<p>Given</p>

<p>$$\max_{x_2^h} \quad U^h(\phi(\bar z + y^h - x_2^h), x_2^h)$$</p>

<p>You have to find FOCs by taking the chain rule twice for the first argument. Denote the argument for function $\phi$ as $(\cdot)$.</p>

<p>$$\frac{\partial U^h}{\partial x_2^h} = \frac{\partial U^h}{\partial \phi} \cdot \frac{\partial \phi}{\partial x_2^h}$$
$$\frac{\partial \phi}{\partial x_2^h} = \frac{\partial \phi}{\partial (\cdot)} \cdot \frac{\partial (\cdot)}{\partial x_2^h}$$
$$\implies \frac{\partial U^h}{\partial x_2^h} = \frac{\partial U^h}{\partial \phi} \cdot \frac{\partial \phi}{\partial (\cdot)} \cdot \frac{\partial (\cdot)}{\partial x_2^h}$$</p>

<p>Thus, letting $U^h_{1,2}$ represent the partial derivative with respect to the first and second arguments of $U^h$ respectively:</p>

<p>$$\begin{align}
\frac{\partial U^h}{\partial x_2^h} &amp; = U_1^h(\phi(\bar z + y^h - x_2^h), x_2^h) \cdot \phi_z(\bar z + y^h - x_2^h) \cdot -1 \\
&amp; + U_2^h(\phi(\bar z + y^h - x_2^h), x_2^h) \cdot 1
\end{align}$$</p>

<p>The second argument in $U$ has no $\phi$, so only chain rule should be applied there.</p>

CC BY-SA 3.0
2016-12-03T00:39:58.187Z
isnt that what i did?
user10699
CC BY-SA 3.0
2016-12-05T15:31:16.787Z
More or less. I'm merely adding some more detail @user10699
Kitsune Cavalry
CC BY-SA 3.0
2016-12-02T22:59:04.110Z
2016-12-02T22:59:04.110Z
Kitsune Cavalry
<p>Cant you just apply this</p>
<p><a href="https://i.sstatic.net/BfKId.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/BfKId.gif" alt="enter image description here" /></a></p>
<p>Obviously different values but same principle ?</p>

CC BY-SA 3.0
2016-12-02T23:49:52.803Z
2016-12-02T23:49:52.803Z
2020-06-17T09:40:54.643Z
Asdf
<p>I am currently reading the book "Microeconomics: Principles and Analysis" by Cowell (2006), page 452-453. e have a two-commodity world, in which there are
${n_h}$ agents (households): commodity 1 is a pure public good and commodity 2 is
purely private. </p>

<p>Each agent has an exogenously given income ${y^h}$ , denominated in units of the
private good 2. We imagine that the public good is to be financed voluntarily,
each household makes a contribution ${z^h}$ which leaves</p>

<p>$$x_2^h = {y^h} - {z^h}$$ of the private good available for h’s own consumption
and
$${x_1} = \phi (\bar z + {y^h} - x_2^h)$$    where z is the total input of the good 2 used in the production process. Each agent realises that the total output of the public good depends upon his
or her own contribution and upon that made by others</p>

<hr>

<p>The expression (13.30)</p>

<p><a href="https://i.sstatic.net/THxdh.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/THxdh.png" alt="enter image description here"></a></p>

<p>The expression (13.31)</p>

<p><a href="https://i.sstatic.net/haJO8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/haJO8.png" alt="enter image description here"></a></p>

<hr>

<p><a href="https://i.sstatic.net/pI4Dg.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/pI4Dg.png" alt="enter image description here"></a></p>

<p>Did he come to expression (13.33) by applying the chain rule as </p>

<p>$$\eqalign{
  &amp; z = U({x_1},x_2^h)  \cr 
  &amp; {x_1} = g({x_2})  \cr 
  &amp; {{dz} \over {d{x_2}}} = {{\partial f} \over {\partial {x_1}}}{{d{x_1}} \over {d{x_2}}} + {{\partial f} \over {\partial {x_2}}}{{d{x_2}} \over {d{x_2}}} = {{\partial f} \over {\partial {x_1}}}{{d{x_1}} \over {d{x_2}}} + {{\partial f} \over {\partial {x_2}}} \cr} $$</p>

<p>since ${x_1}$ is a function of $x_2^h$ ? </p>

<microeconomics><mathematical-economics><public-goods>
Public Goods - Voluntary provision
user10699
# Public Goods - Voluntary provision

(Asked by: username_0 on 2016-12-02)

I am currently reading the book "Microeconomics: Principles and Analysis" by Cowell (2006), page 452-453. e have a two-commodity world, in which there are
${n\_h}$ agents (households): commodity 1 is a pure public good and commodity 2 is
purely private.

