Partial derivatives of the dependent variable.
$begingroup$
I have a function of the form $$f(x,y)=frac{1}{exp((y-mu)/tau)+1 }$$ where $mu,tau,m$ are constants. Also $y^2=x^2+m^2$. How do I calculate $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$. Can I just not substitute the value of y in the expression?
multivariable-calculus partial-derivative
edited Dec 29 '18 at 10:55
Atheesh Rathnaweera
31
asked Dec 29 '18 at 9:51
IcchyamoyIcchyamoy
11
$endgroup$
add a comment |
$begingroup$
I have a function of the form $$f(x,y)=frac{1}{exp((y-mu)/tau)+1 }$$ where $mu,tau,m$ are constants. Also $y^2=x^2+m^2$. How do I calculate $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$. Can I just not substitute the value of y in the expression?
multivariable-calculus partial-derivative
edited Dec 29 '18 at 10:55
Atheesh Rathnaweera
31
asked Dec 29 '18 at 9:51
IcchyamoyIcchyamoy
11
$endgroup$
$begingroup$
Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:02
$begingroup$
I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
$endgroup$
– Icchyamoy
Dec 29 '18 at 10:05
$begingroup$
I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:06
add a comment |
$begingroup$
I have a function of the form $$f(x,y)=frac{1}{exp((y-mu)/tau)+1 }$$ where $mu,tau,m$ are constants. Also $y^2=x^2+m^2$. How do I calculate $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$. Can I just not substitute the value of y in the expression?
multivariable-calculus partial-derivative
edited Dec 29 '18 at 10:55
Atheesh Rathnaweera
31
asked Dec 29 '18 at 9:51
IcchyamoyIcchyamoy
11
$endgroup$
I have a function of the form $$f(x,y)=frac{1}{exp((y-mu)/tau)+1 }$$ where $mu,tau,m$ are constants. Also $y^2=x^2+m^2$. How do I calculate $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$. Can I just not substitute the value of y in the expression?
multivariable-calculus partial-derivative
multivariable-calculus partial-derivative
edited Dec 29 '18 at 10:55
Atheesh Rathnaweera
31
asked Dec 29 '18 at 9:51
IcchyamoyIcchyamoy
11
edited Dec 29 '18 at 10:55
Atheesh Rathnaweera
31
asked Dec 29 '18 at 9:51
IcchyamoyIcchyamoy
11
edited Dec 29 '18 at 10:55
Atheesh Rathnaweera
31
edited Dec 29 '18 at 10:55
Atheesh Rathnaweera
31
edited Dec 29 '18 at 10:55
Atheesh Rathnaweera
31
31
asked Dec 29 '18 at 9:51
IcchyamoyIcchyamoy
11
asked Dec 29 '18 at 9:51
IcchyamoyIcchyamoy
11
asked Dec 29 '18 at 9:51
IcchyamoyIcchyamoy
11
11
$begingroup$
Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:02
$begingroup$
I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
$endgroup$
– Icchyamoy
Dec 29 '18 at 10:05
$begingroup$
I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:06
add a comment |
$begingroup$
Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:02
$begingroup$
I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
$endgroup$
– Icchyamoy
Dec 29 '18 at 10:05
$begingroup$
I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:06
$begingroup$
Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:02
$begingroup$
Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:02
$begingroup$
I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
$endgroup$
– Icchyamoy
Dec 29 '18 at 10:05
$begingroup$
I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
$endgroup$
– Icchyamoy
Dec 29 '18 at 10:05
$begingroup$
I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:06
$begingroup$
I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.
$$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$
The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.
answered Dec 29 '18 at 10:11
Devashish KaushikDevashish Kaushik
638220
$endgroup$
add a comment |
Your Answer
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%2f3055692%2fpartial-derivatives-of-the-dependent-variable%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 $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.
$$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$
The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.
answered Dec 29 '18 at 10:11
Devashish KaushikDevashish Kaushik
638220
$endgroup$
add a comment |
$begingroup$
If $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.
$$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$
The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.
answered Dec 29 '18 at 10:11
Devashish KaushikDevashish Kaushik
638220
$endgroup$
add a comment |
$begingroup$
If $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.
$$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$
The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.
answered Dec 29 '18 at 10:11
Devashish KaushikDevashish Kaushik
638220
$endgroup$
If $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.
$$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$
The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.
answered Dec 29 '18 at 10:11
Devashish KaushikDevashish Kaushik
638220
answered Dec 29 '18 at 10:11
Devashish KaushikDevashish Kaushik
638220
answered Dec 29 '18 at 10:11
Devashish KaushikDevashish Kaushik
638220
answered Dec 29 '18 at 10:11
Devashish KaushikDevashish Kaushik
638220
638220
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%2f3055692%2fpartial-derivatives-of-the-dependent-variable%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
$begingroup$
Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:02
$begingroup$
I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
$endgroup$
– Icchyamoy
Dec 29 '18 at 10:05
$begingroup$
I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:06