First order condition of log functions in general and interpretation 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.
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.