Each agent has an exogenously given income ${y^h}$ , denominated in units of the
private good 2. We imagine that the public good is to be financed voluntarily,
each household makes a contribution ${z^h}$ which leaves

$$x\_2^h = {y^h} - {z^h}$$ of the private good available for h’s own consumption
and
$${x\_1} = \phi (\bar z + {y^h} - x\_2^h)$$ where z is the total input of the good 2 used in the production process. Each agent realises that the total output of the public good depends upon his
or her own contribution and upon that made by others

---

The expression (13.30)

[![enter image description here](https://i.sstatic.net/THxdh.png)](https://i.sstatic.net/THxdh.png)

The expression (13.31)

[![enter image description here](https://i.sstatic.net/haJO8.png)](https://i.sstatic.net/haJO8.png)

---

[![enter image description here](https://i.sstatic.net/pI4Dg.png)](https://i.sstatic.net/pI4Dg.png)

Did he come to expression (13.33) by applying the chain rule as

$$\eqalign{
& z = U({x\_1},x\_2^h) \cr
& {x\_1} = g({x\_2}) \cr
& {{dz} \over {d{x\_2}}} = {{\partial f} \over {\partial {x\_1}}}{{d{x\_1}} \over {d{x\_2}}} + {{\partial f} \over {\partial {x\_2}}}{{d{x\_2}} \over {d{x\_2}}} = {{\partial f} \over {\partial {x\_1}}}{{d{x\_1}} \over {d{x\_2}}} + {{\partial f} \over {\partial {x\_2}}} \cr} $$

since ${x\_1}$ is a function of $x\_2^h$ ?

**Tags:** microeconomics, mathematical-economics, public-goods

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## 2 answers

### Answer by username_1 on 2016-12-02. Score: 1

Given

$$\max\_{x\_2^h} \quad U^h(\phi(\bar z + y^h - x\_2^h), x\_2^h)$$

You have to find FOCs by taking the chain rule twice for the first argument. Denote the argument for function $\phi$ as $(\cdot)$.

$$\frac{\partial U^h}{\partial x\_2^h} = \frac{\partial U^h}{\partial \phi} \cdot \frac{\partial \phi}{\partial x\_2^h}$$
$$\frac{\partial \phi}{\partial x\_2^h} = \frac{\partial \phi}{\partial (\cdot)} \cdot \frac{\partial (\cdot)}{\partial x\_2^h}$$
$$\implies \frac{\partial U^h}{\partial x\_2^h} = \frac{\partial U^h}{\partial \phi} \cdot \frac{\partial \phi}{\partial (\cdot)} \cdot \frac{\partial (\cdot)}{\partial x\_2^h}$$

Thus, letting $U^h\_{1,2}$ represent the partial derivative with respect to the first and second arguments of $U^h$ respectively:

$$\begin{align}
\frac{\partial U^h}{\partial x\_2^h} & = U\_1^h(\phi(\bar z + y^h - x\_2^h), x\_2^h) \cdot \phi\_z(\bar z + y^h - x\_2^h) \cdot -1 \\
& + U\_2^h(\phi(\bar z + y^h - x\_2^h), x\_2^h) \cdot 1
\end{align}$$

The second argument in $U$ has no $\phi$, so only chain rule should be applied there.

> **Comments:**
> **username_0**: isnt that what i did?  
> **username_1**: More or less. I'm merely adding some more detail @user10699  

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### Answer by username_2 on 2016-12-02. Score: 0

Cant you just apply this

[![enter image description here](https://i.sstatic.net/BfKId.gif)](https://i.sstatic.net/BfKId.gif)

Obviously different values but same principle ?


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