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

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

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

<hr>

<p>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) $:</p>

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

<hr>

<p>Hence concluding the time derivative is:</p>

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

CC BY-SA 3.0
2016-02-23T15:51:09.757Z
Yes, it did, but i figured out the solution after posting the problem. Many thanks!
london
CC BY-SA 3.0
2016-02-23T15:59:45.903Z
ah I see, okay then(:
Sunhwa
CC BY-SA 3.0
2016-02-23T15:24:45.220Z
2016-02-23T15:24:45.220Z
Sunhwa
<p>Can anyone please help with taking the derivative of the following equation with respect to time?</p>

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

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

<p>Thanks</p>

CC BY-SA 3.0
2016-02-23T13:04:56.907Z
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.
Alecos Papadopoulos
CC BY-SA 3.0
2016-02-23T13:27:28.877Z
@Alecos thanks and never mind, I figured it out.  $t$ - indexed elements are variables and the rest are constants.
london
<mathematical-economics><economic-growth>
Taking time derivative, growth equations
london
# Taking time derivative, growth equations

(Asked by: username_0 on 2016-02-23)

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.

Thanks

**Tags:** mathematical-economics, economic-growth
> **Comments:**
> **username_1**: 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.  
> **username_0**: @Alecos thanks and never mind, I figured it out.  $t$ - indexed elements are variables and the rest are constants.  

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## 1 answers

### Answer by username_2 on 2016-02-23. Score: 3 | Accepted answer

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

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

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

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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) $:

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

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Hence concluding the time derivative is:

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

> **Comments:**
> **username_0**: Yes, it did, but i figured out the solution after posting the problem. Many thanks!  
> **username_2**: ah I see, okay then(:  

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