Partial differentiation of an integral

[lfdahl]

Hello MHB members and friends!(Callme)

An economy student asked me, if I could explain the following partial differentiation:

\[\frac{\partial}{\partial C(i)}\int_{i\in[0;1]}[C(i)]^\frac{\eta - 1}{\eta}di
=\int_{j\in[0;1]}[C(j)]^\frac{\eta - 1}{\eta}dj\frac{\eta - 1}{\eta}[C(i)]^{-\frac{1}{\eta}}
\]

I am not sure, why the differentiation is performed as shown above ($\eta$ is a constant).

If it can be of any help in understanding the identity, the following should be added:

The function C(i) may or may not take a specific form. Whether or not, the C(i) is usually implicitly defined by the so called “felicity function”, which in this case takes the form:

$ u(C)=\frac{[C(i)]^{1-a}-1}{1-a}$, where a is a constant.

The function u(C) is a measure of the instantaneous utility a consumer has of the consumption amount C. The variable i is a time measure. The theory states, that the consumer prefers consumption instantaneously (“here and now”) instead of saving up for the future.

I presume, that the appearance of the partial derivative is a part of Lagranges optimization.

Thankyou in advance for any help in the matter. I´d also like to thank the MHB staff for a very exciting and interesting homepage!

 

[Fantini]

Hello (Wave) This is called differentiation under the integral sign. The notation might obscure this but I'll try to reveal it.

Although $\eta$ is a constant, you can consider $C(i)^{\frac{\eta -1}{\eta}}$ a function of two variables, that is

$$f(\eta, C(i)) = [C(i)]^{\frac{\eta -1}{\eta}}.$$

With this in mind, we can apply differentiation under the integral sign. Specifying

$$G(C(i)) = \int\limits_{i \in [0,1]} [C(i)]^{\frac{\eta -1}{\eta}} \, di$$

we have

$$
\begin{align}
\frac{d}{dC(i)}G(C(i)) & = \frac{\partial}{\partial C(i)} \int\limits_{i \in [0,1]} [C(i)]^{\frac{eta -1}{\eta}} \, di \\
& = \int\limits_{i \in [0,1]} \frac{\partial}{\partial C(i)} [C(i)]^{\frac{\eta -1}{\eta}} \, di \\
& = \int\limits_{i \in [0,1]} \left( \frac{\eta -1}{\eta} \right) [C(i)]^{\frac{\eta -1}{\eta} -1} \, di \\
& = \int\limits_{i \in [0,1]} \left( \frac{\eta -1}{\eta} \right) [C(i)]^{- \frac{1}{\eta}} \, di.
\end{align}
$$

The letter $j$ is irrelevant because the letter of integration does not matter. :) Notice that

$$\frac{\eta -1}{\eta} - 1 = \frac{\eta -1 - \eta}{\eta} = - \frac{1}{\eta}.$$

What I have done in

$$\frac{\partial}{\partial C(i)} [C(i)]^{\frac{\eta -1}{\eta}}$$

is differentiate as if it were $x^k$, where $x = C(i)$ and $k = (\eta -1)/\eta$.

Best wishes,

Fantini.

 

[lfdahl]

Hi, Fantini

Thankyou for your contribution. I didn´t realize, that you can consider $ C(i)^\frac{\eta - 1}{\eta}$ as a two-dimensional function, which is being partially differentiated.This makes sense! Thankyou.
I totally agree with you in the way you perform the differentiation (power function). In the beginning, I expected it to be that way too. But if you take a closer look at #1, you´ll notice that the derivative is outside the integral:
\[\frac{\partial}{\partial C(i)}\int_{i\in[0;1]}[C(i)]^\frac{\eta - 1}{\eta}di
=\int_{j\in[0;1]}[C(j)]^\frac{\eta - 1}{\eta}dj * \mathbf{\frac{\eta - 1}{\eta}[C(i)]^{-\frac{1}{\eta}}}
\]

Why is that so? Thus, it does make sense to me, that the author has chosen j as integration variable, in order to distinguish from the specific value i for which the differentiation takes place. But what exactly is going on when you perfom the differentiation of the integral?

 

[Fantini]

I don't think it makes sense to simply pop it out of the integral while it still depends on $i$. I guess I'm lost like you in that respect.

 

[CellCoree]

Hello! It's great to see that you are interested in understanding the partial differentiation of an integral in relation to economics. I'll try my best to explain it to you.

Firstly, the notation $\frac{\partial}{\partial C(i)}$ means that we are taking the partial derivative with respect to the variable $C(i)$, while holding all other variables constant. In this case, our integral is dependent on the variable $i$, but we are interested in how it changes with respect to $C(i)$.

Now, let's break down the expression:

\[\frac{\partial}{\partial C(i)}\int_{i\in[0;1]}[C(i)]^\frac{\eta - 1}{\eta}di\]

The integral is taken over the range $i\in[0;1]$, meaning that for each value of $i$ within this range, we are taking the value of $C(i)$ and raising it to the power of $\frac{\eta - 1}{\eta}$. We can think of this as a function of $C(i)$, which we will call $f(C(i))$. So, the integral can be rewritten as:

\[\int_{i\in[0;1]}f(C(i))di\]

Now, when we take the partial derivative with respect to $C(i)$, we are essentially finding the rate of change of the function $f(C(i))$ with respect to $C(i)$. In other words, we are finding how much the value of $f(C(i))$ changes as we change the value of $C(i)$. This can be represented as:

\[\frac{\partial}{\partial C(i)}f(C(i))\]

To find this rate of change, we can use the chain rule of differentiation, which states that:

\[\frac{d}{dx}f(g(x))=f'(g(x))g'(x)\]

Applying this to our expression, we get:

\[\frac{\partial}{\partial C(i)}f(C(i))=\frac{\eta - 1}{\eta}[C(i)]^{-\frac{1}{\eta}}\]

This is the first part of the expression that you were given. Now, we also need to consider the fact that we are integrating over the range $i\in[0;1]$. This means that we need to multiply