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.



closed as off-topic by Giskard, optimal control, FooBar, VicAche, BKay Feb 26 '16 at 18:31

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  • "This question does not meet the standards for homework questions as spelled out in the relevant meta posts. For more information, see our policy on homework question and the general FAQ." – Giskard, optimal control, FooBar, VicAche, BKay

  • $\begingroup$ 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. $\endgroup$ – Alecos Papadopoulos 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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