Taking derivative of an integral with respect to a function
$begingroup$
I'm asked to do the following exercise: Find the first-order-condition with respect to $C(i)$:
$L=(int_{0}^1 C(i)^{1-frac{1}{epsilon}}di)^{frac{epsilon}{epsilon-1}}-lambda({int_{0}^1 P(i)C(i)di}-Z)$
I was confused when I came to $int_{0}^1 C(i)^{1-frac{1}{epsilon}}di$. I tried to apply Fundamental Theorem of Calculus, which says that integration is the inverse operation of differentiation. However, here I'm differentiating with respect to $C(i)$, not $i$, so it does not help.
Then I tried ${frac{dL}{di}{frac{di}{dC(i)}}}$. I think that if $C(i)$ is not constant, $frac{dC(i)}{di}neq0$, the first-order-condition is equivalent to $frac{dL}{di}=0$. When I differentiated $int_{0}^1 C(i)^{1-frac{1}{epsilon}}di$ with respect to $i$, I applied Fundamental Theorem of Calculus, got rid of the integration symbol, and kept $C(i)^{1-frac{1}{epsilon}}$.In this way, my result is:
${frac{epsilon}{epsilon-1}}{left(int_{0}^1 C(i)^{1-frac{1}{epsilon}}diright)^{frac{1}{epsilon-1}}}{C(i)^{-frac{1}{epsilon}}}=lambda P(i)$
However, my teacher's answer is:
${left(int_{0}^1 C(i)^{1-frac{1}{epsilon}}diright)^{frac{1}{epsilon-1}}}{C(i)^{-frac{1}{epsilon}}}=lambda P(i)$
I don't know where I made a mistake. I guess I couldn't apply Fundamental Theorem of Calculus here because I'm differentiating a definite integral, not an indefinite integral.
If you know how to do it in a more direct way, please write it down. Thank you for your time!
integration derivatives
asked Dec 5 '18 at 14:52
Yijia ChenYijia Chen
1
$endgroup$
add a comment |
$begingroup$
I'm asked to do the following exercise: Find the first-order-condition with respect to $C(i)$:
$L=(int_{0}^1 C(i)^{1-frac{1}{epsilon}}di)^{frac{epsilon}{epsilon-1}}-lambda({int_{0}^1 P(i)C(i)di}-Z)$
I was confused when I came to $int_{0}^1 C(i)^{1-frac{1}{epsilon}}di$. I tried to apply Fundamental Theorem of Calculus, which says that integration is the inverse operation of differentiation. However, here I'm differentiating with respect to $C(i)$, not $i$, so it does not help.
Then I tried ${frac{dL}{di}{frac{di}{dC(i)}}}$. I think that if $C(i)$ is not constant, $frac{dC(i)}{di}neq0$, the first-order-condition is equivalent to $frac{dL}{di}=0$. When I differentiated $int_{0}^1 C(i)^{1-frac{1}{epsilon}}di$ with respect to $i$, I applied Fundamental Theorem of Calculus, got rid of the integration symbol, and kept $C(i)^{1-frac{1}{epsilon}}$.In this way, my result is:
${frac{epsilon}{epsilon-1}}{left(int_{0}^1 C(i)^{1-frac{1}{epsilon}}diright)^{frac{1}{epsilon-1}}}{C(i)^{-frac{1}{epsilon}}}=lambda P(i)$
However, my teacher's answer is:
${left(int_{0}^1 C(i)^{1-frac{1}{epsilon}}diright)^{frac{1}{epsilon-1}}}{C(i)^{-frac{1}{epsilon}}}=lambda P(i)$
I don't know where I made a mistake. I guess I couldn't apply Fundamental Theorem of Calculus here because I'm differentiating a definite integral, not an indefinite integral.
If you know how to do it in a more direct way, please write it down. Thank you for your time!
integration derivatives
asked Dec 5 '18 at 14:52
Yijia ChenYijia Chen
1
$endgroup$
$begingroup$
Now I realize what's wrong. The Lagrangian $L$ is a functional, so it can only be differentiated by applying Frechet derivative.
$endgroup$
– Yijia Chen
Dec 6 '18 at 19:26
add a comment |
$begingroup$
I'm asked to do the following exercise: Find the first-order-condition with respect to $C(i)$:
$L=(int_{0}^1 C(i)^{1-frac{1}{epsilon}}di)^{frac{epsilon}{epsilon-1}}-lambda({int_{0}^1 P(i)C(i)di}-Z)$
I was confused when I came to $int_{0}^1 C(i)^{1-frac{1}{epsilon}}di$. I tried to apply Fundamental Theorem of Calculus, which says that integration is the inverse operation of differentiation. However, here I'm differentiating with respect to $C(i)$, not $i$, so it does not help.
