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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

4

$begingroup$

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 Apr 3 at 1:07

mathenthusiast

asked Apr 3 at 0:52

mathenthusiastmathenthusiast

808

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 3 at 1:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I also will note that if $2xfracpartial zpartial y =0$, then either $x=0$ or $fracpartial zpartial y =0$.
  $endgroup$
  – BSplitter
  Apr 3 at 1:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BSplitter the problem itself poses this argument as incorrect, the object being to find out why
  $endgroup$
  – mathenthusiast
  Apr 3 at 1:17
  

  
  
  
  

add a comment | 

4

$begingroup$

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 Apr 3 at 1:07

mathenthusiast

asked Apr 3 at 0:52

mathenthusiastmathenthusiast

808

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 3 at 1:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I also will note that if $2xfracpartial zpartial y =0$, then either $x=0$ or $fracpartial zpartial y =0$.
  $endgroup$
  – BSplitter
  Apr 3 at 1:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BSplitter the problem itself poses this argument as incorrect, the object being to find out why
  $endgroup$
  – mathenthusiast
  Apr 3 at 1:17
  

  
  
  
  

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 Apr 3 at 1:07

mathenthusiast

asked Apr 3 at 0:52

mathenthusiastmathenthusiast

808

$endgroup$

share|cite|improve this question

edited Apr 3 at 1:07

mathenthusiast

asked Apr 3 at 0:52

mathenthusiastmathenthusiast

808

share|cite|improve this question

edited Apr 3 at 1:07

mathenthusiast

asked Apr 3 at 0:52

mathenthusiastmathenthusiast

808

asked Apr 3 at 0:52

mathenthusiastmathenthusiast

808

asked Apr 3 at 0:52

mathenthusiastmathenthusiast

808

1 Answer
1

active

oldest

votes

5

$begingroup$

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 Apr 3 at 1:21

Holding ArthurHolding Arthur

1,441417

$endgroup$

add a comment | 

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5

$begingroup$

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 Apr 3 at 1:21

Holding ArthurHolding Arthur

1,441417

$endgroup$

add a comment | 

5

$begingroup$

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 Apr 3 at 1:21

Holding ArthurHolding Arthur

1,441417

$endgroup$

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 Apr 3 at 1:21

Holding ArthurHolding Arthur

1,441417

$endgroup$

share|cite|improve this answer

answered Apr 3 at 1:21

Holding ArthurHolding Arthur

1,441417

answered Apr 3 at 1:21

Holding ArthurHolding Arthur

1,441417

answered Apr 3 at 1:21

Holding ArthurHolding Arthur

1,441417