Understanding operator under a subtitution
$begingroup$
in my notes, I have the following phrase:
With $x = e^t$ and $x frac{d}{dx} = frac{d}{dt}$
How, how come we get $x frac{d}{dx} = frac{d}{dt} $? I know if I diferentiate with respect to $t$ I obtain
$$ frac{d}{dt} x = x $$
I do not understand how this operators are defined? maybe I am misunderstading the notation?
calculus
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,171319
$endgroup$
add a comment |
$begingroup$
in my notes, I have the following phrase:
With $x = e^t$ and $x frac{d}{dx} = frac{d}{dt}$
How, how come we get $x frac{d}{dx} = frac{d}{dt} $? I know if I diferentiate with respect to $t$ I obtain
$$ frac{d}{dt} x = x $$
I do not understand how this operators are defined? maybe I am misunderstading the notation?
calculus
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,171319
$endgroup$
add a comment |
$begingroup$
in my notes, I have the following phrase:
With $x = e^t$ and $x frac{d}{dx} = frac{d}{dt}$
How, how come we get $x frac{d}{dx} = frac{d}{dt} $? I know if I diferentiate with respect to $t$ I obtain
$$ frac{d}{dt} x = x $$
I do not understand how this operators are defined? maybe I am misunderstading the notation?
calculus
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,171319
$endgroup$
in my notes, I have the following phrase:
With $x = e^t$ and $x frac{d}{dx} = frac{d}{dt}$
How, how come we get $x frac{d}{dx} = frac{d}{dt} $? I know if I diferentiate with respect to $t$ I obtain
$$ frac{d}{dt} x = x $$
I do not understand how this operators are defined? maybe I am misunderstading the notation?
calculus
calculus
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,171319
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,171319
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,171319
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,171319
asked Jan 9 at 5:14
Jimmy SabaterJimmy Sabater
2,171319
2,171319
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
With
$x = e^t tag 1$
we have
$dfrac{dx}{dt} = e^t = x; tag 2$
thus for any function $f$
$xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$
by the chain rule. Thus,
$xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$
answered Jan 9 at 5:33
Robert LewisRobert Lewis
44.6k22964
$endgroup$
add a comment |
$begingroup$
$frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
$endgroup$
add a comment |
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%2f3067102%2funderstanding-operator-under-a-subtitution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
With
$x = e^t tag 1$
we have
$dfrac{dx}{dt} = e^t = x; tag 2$
thus for any function $f$
$xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$
by the chain rule. Thus,
$xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$
answered Jan 9 at 5:33
Robert LewisRobert Lewis
44.6k22964
$endgroup$
add a comment |
$begingroup$
With
$x = e^t tag 1$
we have
$dfrac{dx}{dt} = e^t = x; tag 2$
thus for any function $f$
$xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$
by the chain rule. Thus,
$xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$
answered Jan 9 at 5:33
Robert LewisRobert Lewis
44.6k22964
$endgroup$
add a comment |
$begingroup$
With
$x = e^t tag 1$
we have
$dfrac{dx}{dt} = e^t = x; tag 2$
thus for any function $f$
$xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$
by the chain rule. Thus,
$xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$
answered Jan 9 at 5:33
Robert LewisRobert Lewis
44.6k22964
$endgroup$
With
$x = e^t tag 1$
we have
$dfrac{dx}{dt} = e^t = x; tag 2$
thus for any function $f$
$xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$
by the chain rule. Thus,
$xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$
answered Jan 9 at 5:33
Robert LewisRobert Lewis
44.6k22964
answered Jan 9 at 5:33
Robert LewisRobert Lewis
44.6k22964
answered Jan 9 at 5:33
Robert LewisRobert Lewis
44.6k22964
answered Jan 9 at 5:33
Robert LewisRobert Lewis
44.6k22964
44.6k22964
add a comment |
add a comment |
$begingroup$
$frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
$endgroup$
add a comment |
$begingroup$
$frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
$endgroup$
add a comment |
$begingroup$
$frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
$endgroup$
$frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
answered Jan 9 at 5:22
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
53.9k32055
add a comment |
add a comment |
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%2f3067102%2funderstanding-operator-under-a-subtitution%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
