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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 $f(x, y, z) = x^{frac{y}{z}}$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
$frac{partial z}{partial x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$
Therefore
$2xfrac{partial z}{partial y}=0$
and
$frac{partial z}{partial y}=0$
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.) $frac{partial z}{partial y}$ 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 2 hours ago
mathenthusiastmathenthusiast
758
$endgroup$
add a comment |
$begingroup$
If $z=f(x,y)$ and $y=x^2$, then by the chain rule
$frac{partial z}{partial x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$
Therefore
$2xfrac{partial z}{partial y}=0$
and
$frac{partial z}{partial y}=0$
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.) $frac{partial z}{partial y}$ 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 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
2 hours ago
1
$begingroup$
I also will note that if $2xfrac{partial z}{partial y} =0$, then either $x=0$ or $frac{partial z}{partial y} =0$.
$endgroup$
– 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
$frac{partial z}{partial x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$
Therefore
$2xfrac{partial z}{partial y}=0$
and
$frac{partial z}{partial y}=0$
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.) $frac{partial z}{partial y}$ 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 2 hours ago
mathenthusiastmathenthusiast
758
$endgroup$
If $z=f(x,y)$ and $y=x^2$, then by the chain rule
$frac{partial z}{partial x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$
Therefore
$2xfrac{partial z}{partial y}=0$
and
$frac{partial z}{partial y}=0$
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.) $frac{partial z}{partial y}$ 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 2 hours ago
mathenthusiastmathenthusiast
758
asked 2 hours ago
mathenthusiastmathenthusiast
758
edited 2 hours ago
mathenthusiast
asked 2 hours ago
mathenthusiastmathenthusiast
758
asked 2 hours ago
mathenthusiastmathenthusiast
758
asked 2 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
2 hours ago
1
$begingroup$
I also will note that if $2xfrac{partial z}{partial y} =0$, then either $x=0$ or $frac{partial z}{partial y} =0$.
$endgroup$
– 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 |
$begingroup$
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
2 hours ago
1
$begingroup$
I also will note that if $2xfrac{partial z}{partial y} =0$, then either $x=0$ or $frac{partial z}{partial y} =0$.
$endgroup$
– 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
$begingroup$
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
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
2 hours ago
1
1
$begingroup$
I also will note that if $2xfrac{partial z}{partial y} =0$, then either $x=0$ or $frac{partial z}{partial y} =0$.
$endgroup$
– BSplitter
1 hour ago
$begingroup$
I also will note that if $2xfrac{partial z}{partial y} =0$, then either $x=0$ or $frac{partial z}{partial y} =0$.
$endgroup$
– 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
$begingroup$
@BSplitter the problem itself poses this argument as incorrect, the object being to find out why
$endgroup$
– mathenthusiast
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Nothing wrong. Just change it into
$$frac{d z}{d x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$$
Note that that the first term is $frac{d z}{d x}$, which is different from $frac{partial z}{partial x}$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $frac{d z}{d x}$, which is NOT a partial derivative.
Actually, a better way to say this is that
$$left[frac{partial z}{partial x}right]_{y=x^2}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}.$$
Where I have clearly written down the restriction $y=x^2$.
answered 1 hour ago
Holding ArthurHolding Arthur
1,360417
$endgroup$
add a comment |
Your Answer
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1 Answer
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oldest
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1 Answer
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votes
$begingroup$
Nothing wrong. Just change it into
$$frac{d z}{d x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$$
Note that that the first term is $frac{d z}{d x}$, which is different from $frac{partial z}{partial x}$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $frac{d z}{d x}$, which is NOT a partial derivative.
Actually, a better way to say this is that
$$left[frac{partial z}{partial x}right]_{y=x^2}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}.$$
Where I have clearly written down the restriction $y=x^2$.
answered 1 hour ago
Holding ArthurHolding Arthur
1,360417
$endgroup$
add a comment |
$begingroup$
Nothing wrong. Just change it into
$$frac{d z}{d x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$$
Note that that the first term is $frac{d z}{d x}$, which is different from $frac{partial z}{partial x}$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $frac{d z}{d x}$, which is NOT a partial derivative.
Actually, a better way to say this is that
$$left[frac{partial z}{partial x}right]_{y=x^2}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}.$$
Where I have clearly written down the restriction $y=x^2$.
answered 1 hour ago
Holding ArthurHolding Arthur
1,360417
$endgroup$
add a comment |
$begingroup$
Nothing wrong. Just change it into
$$frac{d z}{d x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$$
Note that that the first term is $frac{d z}{d x}$, which is different from $frac{partial z}{partial x}$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $frac{d z}{d x}$, which is NOT a partial derivative.
Actually, a better way to say this is that
$$left[frac{partial z}{partial x}right]_{y=x^2}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}.$$
Where I have clearly written down the restriction $y=x^2$.
answered 1 hour ago
Holding ArthurHolding Arthur
1,360417
$endgroup$
Nothing wrong. Just change it into
$$frac{d z}{d x}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}$$
Note that that the first term is $frac{d z}{d x}$, which is different from $frac{partial z}{partial x}$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $frac{d z}{d x}$, which is NOT a partial derivative.
Actually, a better way to say this is that
$$left[frac{partial z}{partial x}right]_{y=x^2}=frac{partial z}{partial x}frac{partial x}{partial x}+frac{partial z}{partial y}frac{partial y}{partial x}=frac{partial z}{partial x}+2xfrac{partial z}{partial y}.$$
Where I have clearly written down the restriction $y=x^2$.
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
answered 1 hour ago
Holding ArthurHolding Arthur
1,360417
1,360417
add a comment |
add a comment |
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$begingroup$
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
2 hours ago
1
$begingroup$
I also will note that if $2xfrac{partial z}{partial y} =0$, then either $x=0$ or $frac{partial z}{partial y} =0$.
$endgroup$
– 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