Given that the per-capita capital $ k = \frac{K}{L}\ $ (total capita divided by labor force), and we want to find $ \dot{k}\ $, it seems that $ \frac{\dot{k}}{k}\ = \frac{\dot{K}}{K}\ - \frac{\dot{L}}{L}\ $

Why is this? Shouldnt't the derivative with respect to time $ \dot{k}\ $yield something like a quotient rule?

Furthermore, why does $ \frac{\dot{L}}{L}\ = n $ ? Shouldn't the derivative with respect to time of the labor force be equal to $ n $ (the growth rate of L, in discrete time, $ L_1 = (1+n)L_0 $ ) only, so that $ \dot{L}\ = n $ ?


4 Answers 4



Taking log both sides, we get

$\ln k(t)=\ln K(t)- \ln L(t)$

Differentiation with respect to $t$, we get


which can also be written as $\dfrac{\dot{k}}{k}=\dfrac{\dot{K}}{K}-\dfrac{\dot{L}}{L}$


This only answers your first question, but the quotient rule does apply here. The right hand side of your relation is more naturally derived using the log formulation (as in other answers), but consider the following:

$k = K/L$

$\dot{k} = \frac{L \dot{K} - K\dot{L}}{L^2}$

$\frac{\dot{k}}{k} = \frac{\dot{k}L}{K} = \frac{L}{K} \left( \frac{L \dot{K} - K\dot{L}}{L^2} \right) = \frac{L^2\dot{K}}{K L^2} - \frac{LK\dot{L}}{KL^2} = \frac{\dot{K}}{K} - \frac{\dot{L}}{L}$

  • $\begingroup$ What is the more natural way of doing that calculation is a subjective matter. I think that it could be a test: If one prefers the quotient rule is a mathematician, if one prefers the logaritm is an economist :). It is a joke, but has a kernel of truth: economist are more used to use logarithms, mathematicians think at first at derivation rules. $\endgroup$ Jan 10, 2023 at 21:04

There is another way to formulate the issue, equivalent to the formulation of the preceding answers, to explain why $\frac {\dot L}{L}=n$.

In the Solow growth model, as in your case, the usual assumption is that the growth rate of the population is a constant, $n$.

The growth rate (the equivalent of the percentage growth rate in discrete time) in continuos time is, by definition, $\frac {\dot L}{L}$.

A system with a constant growth rate is ruled by an equation$^1$ as:

$$ L(t)= L_0 e^{nt}\;\;\;(1),$$

where $L_0$ is the population at time $0$.

Indeed, taking the derivative of $(1)$ with respect to time we have: $$\dot L(t)= nL_0 e^{nt}.$$

Dividing both sides by $L(t)$, we have (neglecting $t$):

$$\frac {\dot L}{L}= \frac{nL_0 e^{nt}}{L_0 e^{nt}}= n.$$

$^1$ This is the formalization of the assumption of constant growth rate of population in the Solow model in continuous time


No, you shouldn't use quotient rule here. These are the steps:

  1. log linearize $$k=K/L \implies \ln k = \ln K - \ln L$$
  2. Take time derivative vrt $t$ (treat any variable $x$ as $x(t)$) use chain rule.

$$ \frac{1}{k}k'(t)= \frac{1}{K} K'(t) -\frac{1}{L} L'(t)$$

  1. Simplify and adjust notation ($x'(t) = \dot{x}$):

$$ \frac{\dot{k}}{k} = \frac{\dot{K}}{K} -\frac{\dot{L}}{L} $$

As you see there is no reason for using quotient rule here.

Regarding to second question. A continuous time equivalent of $(L_1 - L_0)/L_0$ is $\dot{L}/L$ since $\dot{L}= L_1 - L_0$ as $\Delta t \to 0$. So if $L_1=(1+n)L_0$ you would expect that $\dot{L}/L=n$ although note $n$ in discrete and continuous time are not 100% the same since in continuous time it is instantaneous growth and in discrete time its over period of time, but if you think of it in terms of methodological continuous time equivalent then they are equivalent.


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