If $f = f(x,y)$ and $C$ is constant, then is this true: $frac{partial f}{partial (Cx)}=frac{1}{C}frac{partial...
$begingroup$
Suppose $f=f(x,y)$ is any function of 2 variables and $C=mbox{const}$. Can one do
$$frac{partial f}{partial (Cx)}=frac{1}{C}frac{partial f}{partial x}$$
Is this always correct? Or do we need the derivative $f_x'$ to exist, or be continuous before we can write the above?
multivariable-calculus derivatives partial-derivative
$endgroup$
add a comment |
$begingroup$
Suppose $f=f(x,y)$ is any function of 2 variables and $C=mbox{const}$. Can one do
$$frac{partial f}{partial (Cx)}=frac{1}{C}frac{partial f}{partial x}$$
Is this always correct? Or do we need the derivative $f_x'$ to exist, or be continuous before we can write the above?
multivariable-calculus derivatives partial-derivative
$endgroup$
add a comment |
$begingroup$
Suppose $f=f(x,y)$ is any function of 2 variables and $C=mbox{const}$. Can one do
$$frac{partial f}{partial (Cx)}=frac{1}{C}frac{partial f}{partial x}$$
Is this always correct? Or do we need the derivative $f_x'$ to exist, or be continuous before we can write the above?
multivariable-calculus derivatives partial-derivative
$endgroup$
Suppose $f=f(x,y)$ is any function of 2 variables and $C=mbox{const}$. Can one do
$$frac{partial f}{partial (Cx)}=frac{1}{C}frac{partial f}{partial x}$$
Is this always correct? Or do we need the derivative $f_x'$ to exist, or be continuous before we can write the above?
multivariable-calculus derivatives partial-derivative
multivariable-calculus derivatives partial-derivative
edited Dec 18 '18 at 21:52
user142523
asked Dec 18 '18 at 15:56
user142523user142523
18112
18112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If by "constant" you mean $C$ depends on neither $x$ nor $y$, then for $Cne 0$ you only need $partial f/partial x$ to exist. Necessity is trivial; sufficiency is an exercise. (Hint: use $lim_{xto a}kg(x)=klim_{xto a}g(x)$.)
$endgroup$
$begingroup$
Thank you for your answer! What about $frac{partial^2 f}{partial (Cx)partial (Cy)} = frac{1}{C^2}frac{partial^2 f}{partial x partial y}$
$endgroup$
– user142523
Dec 18 '18 at 16:13
1
$begingroup$
@user142553 Again, we only need the (in this case second) derivative to exist.
$endgroup$
– J.G.
Dec 18 '18 at 16:14
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%2f3045324%2fif-f-fx-y-and-c-is-constant-then-is-this-true-frac-partial-f-part%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If by "constant" you mean $C$ depends on neither $x$ nor $y$, then for $Cne 0$ you only need $partial f/partial x$ to exist. Necessity is trivial; sufficiency is an exercise. (Hint: use $lim_{xto a}kg(x)=klim_{xto a}g(x)$.)
$endgroup$
$begingroup$
Thank you for your answer! What about $frac{partial^2 f}{partial (Cx)partial (Cy)} = frac{1}{C^2}frac{partial^2 f}{partial x partial y}$
$endgroup$
– user142523
Dec 18 '18 at 16:13
1
$begingroup$
@user142553 Again, we only need the (in this case second) derivative to exist.
$endgroup$
– J.G.
Dec 18 '18 at 16:14
add a comment |
$begingroup$
If by "constant" you mean $C$ depends on neither $x$ nor $y$, then for $Cne 0$ you only need $partial f/partial x$ to exist. Necessity is trivial; sufficiency is an exercise. (Hint: use $lim_{xto a}kg(x)=klim_{xto a}g(x)$.)
$endgroup$
$begingroup$
Thank you for your answer! What about $frac{partial^2 f}{partial (Cx)partial (Cy)} = frac{1}{C^2}frac{partial^2 f}{partial x partial y}$
$endgroup$
– user142523
Dec 18 '18 at 16:13
1
$begingroup$
@user142553 Again, we only need the (in this case second) derivative to exist.
$endgroup$
– J.G.
Dec 18 '18 at 16:14
add a comment |
$begingroup$
If by "constant" you mean $C$ depends on neither $x$ nor $y$, then for $Cne 0$ you only need $partial f/partial x$ to exist. Necessity is trivial; sufficiency is an exercise. (Hint: use $lim_{xto a}kg(x)=klim_{xto a}g(x)$.)
$endgroup$
If by "constant" you mean $C$ depends on neither $x$ nor $y$, then for $Cne 0$ you only need $partial f/partial x$ to exist. Necessity is trivial; sufficiency is an exercise. (Hint: use $lim_{xto a}kg(x)=klim_{xto a}g(x)$.)
answered Dec 18 '18 at 16:02
J.G.J.G.
28.1k22844
28.1k22844
$begingroup$
Thank you for your answer! What about $frac{partial^2 f}{partial (Cx)partial (Cy)} = frac{1}{C^2}frac{partial^2 f}{partial x partial y}$
$endgroup$
– user142523
Dec 18 '18 at 16:13
1
$begingroup$
@user142553 Again, we only need the (in this case second) derivative to exist.
$endgroup$
– J.G.
Dec 18 '18 at 16:14
add a comment |
$begingroup$
Thank you for your answer! What about $frac{partial^2 f}{partial (Cx)partial (Cy)} = frac{1}{C^2}frac{partial^2 f}{partial x partial y}$
$endgroup$
– user142523
Dec 18 '18 at 16:13
1
$begingroup$
@user142553 Again, we only need the (in this case second) derivative to exist.
$endgroup$
– J.G.
Dec 18 '18 at 16:14
$begingroup$
Thank you for your answer! What about $frac{partial^2 f}{partial (Cx)partial (Cy)} = frac{1}{C^2}frac{partial^2 f}{partial x partial y}$
$endgroup$
– user142523
Dec 18 '18 at 16:13
$begingroup$
Thank you for your answer! What about $frac{partial^2 f}{partial (Cx)partial (Cy)} = frac{1}{C^2}frac{partial^2 f}{partial x partial y}$
$endgroup$
– user142523
Dec 18 '18 at 16:13
1
1
$begingroup$
@user142553 Again, we only need the (in this case second) derivative to exist.
$endgroup$
– J.G.
Dec 18 '18 at 16:14
$begingroup$
@user142553 Again, we only need the (in this case second) derivative to exist.
$endgroup$
– J.G.
Dec 18 '18 at 16:14
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%2f3045324%2fif-f-fx-y-and-c-is-constant-then-is-this-true-frac-partial-f-part%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