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Business Math Partial differentiation of an integral

lfdahl

Well-known member
Nov 26, 2013
722
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

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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

Well-known member
Nov 26, 2013
722
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

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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. :confused: