<p>This can be solved using Feynman’s differentiation under the integral sign.</p>
<p>By the chain rule,</p>
<p><span class="math-container">$\frac{d}{dC_i} [\int_{i=0}^{1} C_i^\frac{\eta -1}{\eta} di ]^\frac{\eta}{\eta-1}$</span></p>
<p><span class="math-container">$ = \frac{\eta}{\eta -1} [\int_{i=0}^{1} C_i^\frac{\eta -1}{\eta} di]^\frac{1}{\eta -1} \cdot \frac{d}{dC_i} [\int_{i=0}^{1} C_i^\frac{\eta -1}{\eta} di] $</span></p>
<p>Here is where we use Feynman’s technique (move the differential operator inside the integral):</p>
<p><span class="math-container">$ = \frac{\eta}{\eta -1} [\int_{i=0}^{1} C_i^\frac{\eta -1}{\eta} di]^\frac{1}{\eta - 1}  \cdot \int_{i=0}^{1} \frac{\partial}{\partial C_i} C_{i}^\frac{\eta - 1}{\eta} di$</span></p>
<p>After differentiating, we get the answer:</p>
<p><span class="math-container">$ = \frac{\eta}{\eta - 1} [\int_{i=0}^{1} C_i^\frac{\eta -1}{\eta} di]^\frac{1}{\eta - 1} \cdot \int_{i=0}^{1} \frac{\eta - 1}{\eta} C_{i}^{-\frac{1}{\eta}} di $</span></p>
<p>Taking the constant multiple outside the integral, which is the inverse of the first constant,</p>
<p><span class="math-container">$= [\int_{i=0}^{1} C_i^\frac{\eta -1}{\eta} di]^\frac{1}{\eta - 1} \cdot \int_{i=0}^{1} C_{i}^{-\frac{1}{\eta}} di $</span></p>

CC BY-SA 4.0
2023-03-30T13:40:28.053Z
I didn’t find changing one subscript from $i$ to $j$ necessary. In my solution it doesn’t matter as it’s just a dummy variable (a variable only used to integrate).
Nicolas Torres
CC BY-SA 4.0
2023-03-30T13:23:44.540Z
2023-03-30T13:31:22.893Z
2023-03-30T13:31:22.893Z
Nicolas Torres
<p>I need to solve the partial derivative to C_i of the integral provided below. I'd be more than happy for some help.</p>
<p><a href="https://i.sstatic.net/YznDt.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YznDt.png" alt="enter image description here" /></a></p>
<p>Based on the book's answer, we get:</p>
<p><a href="https://i.sstatic.net/MkWew.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MkWew.png" alt="enter image description here" /></a></p>
<p>Can someone explain how do we get there? And why does the subscript change from i to j?</p>

CC BY-SA 4.0
2023-03-29T22:41:05.533Z
You should use the chain rule: $[f(g(x))]' = f'(g(x))*g'(x)$. In your case, $g(x) = \int_{i=0}^{1} C_i^{\frac{\eta - 1}{\eta}} di$ and $f(g(x)) = g(x)^{\frac{\eta}{\eta - 1}}$.
BB King
CC BY-SA 4.0
2023-03-30T13:33:09.233Z
Didn’t you miss another integral sign when typing the books answer?
Nicolas Torres
CC BY-SA 4.0
2023-03-30T13:37:03.930Z
You meant $C_i$ and $i$ to be different things, right? I interpret it as $i$ being the integration variable and $C_i$ being some function of $i$.
Nicolas Torres
CC BY-SA 4.0
2023-03-30T18:43:13.277Z
Hi Nicolas, thanks for providing with an answer :)
But I didn't type the solution myself, it is a screenshot from Romer's book on Advanced Macroeconomics and this is the solution the author provides when solving the maximization of the Lagrangian under the Menu Cost model
Delia
<macroeconomics><utility><economic-growth><consumption><model>
Lagrangian for the utility-mazimization of a household in a Menu Cost model, first order condition to Ci?
Delia
# Lagrangian for the utility-mazimization of a household in a Menu Cost model, first order condition to Ci?

(Asked by: username_0 on 2023-03-29)

I need to solve the partial derivative to C\_i of the integral provided below. I'd be more than happy for some help.

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

Based on the book's answer, we get:

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

Can someone explain how do we get there? And why does the subscript change from i to j?

**Tags:** macroeconomics, utility, economic-growth, consumption, model

> **Comments:**
> **username_1**: You should use the chain rule: $[f(g(x))]' = f'(g(x))*g'(x)$. In your case, $g(x) = \int_{i=0}^{1} C_i^{\frac{\eta - 1}{\eta}} di$ and $f(g(x)) = g(x)^{\frac{\eta}{\eta - 1}}$.  
> **username_2**: Didn’t you miss another integral sign when typing the books answer?  
> **username_2**: You meant $C_i$ and $i$ to be different things, right? I interpret it as $i$ being the integration variable and $C_i$ being some function of $i$.  
> **username_0**: Hi Nicolas, thanks for providing with an answer :) But I didn't type the solution myself, it is a screenshot from Romer's book on Advanced Macroeconomics and this is the solution the author provides when solving the maximization of the Lagrangian under the Menu Cost model  

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## 1 answers

### Answer by username_2 on 2023-03-30. Score: 2

This can be solved using Feynman’s differentiation under the integral sign.

By the chain rule,

$\frac{d}{dC\_i} [\int\_{i=0}^{1} C\_i^\frac{\eta -1}{\eta} di ]^\frac{\eta}{\eta-1}$

$ = \frac{\eta}{\eta -1} [\int\_{i=0}^{1} C\_i^\frac{\eta -1}{\eta} di]^\frac{1}{\eta -1} \cdot \frac{d}{dC\_i} [\int\_{i=0}^{1} C\_i^\frac{\eta -1}{\eta} di] $

Here is where we use Feynman’s technique (move the differential operator inside the integral):

$ = \frac{\eta}{\eta -1} [\int\_{i=0}^{1} C\_i^\frac{\eta -1}{\eta} di]^\frac{1}{\eta - 1} \cdot \int\_{i=0}^{1} \frac{\partial}{\partial C\_i} C\_{i}^\frac{\eta - 1}{\eta} di$

After differentiating, we get the answer:

$ = \frac{\eta}{\eta - 1} [\int\_{i=0}^{1} C\_i^\frac{\eta -1}{\eta} di]^\frac{1}{\eta - 1} \cdot \int\_{i=0}^{1} \frac{\eta - 1}{\eta} C\_{i}^{-\frac{1}{\eta}} di $

Taking the constant multiple outside the integral, which is the inverse of the first constant,

$= [\int\_{i=0}^{1} C\_i^\frac{\eta -1}{\eta} di]^\frac{1}{\eta - 1} \cdot \int\_{i=0}^{1} C\_{i}^{-\frac{1}{\eta}} di $

> **Comments:**
> **username_2**: I didn’t find changing one subscript from $i$ to $j$ necessary. In my solution it doesn’t matter as it’s just a dummy variable (a variable only used to integrate).  

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