$ f'(x)+g'(y)=h(x)+t(y)$ The Next CEO of Stack OverflowAbout the number of functions passing for N pointsWhat are functions with the property $f(f(x)) = x$ called?Composite FunctionsIs it correct to say that since the derivative of a function is zero at a certain point imply that this point is an inflection point?What is the difference between a function and a formula?Is a function changed into another function by a change of variables?Repeating/“Periodic” Derivatives?Inner Function IdentificationDefining functions in terms of relationsCondition on functions satisfying certain inequality
Why didn't Khan get resurrected in the Genesis Explosion?
How to get from Geneva Airport to Metabief?
Why the difference in type-inference over the as-pattern in two similar function definitions?
Would a grinding machine be a simple and workable propulsion system for an interplanetary spacecraft?
Why do remote US companies require working in the US?
Can you be charged for obstruction for refusing to answer questions?
Is it my responsibility to learn a new technology in my own time my employer wants to implement?
How do I align (1) and (2)?
Easy to read palindrome checker
Powershell. How to parse gci Name?
What did we know about the Kessel run before the prequels?
Find non-case sensitive string in a mixed list of elements?
Method for adding error messages to a dictionary given a key
Is micro rebar a better way to reinforce concrete than rebar?
Why is the US ranked as #45 in Press Freedom ratings, despite its extremely permissive free speech laws?
Why is my new battery behaving weirdly?
Would this house-rule that treats advantage as a +1 to the roll instead (and disadvantage as -1) and allows them to stack be balanced?
Running a General Election and the European Elections together
Proper way to express "He disappeared them"
Which one is the true statement?
INSERT to a table from a database to other (same SQL Server) using Dynamic SQL
Can a Bladesinger Wizard use Bladesong with a Hand Crossbow?
Chain wire methods together in Lightning Web Components
Is there a difference between "Fahrstuhl" and "Aufzug"
$ f'(x)+g'(y)=h(x)+t(y)$
The Next CEO of Stack OverflowAbout the number of functions passing for N pointsWhat are functions with the property $f(f(x)) = x$ called?Composite FunctionsIs it correct to say that since the derivative of a function is zero at a certain point imply that this point is an inflection point?What is the difference between a function and a formula?Is a function changed into another function by a change of variables?Repeating/“Periodic” Derivatives?Inner Function IdentificationDefining functions in terms of relationsCondition on functions satisfying certain inequality
$begingroup$
$f$ and $h$ are two functions of $x$
$g$ and $t$ are two functions of $y$
And $f'$ derivative of $f$ ,$g'$ derivative of $g$
Is it correct to say that by identification $f'(x)=h(x)$ and $g'(y)=t(y)$ ?
functions
asked Mar 27 at 18:45
Pedro AlvarèsPedro Alvarès
756
$endgroup$
add a comment |
$begingroup$
$f$ and $h$ are two functions of $x$
$g$ and $t$ are two functions of $y$
And $f'$ derivative of $f$ ,$g'$ derivative of $g$
Is it correct to say that by identification $f'(x)=h(x)$ and $g'(y)=t(y)$ ?
functions
asked Mar 27 at 18:45
Pedro AlvarèsPedro Alvarès
756
$endgroup$
add a comment |
$begingroup$
$f$ and $h$ are two functions of $x$
$g$ and $t$ are two functions of $y$
And $f'$ derivative of $f$ ,$g'$ derivative of $g$
Is it correct to say that by identification $f'(x)=h(x)$ and $g'(y)=t(y)$ ?
functions
asked Mar 27 at 18:45
Pedro AlvarèsPedro Alvarès
756
$endgroup$
$f$ and $h$ are two functions of $x$
$g$ and $t$ are two functions of $y$
And $f'$ derivative of $f$ ,$g'$ derivative of $g$
Is it correct to say that by identification $f'(x)=h(x)$ and $g'(y)=t(y)$ ?
