<p>You need to use the <a href="http://en.wikipedia.org/wiki/Implicit_function_theorem">implicit function theorem</a>. The long-run relationship implies</p>

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

<p>and this implicitly defines function $k(n)$. The derivative is then</p>

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

<p>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:</p>

<p>$$
\frac{d y}{d n} = f'(k(n)) \frac{d k(n)}{d n}.
$$</p>

CC BY-SA 3.0
2015-02-20T08:34:00.820Z
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?
Tony
CC BY-SA 3.0
2015-02-20T10:53:03.093Z
$F(n,k)$ 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).
ivansml
CC BY-SA 3.0
2015-02-20T01:00:30.707Z
2015-02-20T01:00:30.707Z
ivansml
<p>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:</p>

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

<p>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?</p>

CC BY-SA 3.0
2015-02-19T20:26:54.817Z
Please list the initial equation in your version of the Solow model.
FooBar
CC BY-SA 3.0
2015-02-19T20:58:55.710Z
@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.
Tony
CC BY-SA 3.0
2015-02-19T21:16:00.100Z
So you want to find out how steady state capital changes when we change the growth rate $n$?
FooBar
CC BY-SA 3.0
2015-02-19T21:24:57.400Z
@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 $\frac{\partial k}{\partial n}$.
Tony
CC BY-SA 3.0
2015-02-19T22:21:16.507Z
Immediately, nothing will happen with production, because $n$ does not instantly affect $N$.
FooBar
CC BY-SA 3.0
2015-02-19T22:47:16.360Z
@FooBar Sorry, I should have specified that this question is about the long term equilibrium state. Anyway do you know how to solve the derivative k/n?
Tony
<macroeconomics><economic-growth>
Solow model: Partial derivative of y with respect to n
Tony
# Solow model: Partial derivative of y with respect to n

(Asked by: username_0 on 2015-02-19)

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?

**Tags:** macroeconomics, economic-growth
> **Comments:**
> **username_1**: Please list the initial equation in your version of the Solow model.  
> **username_0**: @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.  
> **username_1**: So you want to find out how steady state capital changes when we change the growth rate $n$?  
> **username_0**: @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 $\frac{\partial k}{\partial n}$.  
> **username_1**: Immediately, nothing will happen with production, because $n$ does not instantly affect $N$.  
> **username_0**: @FooBar Sorry, I should have specified that this question is about the long term equilibrium state. Anyway do you know how to solve the derivative k/n?  

---

## 1 answers

### Answer by username_2 on 2015-02-20. Score: 6 | Accepted answer

You need to use the [implicit function theorem](http://en.wikipedia.org/wiki/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:

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

> **Comments:**
> **username_0**: 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?  
> **username_2**: $F(n,k)$ 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).  

---

