Crthuky

how do we prove that a sum of two periods is still a period?

Finitely generated matrix groups whose eigenvalues are all algebraic

Using "tail" to follow a file without displaying the most recent lines

Processor speed limited at 0.4 Ghz

How to find if SQL server backup is encrypted with TDE without restoring the backup

Does int main() need a declaration on C++?

Why are UK visa biometrics appointments suspended at USCIS Application Support Centers?

How to Prove P(a) → ∀x(P(x) ∨ ¬(x = a)) using Natural Deduction

How exploitable/balanced is this homebrew spell: Spell Permanency?

How to travel to Japan while expressing milk?

Different meanings of こわい

Was the old ablative pronoun "med" or "mēd"?

Can a virus destroy the BIOS of a modern computer?

Send out email when Apex Queueable fails and test it

Is there a hemisphere-neutral way of specifying a season?

Fair gambler's ruin problem intuition

OP Amp not amplifying audio signal

How to install cross-compiler on Ubuntu 18.04?

How does a refinance allow a mortgage to be repaid?

Implication of namely

Getting extremely large arrows with tikzcd

Why do I get negative height?

Is this draw by repetition?

Can someone clarify Hamming's notion of important problems in relation to modern academia?

Finding the error in an argument

Chain rule notation for function with two variablesThe multivariable chain rule and functions that depend on themselvesCalculate partial derivative $f'_x, f'_y, f'_z$ where $f(x, y, z) = x^fracyz$Simple Chain Rule for PartialsChain rule for partial derivativesQuestion regarding the proof of the directional derivativePartial derivative of a function w.r.t an argument that occurs multiple timesDerivative of function of matrices using the product ruleWhen to use Partial derivatives and chain rulePartial derivative with dependent variables

3

$begingroup$

If $z=f(x,y)$ and $y=x^2$, then by the chain rule

$fracpartial zpartial x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$

Therefore

$2xfracpartial zpartial y=0$

and

$fracpartial zpartial y=0$

What is wrong with this argument?

I have a feeling that

1.) $x$ and $y$ do not have partial derivatives because they are single-variable, and

2.) $fracpartial zpartial y$ cannot be zero, because $y=x^2$ and therefore the derivative of any $y$ term exists.

How is my reasoning? I am pretty confused by this question.

calculus multivariable-calculus partial-derivative

share|cite|improve this question

edited 3 hours ago

mathenthusiast

asked 3 hours ago

mathenthusiastmathenthusiast

758

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My first question for you is why do you think this line of reasoning is incorrect? Did someone tell you it wasn't right?
  $endgroup$
  – BSplitter
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I also will note that if $2xfracpartial zpartial y =0$, then either $x=0$ or $fracpartial zpartial y =0$.
  $endgroup$
  – BSplitter
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BSplitter the problem itself poses this argument as incorrect, the object being to find out why
  $endgroup$
  – mathenthusiast
  3 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

If $z=f(x,y)$ and $y=x^2$, then by the chain rule

$fracpartial zpartial x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$

Therefore

$2xfracpartial zpartial y=0$

and

$fracpartial zpartial y=0$

What is wrong with this argument?

I have a feeling that

1.) $x$ and $y$ do not have partial derivatives because they are single-variable, and

2.) $fracpartial zpartial y$ cannot be zero, because $y=x^2$ and therefore the derivative of any $y$ term exists.

How is my reasoning? I am pretty confused by this question.

calculus multivariable-calculus partial-derivative

share|cite|improve this question

edited 3 hours ago

mathenthusiast

asked 3 hours ago

mathenthusiastmathenthusiast

758

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My first question for you is why do you think this line of reasoning is incorrect? Did someone tell you it wasn't right?
  $endgroup$
  – BSplitter
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I also will note that if $2xfracpartial zpartial y =0$, then either $x=0$ or $fracpartial zpartial y =0$.
  $endgroup$
  – BSplitter
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BSplitter the problem itself poses this argument as incorrect, the object being to find out why
  $endgroup$
  – mathenthusiast
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

If $z=f(x,y)$ and $y=x^2$, then by the chain rule

$fracpartial zpartial x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$

Therefore

$2xfracpartial zpartial y=0$

and

$fracpartial zpartial y=0$

What is wrong with this argument?

I have a feeling that

1.) $x$ and $y$ do not have partial derivatives because they are single-variable, and

2.) $fracpartial zpartial y$ cannot be zero, because $y=x^2$ and therefore the derivative of any $y$ term exists.

How is my reasoning? I am pretty confused by this question.

calculus multivariable-calculus partial-derivative

share|cite|improve this question

edited 3 hours ago

mathenthusiast

asked 3 hours ago

mathenthusiastmathenthusiast

758

$endgroup$

share|cite|improve this question

edited 3 hours ago

mathenthusiast

asked 3 hours ago

mathenthusiastmathenthusiast

758

share|cite|improve this question

edited 3 hours ago

mathenthusiast

asked 3 hours ago

mathenthusiastmathenthusiast

758

asked 3 hours ago

mathenthusiastmathenthusiast

758

asked 3 hours ago

mathenthusiastmathenthusiast

758

1 Answer
1

active

oldest

votes

4

$begingroup$

Nothing wrong. Just change it into

$$fracd zd x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$$

Note that that the first term is $fracd zd x$, which is different from $fracpartial zpartial x$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $fracd zd x$, which is NOT a partial derivative.

Actually, a better way to say this is that

$$left[fracpartial zpartial xright]_y=x^2=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y.$$

Where I have clearly written down the restriction $y=x^2$.

share|cite|improve this answer

answered 3 hours ago

Holding ArthurHolding Arthur

1,360417

$endgroup$

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
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');

var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);

;
$window.on('scroll', onScroll);

);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3172680%2ffinding-the-error-in-an-argument%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

Nothing wrong. Just change it into

$$fracd zd x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$$

Note that that the first term is $fracd zd x$, which is different from $fracpartial zpartial x$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $fracd zd x$, which is NOT a partial derivative.

Actually, a better way to say this is that

$$left[fracpartial zpartial xright]_y=x^2=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y.$$

Where I have clearly written down the restriction $y=x^2$.

share|cite|improve this answer

answered 3 hours ago

Holding ArthurHolding Arthur

1,360417

$endgroup$

add a comment | 

4

$begingroup$

Nothing wrong. Just change it into

$$fracd zd x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$$

Note that that the first term is $fracd zd x$, which is different from $fracpartial zpartial x$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $fracd zd x$, which is NOT a partial derivative.

Actually, a better way to say this is that

$$left[fracpartial zpartial xright]_y=x^2=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y.$$

Where I have clearly written down the restriction $y=x^2$.

share|cite|improve this answer

answered 3 hours ago

Holding ArthurHolding Arthur

1,360417

$endgroup$

add a comment | 

$begingroup$

Nothing wrong. Just change it into

$$fracd zd x=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y$$

Note that that the first term is $fracd zd x$, which is different from $fracpartial zpartial x$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $fracd zd x$, which is NOT a partial derivative.

Actually, a better way to say this is that

$$left[fracpartial zpartial xright]_y=x^2=fracpartial zpartial xfracpartial xpartial x+fracpartial zpartial yfracpartial ypartial x=fracpartial zpartial x+2xfracpartial zpartial y.$$

Where I have clearly written down the restriction $y=x^2$.

share|cite|improve this answer

answered 3 hours ago

Holding ArthurHolding Arthur

1,360417

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Holding ArthurHolding Arthur

1,360417

answered 3 hours ago

Holding ArthurHolding Arthur

1,360417

answered 3 hours ago

Holding ArthurHolding Arthur

1,360417