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


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.


1 Answer 1


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.

  • $\begingroup$ Very thorough answer - thank you. So if I understand it correctly, we actually have $\frac{\partial k^\star}{\partial n}=\frac{\partial k^\star}{\partial g}=\frac{\partial k^\star}{\partial \delta}$ $\endgroup$
    – Sirmimer
    Feb 13, 2021 at 17:07
  • $\begingroup$ @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. $\endgroup$
    – 1muflon1
    Feb 13, 2021 at 17:13
  • $\begingroup$ I just did :) I think this answer is great for the achieve as well $\endgroup$
    – Sirmimer
    Feb 13, 2021 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.