<p>The <span class="math-container">$\gamma$</span> on the RHS comes from applying chain rule when differentiating the second term with respect to <span class="math-container">$a_i$</span>.</p>
<p>Regarding elasticity, note that with a differentiable function <span class="math-container">$f$</span>, the ratio <span class="math-container">$f'(x)/f(x)$</span> can be interpreted as the <em>percentage change in the value of <span class="math-container">$f$</span></em> around <span class="math-container">$x$</span>. So isolating <span class="math-container">$\gamma$</span> from the FOC, you'll get an expression for elasticity, which you can interpret in the context of you problem.</p>

CC BY-SA 4.0
2021-02-02T21:52:01.910Z
2021-02-02T21:52:01.910Z
Herr K.
<p>It just comes from the derivative of profit function. I assume that <span class="math-container">$a_i$</span> is the choice variable here so the derivative of <span class="math-container">$\pi$</span> wrt <span class="math-container">$a_i$</span> is (step by step):</p>
<p><span class="math-container">$$\frac{\partial \pi}{ \partial a_i} = \frac{\partial \pi}{ \partial a_i} [ \ln R(a_i) ]  +  \frac{\partial \pi}{ \partial a_i} [ \ln N_i(\gamma a_i, \gamma a_{-i}) ]  \\ 
= \frac{1}{ R(a_i)}   R'(a_i)  +  \frac{1}{ N_i(\gamma a_i, \gamma a_{-i})}  N_i'(\gamma a_i, \gamma a_{-i}) \gamma $$</span></p>
<p>The gamma at the end pops out because of the chain rule. If you have <span class="math-container">$G(H(F(x)))$</span> then <span class="math-container">$\frac{dG}{dx} = \frac{dG}{dH} \frac{dH}{dF}\frac{dF}{dx}$</span>. You just simply have to realize that <span class="math-container">$\gamma a_i = F(a_i) $</span>.</p>
<p>Lastly to derive FOC you just set the last equation above to zero and you are there:</p>
<p><span class="math-container">$$\frac{1}{ R(a_i)}   R'(a_i)  +  \frac{1}{ N_i(\gamma a_i, \gamma a_{-i})}  N_i'(\gamma a_i, \gamma a_{-i}) \gamma = 0 \\ \frac{ R'(a_i)}{ R(a_i)}    = - \gamma \frac{ N_i'(\gamma a_i, \gamma a_{-i})}{ N_i(\gamma a_i, \gamma a_{-i})} $$</span></p>
<p>To sum up, the <span class="math-container">$-\gamma$</span> is there because of the chain rule, nothing special aside of taking derivative is going on here.</p>
<p>PS: I just realized that I forgot to explain why you can interpret them as elasticities but great +1 answer of Herr K. already did that so I wont unnecessarily repeat it here.</p>

CC BY-SA 4.0
2021-02-04T18:10:55.953Z
Thank you for the thorough answers. A quick follow up question: elasticity should be normally defined as (%change in N)/(%change in a), however, I assume it is not the case here, is this correct? Here it should be interpreted as the %change in N when a increases by one UNIT? Is this correct?
randomname
CC BY-SA 4.0
2021-02-04T18:54:08.067Z
@randomname you are welcome, a rigorous definition of elasticity of y wrt x is $dY/dx(x/y)$ (%change in Y over % change in x is just an approximation **not** the definition). As mentioned in the other answer if you are interested in the elasticity around fixed x then as mentioned in the other answer you will get the elasticity already from the fraction.
1muflon1
CC BY-SA 4.0
2021-02-02T21:42:12.943Z
2021-02-02T22:00:31.853Z
2021-02-02T22:00:31.853Z
1muflon1
<p>In the following, a first order condition for a log function is calculated.</p>
<p>I know how the left part was calculated, however, I am a little bit confused where the gamma on the right site comes from without further knowledge about the concrete function of N. Could someone explain?</p>
<p>And why can one interpret these findings as elasticity?</p>
<p>Info: R(a) is the ad revenue for a platform i dependent on ad level a. N is the number of viewers which depends on the ad nuisance of platform i and the other platforms -i. The maximisation problem is to choose optimal ad level a.</p>
<p><a href="https://i.sstatic.net/eCX08.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/eCX08.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2021-02-02T18:57:30.530Z
Pls. consider fixing the grammar/language of your post. I realize english may not be your first language but your other posts do not seem to be this poorly formulated.
Jesper Hybel
<supply-and-demand><elasticity><market-failure><log-linearization><networks>
First order condition of log functions in general and interpretation
randomname
# First order condition of log functions in general and interpretation

(Asked by: username_0 on 2021-02-02)

In the following, a first order condition for a log function is calculated.

