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I am trying to find the derivative of $y$ with respect to $n$. All I need to do is take the partial derivative $\frac{\partial y}{\partial n}$ of the function: $sf(k) = (n+g+\delta)k$ Here is what I get using the chain rule, but it's wrong:

$$sf'(k) \frac{\partial k}{\partial n} = (n+g+\delta)\frac{\partial k}{\partial n}$$

To solve $\frac{\partial y}{\partial n}$ some more steps are required, but as far as I know the equation is already wrong. From what I have seen the right side should be augmented by k and then it would be right. Any ideas where the derivation went wrong?

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  • $\begingroup$ Please list the initial equation in your version of the Solow model. $\endgroup$ – FooBar Feb 19 '15 at 20:26
  • $\begingroup$ @FooBar Initial equations: $Y= K^a(AL)^(a-1)$ and $k = K/L$ and Growth of k the equation is: (sf(k)-(n+g+delta)k, which is set to 0 (equilibrium). n = population growth, g = technological progress and delta is depreciation, a<1. It's the most basic of solow model with technology and population growth. $\endgroup$ – Tony Feb 19 '15 at 20:58
  • $\begingroup$ So you want to find out how steady state capital changes when we change the growth rate $n$? $\endgroup$ – FooBar Feb 19 '15 at 21:16
  • $\begingroup$ @FooBar Yes, or actually I want to find what happens to production when n is changed by x %, when the a, g, n and delta are given. It should be straight forward, but I am having difficulties solving the partial derivative $\frac{\partial k}{\partial n}$. $\endgroup$ – Tony Feb 19 '15 at 21:24
  • $\begingroup$ Immediately, nothing will happen with production, because $n$ does not instantly affect $N$. $\endgroup$ – FooBar Feb 19 '15 at 22:21
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You need to use the implicit function theorem. The long-run relationship implies

$$ 0 = F(n,k) = s f(k) - (n+g+\delta)k, $$

and this implicitly defines function $k(n)$. The derivative is then

$$ \frac{d k(n)}{d n} = -\frac{\partial F / \partial n}{\partial F / \partial k} = \frac{k}{s f'(k) - (n+g+\delta)}, $$

which can be evaluated once one has solved for particular long-run (n,k) combination. To finally obtain derivative of output $y = f(k)$ wrt. $n$, simply use chain rule:

$$ \frac{d y}{d n} = f'(k(n)) \frac{d k(n)}{d n}. $$

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  • $\begingroup$ Thank you, this is indeed the correct answer! I had it wrong because in the provided answer a partial derivative symbol was (accidentally?) used. Relating to that I would like to inquire, why is the function = F(n,k), where does the k come from and why no other symbols? And on the other hand why is the whole derivative needed to figure out the change instead of a partial one? $\endgroup$ – Tony Feb 20 '15 at 8:34
  • $\begingroup$ $F(n,k)$ is just a piece of notation, expressing that we have an equation involving the two variables we're interested in (other symbols we can take as constant parameters, so don't need to have them as arguments). Partial derivative is not sufficient because $k$ is not independent variable, but a function of $n$, and this must be taken into account (any textbook on calculus or mathematical methods for economics should discuss implicit function theorem in more detail). $\endgroup$ – ivansml Feb 20 '15 at 10:53

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