2
$\begingroup$

Consider the following integral

$$I = \int^b_af[x[t];c]dt$$

where $a,b$ and $c$ are scalers and $x[t]$ is a function of $t$. I want to know $\frac{\partial I}{\partial x}$.

If I apply Leibniz Rule, I get

$$\frac{\partial I}{\partial x} = \int^b_a\frac{\partial f}{\partial x}dt$$

I feel this is an incorrect application of the rule, since $x$ is a function of $t$.

Alternatively, I can think of re-defining $f$ as $g = f[x+\epsilon;c]$, where for $\epsilon = 0$ we have the original function. Now if I apply Leibniz Rule by differentiating with respect to $\epsilon$ (and then evaluate the derivative at $\epsilon = 0$), I get,

$$\frac{\partial I}{\partial \epsilon} = \int^b_a\frac{\partial g}{\partial x}|_{\epsilon=0} dt$$

Questions:

  1. Am I correct in saying that my first method is incorrect?
  2. If the alternative method is correct, does it correctly capture $\frac{\partial I}{\partial x}$?

Edit:

It might help to explain what I was thinking of when I wrote the question:

Suppose $x$ represents monetary values at different points of time ($t$) and $f$ is a function of $x$. $I$ is adding $f$ from $t=a$ to $t=b$. I would like to calculate, how the sum ($I$) changes if the entire sequence $x$, say, goes up or down.

$\endgroup$
11
  • $\begingroup$ Regarding 1) note that you are taking the partial derivative. $\endgroup$ Jul 31 '20 at 15:11
  • $\begingroup$ @MathsWizzard I see. But $x$ is the only variable argument, so would it matter if I wrote partial instead of total derivative? $\endgroup$
    – erik
    Jul 31 '20 at 15:15
  • 1
    $\begingroup$ $I$ is a number since bounds do not depend on $x$ so derivative is $0$. $\endgroup$
    – zwim
    Jul 31 '20 at 15:18
  • 1
    $\begingroup$ @erik I would still write it as partial derivative because remember that one could also take the derivative with respect to $t$ (by the fundamental theorem of calculus) $\endgroup$ Jul 31 '20 at 15:23
  • 1
    $\begingroup$ @MathsWizzard: $I$ is a function on the space of functions $x(t)$ (with domain $[a,b]$ and presumably some continuity/differentiability restrictions). This is called a functional in the calculus of variations literature. In no way is an integral $\int_a^b g(t)\,dt$ a function of $t$. At any rate, NO partial derivatives involved, but recognize that you're differentiating in an infinite-dimensional space. $\endgroup$ Jul 31 '20 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.