I know how the left part was calculated, however, I am a little bit confused where the gamma on the right site comes from without further knowledge about the concrete function of N. Could someone explain?

And why can one interpret these findings as elasticity?

Info: R(a) is the ad revenue for a platform i dependent on ad level a. N is the number of viewers which depends on the ad nuisance of platform i and the other platforms -i. The maximisation problem is to choose optimal ad level a.

[![enter image description here](https://i.sstatic.net/eCX08.jpg)](https://i.sstatic.net/eCX08.jpg)

**Tags:** supply-and-demand, elasticity, market-failure, log-linearization, networks
> **Comments:**
> **username_1**: Pls. consider fixing the grammar/language of your post. I realize english may not be your first language but your other posts do not seem to be this poorly formulated.  

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## 2 answers

### Answer by username_3 on 2021-02-02. Score: 6 | Accepted answer

It just comes from the derivative of profit function. I assume that $a\_i$ is the choice variable here so the derivative of $\pi$ wrt $a\_i$ is (step by step):

$$\frac{\partial \pi}{ \partial a\_i} = \frac{\partial \pi}{ \partial a\_i} [ \ln R(a\_i) ] + \frac{\partial \pi}{ \partial a\_i} [ \ln N\_i(\gamma a\_i, \gamma a\_{-i}) ] \\
= \frac{1}{ R(a\_i)} R'(a\_i) + \frac{1}{ N\_i(\gamma a\_i, \gamma a\_{-i})} N\_i'(\gamma a\_i, \gamma a\_{-i}) \gamma $$

The gamma at the end pops out because of the chain rule. If you have $G(H(F(x)))$ then $\frac{dG}{dx} = \frac{dG}{dH} \frac{dH}{dF}\frac{dF}{dx}$. You just simply have to realize that $\gamma a\_i = F(a\_i) $.

Lastly to derive FOC you just set the last equation above to zero and you are there:

$$\frac{1}{ R(a\_i)} R'(a\_i) + \frac{1}{ N\_i(\gamma a\_i, \gamma a\_{-i})} N\_i'(\gamma a\_i, \gamma a\_{-i}) \gamma = 0 \\ \frac{ R'(a\_i)}{ R(a\_i)} = - \gamma \frac{ N\_i'(\gamma a\_i, \gamma a\_{-i})}{ N\_i(\gamma a\_i, \gamma a\_{-i})} $$

To sum up, the $-\gamma$ is there because of the chain rule, nothing special aside of taking derivative is going on here.

PS: I just realized that I forgot to explain why you can interpret them as elasticities but great +1 answer of Herr K. already did that so I wont unnecessarily repeat it here.

> **Comments:**
> **username_0**: Thank you for the thorough answers. A quick follow up question: elasticity should be normally defined as (%change in N)/(%change in a), however, I assume it is not the case here, is this correct? Here it should be interpreted as the %change in N when a increases by one UNIT? Is this correct?  
> **username_3**: @randomname you are welcome, a rigorous definition of elasticity of y wrt x is $dY/dx(x/y)$ (%change in Y over % change in x is just an approximation **not** the definition). As mentioned in the other answer if you are interested in the elasticity around fixed x then as mentioned in the other answer you will get the elasticity already from the fraction.  

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### Answer by username_2 on 2021-02-02. Score: 6

The $\gamma$ on the RHS comes from applying chain rule when differentiating the second term with respect to $a\_i$.

Regarding elasticity, note that with a differentiable function $f$, the ratio $f'(x)/f(x)$ can be interpreted as the *percentage change in the value of $f$* around $x$. So isolating $\gamma$ from the FOC, you'll get an expression for elasticity, which you can interpret in the context of you problem.


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