<p>When you are using implicit differentiation along a level curve you will treat the variable with respect to which you are differentiating as a single variable, rather than function.  This is because the formula for implicit differentiation along level curve is already based previous derivation where you already solve for <span class="math-container">$y'$</span>.</p>
<p>For example, for general function <span class="math-container">$F(x,y)=c$</span> where <span class="math-container">$y=f(x)$</span>, we get by implicit differentiation:</p>
<p><span class="math-container">$$v(x) = F(x,y) = c \implies v'(x) = F_x' + F_y' y' = 0 $$</span></p>
<p>Consequently we get the result that:</p>
<p><span class="math-container">$$y' = - \frac{F_x'}{F_y'} \tag{1}$$</span></p>
<p>Hence, if you just apply formula given by 1 you will not use chain rule anymore, since it was already used in derivation of the formula. For example:</p>
<p><span class="math-container">$$xy=5 \implies y'= -\frac{y}{x}  $$</span></p>
<p>so in your case the correct implicit differentiation along a level curve would yield:</p>
<p><span class="math-container">$$\frac{\partial k^*}{\partial n}=  -\frac{F_n'}{F_k'} =\frac{k^*}{sf′(k^⋆)−(n+g+\delta)}$$</span></p>
<p>You could verify it by actually implicitly differentiating the original function which would yield:</p>
<p><span class="math-container">$$\frac{\partial k^*}{\partial n}[sf(k^*)=(n+g+\delta)k^*] \implies sf'(k^*)\frac{\partial k^*}{\partial n} = k^* + (n +g + \delta )\frac{\partial k^*}{\partial n}$$</span></p>
<p>Solving for <span class="math-container">$\frac{\partial k^*}{\partial n}$</span> yields:</p>
<p><span class="math-container">$$\frac{\partial k^*}{\partial n} =\frac{k^*}{sf′(k^⋆)−(n+g+\delta)}$$</span></p>
<p>so it clearly works.</p>

CC BY-SA 4.0
2021-02-13T17:07:24.947Z
Very thorough answer - thank you. So if I understand it correctly, we actually have $\frac{\partial k^\star}{\partial n}=\frac{\partial k^\star}{\partial g}=\frac{\partial k^\star}{\partial \delta}$
Sirmimer
CC BY-SA 4.0
2021-02-13T17:13:52.017Z
@Sirmimer you are welcome. Yes for this function $\partial k / \partial n = \partial k / \partial g = \partial k / \partial \delta$ as far as I can see. Also, if you think my answer answered your question consider eventually accepting it.
1muflon1
CC BY-SA 4.0
2021-02-13T17:15:25.990Z
I just did :) I think this answer is great for the achieve as well
Sirmimer
CC BY-SA 4.0
2021-02-13T16:56:42.843Z
2021-02-13T16:56:42.843Z
1muflon1
<p>Given the solow model
<span class="math-container">$sf(k)=(n+g+\delta)k$</span>, written as</p>
<p><span class="math-container">$$F=sf(k^\star)-(n+g+\delta)k^\star=0$$</span></p>
<p>I am trying to find the partial derivative of k* with respect to n</p>
<p>Here is what I've tried: I use the implicit function theorem such that I need to calculate <span class="math-container">$\frac{\partial k^\star}{\partial n}=-\frac{F_n}{F_{k^\star}}$</span>. I've previously found that</p>
<p><span class="math-container">$$F_{k^\star} = sf'(k^\star)-(n+g+\delta)$$</span></p>
<p>Next I want to find <span class="math-container">$F_{n}$</span>. Taking the partial derivative and finding that <span class="math-container">$F_n=-k^\star$</span>. However I am in doubt as to whether or not I should use the chainrule, as <span class="math-container">$k^{\star}$</span> is dependant on <span class="math-container">$n,g,s,\delta$</span>. In that case I find that, because
<span class="math-container">$$\frac{\partial f(k^\star)}{\partial n}=f'(k^\star)\frac{\partial k^\star}{\partial n}$$</span></p>
<p>Then we have that <span class="math-container">$F_n=sf'(k^\star)\frac{\partial k^\star}{\partial n}-k^\star$</span>. Thus we have</p>
<p><span class="math-container">$$\frac{\partial k^\star}{\partial n}=-\frac{F_n}{F_{k^\star}}=\frac{sf'(k^\star)\frac{\partial k^\star}{\partial n}-k^\star}{sf'(k^\star)-(n+g+\delta)}$$</span></p>
<p>It seems odd to find the result of <span class="math-container">$\frac{\partial k^\star}{n}$</span> includes itself.</p>

