Partial derivative of x - is quotient rule necessary?
Let
$$u(x,y)=frac{x}{x^2+y^2}$$
I'm trying to determine if the given function is harmonic. I know that the 2nd partial derivative with respect to $x$ should, when added to the 2nd partial derivative of $y$, equal $0$.
However, I'm kind of stuck. I'm using the quotient rule to solve for the partial derivative of $x$, but is this the right way to take a partial derivative of a quotient?
multivariable-calculus derivatives partial-derivative
edited 2 days ago
Martin Sleziak
44.7k8115271
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
757521
add a comment |
Let
$$u(x,y)=frac{x}{x^2+y^2}$$
I'm trying to determine if the given function is harmonic. I know that the 2nd partial derivative with respect to $x$ should, when added to the 2nd partial derivative of $y$, equal $0$.
However, I'm kind of stuck. I'm using the quotient rule to solve for the partial derivative of $x$, but is this the right way to take a partial derivative of a quotient?
multivariable-calculus derivatives partial-derivative
edited 2 days ago
Martin Sleziak
44.7k8115271
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
757521
1
This is not a harmonic analysis question. Please read tag descriptions before using.
– Matt Rosenzweig
Feb 2 '16 at 2:11
@MattRosenzweig this is not possible on the mobile app, my apologies
– whatwhatwhat
Feb 2 '16 at 2:13
add a comment |
Let
$$u(x,y)=frac{x}{x^2+y^2}$$
I'm trying to determine if the given function is harmonic. I know that the 2nd partial derivative with respect to $x$ should, when added to the 2nd partial derivative of $y$, equal $0$.
However, I'm kind of stuck. I'm using the quotient rule to solve for the partial derivative of $x$, but is this the right way to take a partial derivative of a quotient?
multivariable-calculus derivatives partial-derivative
edited 2 days ago
Martin Sleziak
44.7k8115271
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
757521
Let
$$u(x,y)=frac{x}{x^2+y^2}$$
I'm trying to determine if the given function is harmonic. I know that the 2nd partial derivative with respect to $x$ should, when added to the 2nd partial derivative of $y$, equal $0$.
However, I'm kind of stuck. I'm using the quotient rule to solve for the partial derivative of $x$, but is this the right way to take a partial derivative of a quotient?
multivariable-calculus derivatives partial-derivative
multivariable-calculus derivatives partial-derivative
edited 2 days ago
Martin Sleziak
44.7k8115271
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
757521
edited 2 days ago
Martin Sleziak
44.7k8115271
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
757521
edited 2 days ago
Martin Sleziak
44.7k8115271
edited 2 days ago
Martin Sleziak
44.7k8115271
edited 2 days ago
Martin Sleziak
44.7k8115271
44.7k8115271
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
757521
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
757521
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
757521
757521
1
This is not a harmonic analysis question. Please read tag descriptions before using.
– Matt Rosenzweig
Feb 2 '16 at 2:11
@MattRosenzweig this is not possible on the mobile app, my apologies
– whatwhatwhat
Feb 2 '16 at 2:13
add a comment |
1
This is not a harmonic analysis question. Please read tag descriptions before using.
– Matt Rosenzweig
Feb 2 '16 at 2:11
@MattRosenzweig this is not possible on the mobile app, my apologies
– whatwhatwhat
Feb 2 '16 at 2:13
1
1
This is not a harmonic analysis question. Please read tag descriptions before using.
– Matt Rosenzweig
Feb 2 '16 at 2:11
This is not a harmonic analysis question. Please read tag descriptions before using.
– Matt Rosenzweig
Feb 2 '16 at 2:11
@MattRosenzweig this is not possible on the mobile app, my apologies
– whatwhatwhat
Feb 2 '16 at 2:13
@MattRosenzweig this is not possible on the mobile app, my apologies
– whatwhatwhat
Feb 2 '16 at 2:13
add a comment |
1 Answer
1
active
oldest
votes
When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.
Now, you can use the product rule if you choose; you just have to rewrite:
$$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$
Then
$$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$
which you can simplify.
answered Feb 1 '16 at 20:05
IcemanIceman
739721
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%2f1636373%2fpartial-derivative-of-x-is-quotient-rule-necessary%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
When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.
Now, you can use the product rule if you choose; you just have to rewrite:
$$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$
Then
$$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$
which you can simplify.
answered Feb 1 '16 at 20:05
IcemanIceman
739721
add a comment |
When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.
Now, you can use the product rule if you choose; you just have to rewrite:
$$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$
Then
$$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$
which you can simplify.
answered Feb 1 '16 at 20:05
IcemanIceman
739721
add a comment |
When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.
Now, you can use the product rule if you choose; you just have to rewrite:
$$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$
Then
$$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$
which you can simplify.
answered Feb 1 '16 at 20:05
IcemanIceman
739721
When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.
Now, you can use the product rule if you choose; you just have to rewrite:
$$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$
Then
$$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$
which you can simplify.
answered Feb 1 '16 at 20:05
IcemanIceman
739721
edited Feb 1 '16 at 20:38
answered Feb 1 '16 at 20:05
IcemanIceman
739721
answered Feb 1 '16 at 20:05
IcemanIceman
739721
answered Feb 1 '16 at 20:05
IcemanIceman
739721
739721
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f1636373%2fpartial-derivative-of-x-is-quotient-rule-necessary%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
1
This is not a harmonic analysis question. Please read tag descriptions before using.
– Matt Rosenzweig
Feb 2 '16 at 2:11
@MattRosenzweig this is not possible on the mobile app, my apologies
– whatwhatwhat
Feb 2 '16 at 2:13