# Solow model: Partial derivative of k* with respect to g

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.

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.

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