<macroeconomics><solow>
Solow model: Partial derivative of k* with respect to g
Sirmimer
# Solow model: Partial derivative of k* with respect to g

(Asked by: username_0 on 2021-02-13)

Given the solow model
$sf(k)=(n+g+\delta)k$, written as

$$F=sf(k^\star)-(n+g+\delta)k^\star=0$$

I am trying to find the partial derivative of k\* with respect to n

Here is what I've tried: I use the implicit function theorem such that I need to calculate $\frac{\partial k^\star}{\partial n}=-\frac{F\_n}{F\_{k^\star}}$. I've previously found that

$$F\_{k^\star} = sf'(k^\star)-(n+g+\delta)$$

Next I want to find $F\_{n}$. Taking the partial derivative and finding that $F\_n=-k^\star$. However I am in doubt as to whether or not I should use the chainrule, as $k^{\star}$ is dependant on $n,g,s,\delta$. In that case I find that, because
$$\frac{\partial f(k^\star)}{\partial n}=f'(k^\star)\frac{\partial k^\star}{\partial n}$$

Then we have that $F\_n=sf'(k^\star)\frac{\partial k^\star}{\partial n}-k^\star$. Thus we have

$$\frac{\partial k^\star}{\partial n}=-\frac{F\_n}{F\_{k^\star}}=\frac{sf'(k^\star)\frac{\partial k^\star}{\partial n}-k^\star}{sf'(k^\star)-(n+g+\delta)}$$

It seems odd to find the result of $\frac{\partial k^\star}{n}$ includes itself.

**Tags:** macroeconomics, solow


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## 1 answers

### Answer by username_1 on 2021-02-13. Score: 2 | Accepted answer

When you are using implicit differentiation along a level curve you will treat the variable with respect to which you are differentiating as a single variable, rather than function. This is because the formula for implicit differentiation along level curve is already based previous derivation where you already solve for $y'$.

For example, for general function $F(x,y)=c$ where $y=f(x)$, we get by implicit differentiation:

$$v(x) = F(x,y) = c \implies v'(x) = F\_x' + F\_y' y' = 0 $$

Consequently we get the result that:

$$y' = - \frac{F\_x'}{F\_y'} \tag{1}$$

Hence, if you just apply formula given by 1 you will not use chain rule anymore, since it was already used in derivation of the formula. For example:

$$xy=5 \implies y'= -\frac{y}{x} $$

so in your case the correct implicit differentiation along a level curve would yield:

$$\frac{\partial k^\*}{\partial n}= -\frac{F\_n'}{F\_k'} =\frac{k^\*}{sf′(k^⋆)−(n+g+\delta)}$$

You could verify it by actually implicitly differentiating the original function which would yield:

$$\frac{\partial k^\*}{\partial n}[sf(k^\*)=(n+g+\delta)k^\*] \implies sf'(k^\*)\frac{\partial k^\*}{\partial n} = k^\* + (n +g + \delta )\frac{\partial k^\*}{\partial n}$$

Solving for $\frac{\partial k^\*}{\partial n}$ yields:

$$\frac{\partial k^\*}{\partial n} =\frac{k^\*}{sf′(k^⋆)−(n+g+\delta)}$$

so it clearly works.

> **Comments:**
> **username_0**: Very thorough answer - thank you. So if I understand it correctly, we actually have $\frac{\partial k^\star}{\partial n}=\frac{\partial k^\star}{\partial g}=\frac{\partial k^\star}{\partial \delta}$  
> **username_1**: @username_0 you are welcome. Yes for this function $\partial k / \partial n = \partial k / \partial g = \partial k / \partial \delta$ as far as I can see. Also, if you think my answer answered your question consider eventually accepting it.  
> **username_0**: I just did :) I think this answer is great for the achieve as well  

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