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Finding the error in an argument

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 2 hours ago

mathenthusiast

asked 2 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
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BSplitter the problem itself poses this argument as incorrect, the object being to find out why
  $endgroup$
  – mathenthusiast
  1 hour 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 2 hours ago

mathenthusiast

asked 2 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
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BSplitter the problem itself poses this argument as incorrect, the object being to find out why
  $endgroup$
  – mathenthusiast
  1 hour 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 2 hours ago

mathenthusiast

asked 2 hours ago

mathenthusiastmathenthusiast

758

$endgroup$

share|cite|improve this question

edited 2 hours ago

mathenthusiast

asked 2 hours ago

mathenthusiastmathenthusiast

758

share|cite|improve this question

edited 2 hours ago

mathenthusiast

asked 2 hours ago

mathenthusiastmathenthusiast

758

asked 2 hours ago

mathenthusiastmathenthusiast

758

asked 2 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 1 hour ago

Holding ArthurHolding Arthur

1,360417

$endgroup$

add a comment | 

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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 1 hour 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 1 hour 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 1 hour ago

Holding ArthurHolding Arthur

1,360417

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Holding ArthurHolding Arthur

1,360417

answered 1 hour ago

Holding ArthurHolding Arthur

1,360417

answered 1 hour ago

Holding ArthurHolding Arthur

1,360417