Finding the error in an argumentChain 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
How can a day be of 24 hours?
Why are UK visa biometrics appointments suspended at USCIS Application Support Centers?
How does a dynamic QR code work?
Is it possible to map the firing of neurons in the human brain so as to stimulate artificial memories in someone else?
Does Dispel Magic work on Tiny Hut?
Getting extremely large arrows with tikzcd
Forgetting the musical notes while performing in concert
Car headlights in a world without electricity
Does the Idaho Potato Commission associate potato skins with healthy eating?
How obscure is the use of 令 in 令和?
Finding the error in an argument
Placement of More Information/Help Icon button for Radio Buttons
Why is the sentence "Das ist eine Nase" correct?
Am I breaking OOP practice with this architecture?
Do creatures with a listed speed of "0 ft., fly 30 ft. (hover)" ever touch the ground?
What historical events would have to change in order to make 19th century "steampunk" technology possible?
Mathematica command that allows it to read my intentions
Avoiding the "not like other girls" trope?
What is required to make GPS signals available indoors?
What is the most common color to indicate the input-field is disabled?
Why didn't Boeing produce its own regional jet?
Can I hook these wires up to find the connection to a dead outlet?
OP Amp not amplifying audio signal
What does the same-ish mean?
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
$begingroup$
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
asked 3 hours ago
mathenthusiastmathenthusiast
758
$endgroup$
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
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
2 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
asked 3 hours ago
mathenthusiastmathenthusiast
758
$endgroup$
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
calculus multivariable-calculus partial-derivative
asked 3 hours ago
mathenthusiastmathenthusiast
758
asked 3 hours ago
mathenthusiastmathenthusiast
758
edited 3 hours ago
mathenthusiast
asked 3 hours ago
mathenthusiastmathenthusiast
758
asked 3 hours ago
mathenthusiastmathenthusiast
758
asked 3 hours ago
mathenthusiastmathenthusiast
758
758
$begingroup$
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
2 hours ago
add a comment |
$begingroup$
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
2 hours ago
$begingroup$
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
$begingroup$
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
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
$begingroup$
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
2 hours ago
$begingroup$
@BSplitter the problem itself poses this argument as incorrect, the object being to find out why
$endgroup$
– mathenthusiast
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $y=x^2$.
answered 2 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
);
);
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%2f3172680%2ffinding-the-error-in-an-argument%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
$begingroup$
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 $y=x^2$.
answered 2 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 $y=x^2$.
answered 2 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 $y=x^2$.
answered 2 hours ago
Holding ArthurHolding Arthur
1,360417
$endgroup$
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 $y=x^2$.
answered 2 hours ago
Holding ArthurHolding Arthur
1,360417
answered 2 hours ago
Holding ArthurHolding Arthur
1,360417
answered 2 hours ago
Holding ArthurHolding Arthur
1,360417
answered 2 hours ago
Holding ArthurHolding Arthur
1,360417
1,360417
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%2f3172680%2ffinding-the-error-in-an-argument%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
$begingroup$
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
2 hours ago