Then I tried ${frac{dL}{di}{frac{di}{dC(i)}}}$. I think that if $C(i)$ is not constant, $frac{dC(i)}{di}neq0$, the first-order-condition is equivalent to $frac{dL}{di}=0$. When I differentiated $int_{0}^1 C(i)^{1-frac{1}{epsilon}}di$ with respect to $i$, I applied Fundamental Theorem of Calculus, got rid of the integration symbol, and kept $C(i)^{1-frac{1}{epsilon}}$.In this way, my result is:
${frac{epsilon}{epsilon-1}}{left(int_{0}^1 C(i)^{1-frac{1}{epsilon}}diright)^{frac{1}{epsilon-1}}}{C(i)^{-frac{1}{epsilon}}}=lambda P(i)$
However, my teacher's answer is:
${left(int_{0}^1 C(i)^{1-frac{1}{epsilon}}diright)^{frac{1}{epsilon-1}}}{C(i)^{-frac{1}{epsilon}}}=lambda P(i)$
I don't know where I made a mistake. I guess I couldn't apply Fundamental Theorem of Calculus here because I'm differentiating a definite integral, not an indefinite integral.
If you know how to do it in a more direct way, please write it down. Thank you for your time!
integration derivatives
asked Dec 5 '18 at 14:52
Yijia ChenYijia Chen
1
$endgroup$
I'm asked to do the following exercise: Find the first-order-condition with respect to $C(i)$:
$L=(int_{0}^1 C(i)^{1-frac{1}{epsilon}}di)^{frac{epsilon}{epsilon-1}}-lambda({int_{0}^1 P(i)C(i)di}-Z)$
I was confused when I came to $int_{0}^1 C(i)^{1-frac{1}{epsilon}}di$. I tried to apply Fundamental Theorem of Calculus, which says that integration is the inverse operation of differentiation. However, here I'm differentiating with respect to $C(i)$, not $i$, so it does not help.
Then I tried ${frac{dL}{di}{frac{di}{dC(i)}}}$. I think that if $C(i)$ is not constant, $frac{dC(i)}{di}neq0$, the first-order-condition is equivalent to $frac{dL}{di}=0$. When I differentiated $int_{0}^1 C(i)^{1-frac{1}{epsilon}}di$ with respect to $i$, I applied Fundamental Theorem of Calculus, got rid of the integration symbol, and kept $C(i)^{1-frac{1}{epsilon}}$.In this way, my result is:
${frac{epsilon}{epsilon-1}}{left(int_{0}^1 C(i)^{1-frac{1}{epsilon}}diright)^{frac{1}{epsilon-1}}}{C(i)^{-frac{1}{epsilon}}}=lambda P(i)$
However, my teacher's answer is:
${left(int_{0}^1 C(i)^{1-frac{1}{epsilon}}diright)^{frac{1}{epsilon-1}}}{C(i)^{-frac{1}{epsilon}}}=lambda P(i)$
I don't know where I made a mistake. I guess I couldn't apply Fundamental Theorem of Calculus here because I'm differentiating a definite integral, not an indefinite integral.
If you know how to do it in a more direct way, please write it down. Thank you for your time!
integration derivatives
integration derivatives
asked Dec 5 '18 at 14:52
Yijia ChenYijia Chen
1
asked Dec 5 '18 at 14:52
Yijia ChenYijia Chen
1
asked Dec 5 '18 at 14:52
Yijia ChenYijia Chen
1
asked Dec 5 '18 at 14:52
Yijia ChenYijia Chen
1
asked Dec 5 '18 at 14:52
Yijia ChenYijia Chen
1
1
$begingroup$
Now I realize what's wrong. The Lagrangian $L$ is a functional, so it can only be differentiated by applying Frechet derivative.
$endgroup$
– Yijia Chen
Dec 6 '18 at 19:26
add a comment |
$begingroup$
Now I realize what's wrong. The Lagrangian $L$ is a functional, so it can only be differentiated by applying Frechet derivative.
$endgroup$
– Yijia Chen
Dec 6 '18 at 19:26
$begingroup$
Now I realize what's wrong. The Lagrangian $L$ is a functional, so it can only be differentiated by applying Frechet derivative.
$endgroup$
– Yijia Chen
Dec 6 '18 at 19:26
$begingroup$
Now I realize what's wrong. The Lagrangian $L$ is a functional, so it can only be differentiated by applying Frechet derivative.
$endgroup$
– Yijia Chen
Dec 6 '18 at 19:26
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027152%2ftaking-derivative-of-an-integral-with-respect-to-a-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027152%2ftaking-derivative-of-an-integral-with-respect-to-a-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Now I realize what's wrong. The Lagrangian $L$ is a functional, so it can only be differentiated by applying Frechet derivative.
$endgroup$
– Yijia Chen
Dec 6 '18 at 19:26