functions
functions
asked Mar 27 at 18:45
Pedro AlvarèsPedro Alvarès
756
asked Mar 27 at 18:45
Pedro AlvarèsPedro Alvarès
756
asked Mar 27 at 18:45
Pedro AlvarèsPedro Alvarès
756
asked Mar 27 at 18:45
Pedro AlvarèsPedro Alvarès
756
asked Mar 27 at 18:45
Pedro AlvarèsPedro Alvarès
756
756
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
No, there may be a constant. For example :
$$f'(x)=x^2+1$$
$$g'(y)=y^2+3$$
$$h(x)=x^2$$
$$t(y)=y^2+4$$
:)
answered Mar 27 at 18:49
EurekaEureka
436112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Not quite. Since we have
$$
f'(x) - h(x) = t(y) - g'(y),
$$
we know that each side must be a constant. If you're not convinced, consider what happens if we change $y$ but not $x$. However, this constant need not be zero. The best we can conclude is
$$
f'(x) = h(x) + C;;;;;;;;; g'(y) = t(y) - C
$$
for some constant $C$.
This constant $C$ is very important and definitely can't be neglected. For example, in the hydrogen atom problem in quantum mechanics, these constants appear as the quantum numbers of electron states.
answered Mar 27 at 18:53
eyeballfrogeyeballfrog
7,044633
$endgroup$
add a comment |
$begingroup$
Not quite. If you isolate the x-terms on one side and the y-terms on the other, you get
$$f'(x)-h(x)=t(y)-g'(y)$$
Since the LHS depends only on $x$ and the RHS depends only on $y$, you may conclude that both sides are equal to some constant $C$.
So you have $$f'(x)=h(x) + C$$ and $$g'(y) = t(y)-C$$
where the $C$ is the same in both equations.
Note there may be many values of $C$ for which this is true, or there may be none. Not every value of $C$ necessarily produces equations which have solutions (but they may).
answered Mar 27 at 18:53
MPWMPW
31k12157
$endgroup$
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%2f3164943%2ffxgy-hxty%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, there may be a constant. For example :
$$f'(x)=x^2+1$$
$$g'(y)=y^2+3$$
$$h(x)=x^2$$
$$t(y)=y^2+4$$
:)
answered Mar 27 at 18:49
EurekaEureka
436112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
No, there may be a constant. For example :
$$f'(x)=x^2+1$$
$$g'(y)=y^2+3$$
$$h(x)=x^2$$
$$t(y)=y^2+4$$
:)
answered Mar 27 at 18:49
EurekaEureka
436112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
No, there may be a constant. For example :
$$f'(x)=x^2+1$$
$$g'(y)=y^2+3$$
$$h(x)=x^2$$
$$t(y)=y^2+4$$
:)
answered Mar 27 at 18:49
EurekaEureka
436112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
No, there may be a constant. For example :
$$f'(x)=x^2+1$$
$$g'(y)=y^2+3$$
$$h(x)=x^2$$
$$t(y)=y^2+4$$
:)
answered Mar 27 at 18:49
EurekaEureka
436112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 27 at 18:49
EurekaEureka
436112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 27 at 18:49
EurekaEureka
436112
answered Mar 27 at 18:49
EurekaEureka
436112
436112
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
Not quite. Since we have
$$
f'(x) - h(x) = t(y) - g'(y),
$$
we know that each side must be a constant. If you're not convinced, consider what happens if we change $y$ but not $x$. However, this constant need not be zero. The best we can conclude is
$$
f'(x) = h(x) + C;;;;;;;;; g'(y) = t(y) - C
$$
for some constant $C$.
This constant $C$ is very important and definitely can't be neglected. For example, in the hydrogen atom problem in quantum mechanics, these constants appear as the quantum numbers of electron states.
answered Mar 27 at 18:53
eyeballfrogeyeballfrog
7,044633
$endgroup$
add a comment |
$begingroup$
Not quite. Since we have
$$
f'(x) - h(x) = t(y) - g'(y),
$$
we know that each side must be a constant. If you're not convinced, consider what happens if we change $y$ but not $x$. However, this constant need not be zero. The best we can conclude is
$$
f'(x) = h(x) + C;;;;;;;;; g'(y) = t(y) - C
$$
for some constant $C$.
This constant $C$ is very important and definitely can't be neglected. For example, in the hydrogen atom problem in quantum mechanics, these constants appear as the quantum numbers of electron states.
answered Mar 27 at 18:53
eyeballfrogeyeballfrog
7,044633
$endgroup$
add a comment |
$begingroup$
Not quite. Since we have
$$
f'(x) - h(x) = t(y) - g'(y),
$$
we know that each side must be a constant. If you're not convinced, consider what happens if we change $y$ but not $x$. However, this constant need not be zero. The best we can conclude is
$$
f'(x) = h(x) + C;;;;;;;;; g'(y) = t(y) - C
$$
for some constant $C$.
