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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)

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The expression (13.31)

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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 $x_2^h$ ?

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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.

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  • $\begingroup$ isnt that what i did? $\endgroup$ – user10699 Dec 3 '16 at 0:39
  • $\begingroup$ More or less. I'm merely adding some more detail @user10699 $\endgroup$ – Kitsune Cavalry Dec 5 '16 at 15:31
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Cant you just apply this

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Obviously different values but same principle ?

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