest are simply the time derivative of a log equals its growth rate $ \left( \frac{d}{dt} ln X = \frac{\dot{X}}{X} \right) $:

$$ \frac{d}{dt} \bigg[ (\beta-1) ln k_t + (\alpha + \beta - 1) ln N_t \bigg] = (\beta-1) \frac{\dot{k}_t}{k_t} + (\alpha + \beta - 1) \frac{\dot{N}_t}{N_t} $$

Hence concluding the time derivative is:

$$ \frac{\dot{\gamma}_t}{\gamma_t + (n+g)} = (\beta-1) \frac{\dot{k}_t}{k_t} + (\alpha + \beta - 1) \frac{\dot{N}_t}{N_t} $$

  • $\begingroup$ Yes, it did, but i figured out the solution after posting the problem. Many thanks! $\endgroup$
    – london
    Feb 23 '16 at 15:51
  • $\begingroup$ ah I see, okay then(: $\endgroup$
    – Sunhwa
    Feb 23 '16 at 15:59

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