This constant $C$ is very important and definitely can't be neglected. For example, in the hydrogen atom problem in quantum mechanics, these constants appear as the quantum numbers of electron states.
answered Mar 27 at 18:53
eyeballfrogeyeballfrog
7,044633
$endgroup$
Not quite. Since we have
$$
f'(x) - h(x) = t(y) - g'(y),
$$
we know that each side must be a constant. If you're not convinced, consider what happens if we change $y$ but not $x$. However, this constant need not be zero. The best we can conclude is
$$
f'(x) = h(x) + C;;;;;;;;; g'(y) = t(y) - C
$$
for some constant $C$.
This constant $C$ is very important and definitely can't be neglected. For example, in the hydrogen atom problem in quantum mechanics, these constants appear as the quantum numbers of electron states.
answered Mar 27 at 18:53
eyeballfrogeyeballfrog
7,044633
answered Mar 27 at 18:53
eyeballfrogeyeballfrog
7,044633
answered Mar 27 at 18:53
eyeballfrogeyeballfrog
7,044633
answered Mar 27 at 18:53
eyeballfrogeyeballfrog
7,044633
7,044633
add a comment |
add a comment |
$begingroup$
Not quite. If you isolate the x-terms on one side and the y-terms on the other, you get
$$f'(x)-h(x)=t(y)-g'(y)$$
Since the LHS depends only on $x$ and the RHS depends only on $y$, you may conclude that both sides are equal to some constant $C$.
So you have $$f'(x)=h(x) + C$$ and $$g'(y) = t(y)-C$$
where the $C$ is the same in both equations.
Note there may be many values of $C$ for which this is true, or there may be none. Not every value of $C$ necessarily produces equations which have solutions (but they may).
answered Mar 27 at 18:53
MPWMPW
31k12157
$endgroup$
add a comment |
$begingroup$
Not quite. If you isolate the x-terms on one side and the y-terms on the other, you get
$$f'(x)-h(x)=t(y)-g'(y)$$
Since the LHS depends only on $x$ and the RHS depends only on $y$, you may conclude that both sides are equal to some constant $C$.
So you have $$f'(x)=h(x) + C$$ and $$g'(y) = t(y)-C$$
where the $C$ is the same in both equations.
Note there may be many values of $C$ for which this is true, or there may be none. Not every value of $C$ necessarily produces equations which have solutions (but they may).
answered Mar 27 at 18:53
MPWMPW
31k12157
$endgroup$
add a comment |
$begingroup$
Not quite. If you isolate the x-terms on one side and the y-terms on the other, you get
$$f'(x)-h(x)=t(y)-g'(y)$$
Since the LHS depends only on $x$ and the RHS depends only on $y$, you may conclude that both sides are equal to some constant $C$.
So you have $$f'(x)=h(x) + C$$ and $$g'(y) = t(y)-C$$
where the $C$ is the same in both equations.
Note there may be many values of $C$ for which this is true, or there may be none. Not every value of $C$ necessarily produces equations which have solutions (but they may).
answered Mar 27 at 18:53
MPWMPW
31k12157
$endgroup$
Not quite. If you isolate the x-terms on one side and the y-terms on the other, you get
$$f'(x)-h(x)=t(y)-g'(y)$$
Since the LHS depends only on $x$ and the RHS depends only on $y$, you may conclude that both sides are equal to some constant $C$.
So you have $$f'(x)=h(x) + C$$ and $$g'(y) = t(y)-C$$
where the $C$ is the same in both equations.
Note there may be many values of $C$ for which this is true, or there may be none. Not every value of $C$ necessarily produces equations which have solutions (but they may).
answered Mar 27 at 18:53
MPWMPW
31k12157
answered Mar 27 at 18:53
MPWMPW
31k12157
answered Mar 27 at 18:53
MPWMPW
31k12157
answered Mar 27 at 18:53
MPWMPW
31k12157
31k12157
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%2f3164943%2ffxgy-hxty%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
