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I have a function

$$U = 2A\{\alpha, \beta\} + 3B\{\beta,\gamma\} + 4\alpha$$

where $\alpha, \beta, \gamma$ are exogenous parameters and with curly brackets I mean a functional dependence (that is, $A$ depends from $\alpha$ and $\beta$ and $B$ depends from $\beta$ and $\gamma$).

I has been told that the partial derivative $\frac {\partial U} {\partial \alpha} $ is $2A_\alpha + 4$.

But shouldn't be that the total derivative? I thought that partial derivatives account only for direct effects, so that $\frac {\partial U} {\partial \alpha} $ should be simply $4$.

UPDATED: In the comments I have got that $2A_\alpha + 4\alpha$ should be considered as the partial derivative, not the total one. However in the book "Fundamental methods of Mathematical Economics" of Chiang it is written:

Finding the Total Derivative

To carry on the discussion in a general framework, let us consider any function $y=f(x,w)$ where $x=g(w)$.

The two functions $f$ and $g$ can also be combined into a composite function $y=f[g(w),w]$.

The three variables $y$, $x$, and $w$ are related to one another as shown in Fig. 8.4. In this figure, which we shall refer to as channel map, it is clearly seen that $w$ - the ultimate source of change- can affect $y$ through two separate channels: (1) indirectly, via the function $g$ and then $f$ (the straight arrows), and (2) directly, via the function $f$ (the curved arrow). The direct effect can simply be represented by the partial derivative $f_w$. But the indirect effect can only be expressed by a product of two derivatives, $f_x \frac{dx}{dw}$ , or $\frac{\partial y}{\partial x} \frac{dx}{dy}$, by the chain rule for a composite function. Adding up the two effects gives us the desired total derivative of $y$ with respect to $w$:

$$\frac{dy}{dw} = f_x \frac{dx}{dw} + f_w = \frac{\partial y}{\partial x} \frac{dx}{dw} + \frac{\partial y} {\partial w}$$

Figure 8.4

According to the Chiang book hence in this case the partial derivative $\frac{\partial U}{\partial \alpha}$ should be just 4, while the total derivative $\frac{dU}{d\alpha}$ should be $2A_\alpha + 4$.

So, who is right?

Antonello
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  • If $\alpha$ changes then $A$ may change, hence $U$ may change. – Simon Nov 02 '16 at 14:09
  • @Simon yes, that's correct.. $\alpha$ influence U directly and indirectly trough A. But my question is if partial derivatives are defined to consider only the direct effects or also the indirect ones. – Antonello Nov 02 '16 at 14:35
  • Any effects, including so-called indirect ones. – Simon Nov 02 '16 at 15:19
  • A partial derivative is a derivative in a given direction, so if you think on it as a limit, it is easier to see that all the effects in that direction are counting – Fernando Nov 02 '16 at 19:11
  • @Simon (and Fernando): so, Chiang is wrong? – Antonello Nov 05 '16 at 09:33
  • Is $y=f(x, y)$ correct, or does it say $y=f (x, w)$ ? – Simon Nov 05 '16 at 09:42
  • Also please can you show us fig 8.4 ? – Simon Nov 05 '16 at 09:43
  • @Simon: Thanks, I have corrected the typo and added the img.. – Antonello Nov 06 '16 at 19:57
  • Thank you for the clarification Antonello. It seems to me that the confusion is very largely semantic. The answer by Gus here http://math.stackexchange.com/questions/174270/what-exactly-is-the-difference-between-a-derivative-and-a-total-derivative sums up very well what I think. – Simon Nov 07 '16 at 11:39

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