# Max of a profit function: partial derivative of an integral function?

I am struggling with the maximization of the following profit function in a New-Keynesian model. Here there is the FOC.

$$\frac{\delta}{\delta Y_t(i)} P_tY_t- \int_0^1 P_t(i)Y_t(i)di = \frac{\delta}{\delta Y_t(i)} P_t\left[\int_0^1 {Y_t(i)}^{\frac{\epsilon-1}{\epsilon}} di\right]^{\frac{\epsilon}{\epsilon-1}} - \int_0^1 P_t(i)Y_t(i)di = 0$$

$$Y_t(i)$$ is the demand for the intermediate good and $$Y_t$$ is the demand for the final good. The solution should be:

$$\frac{\epsilon}{\epsilon -1} P_t \frac{Y_t}{Y_t^\frac{\epsilon -1}{\epsilon}} \frac{\epsilon -1}{\epsilon} Y_t(i)^{-\frac{1}{\epsilon}} = P_t(i)$$

I tried to apply the Leibniz integral rule. However, I cannot understand how to behave when taking a derivative of an integral function.

May anyone explain to me the steps (and the mathematical theory) in order to get the correct solution?

You are really asking about the marginal product of a CES production function. The LHS of your second equation is $$P * dY/dY(i)$$ where $$Y_t$$ is a CES aggregator of the $$Y_t(i)$$.

Let's define $$\rho = \frac{\epsilon-1}{\epsilon}$$ so that $$Y=[\int Y(i)^\rho]^ {\frac{1}{\rho}}$$.

Now apply the chain rule to get the derivative (also available for lookup in any micro text book):

$$\partial Y/\partial Y(i) = \frac{1}{\rho} (\int Y(i) di) ^ \left(\frac{1}{\rho}-1\right) \rho Y(i)^{(\rho-1)}$$

• However, I cannot understand how to use the chain rule in case of $\frac{\delta}{\delta Y_t(i)} \int_0^1 P_t(i)Y_t(i)di$. What puzzles me the most is that the solution seems to neglect the fact that $Y_t(i)$ is inside an integral function. If I were to use the Leibniz rule, the derivative of that integral would be $\int_0^1 P_t(i)di$ and not $P_t(i)$ only.
• I don't think Leibniz rule is the right approach here. It deals with the case where you have a function $f(x, \theta)$, you integrate that and take the derivative w.r.to $\theta$. Here, you are basically changing one entry in a sum (the integral being a glorified sum). Aug 10, 2020 at 18:11