Mixing
Chain rule, two different interpretations, two different results
$begingroup$
I have a function $f:mathbb{R}^2 to mathbb{R}$. So we can write $f(x,y)$ where $x$ and $y$ are independant variables.
Now we define the function $g(u,v) = f(u+v, uv)$.
Using the chain rule we have :
$$frac{partial g}{partial u} = frac{partial f}{partial x} + v frac{partial f}{partial y}$$
$$frac{partial g}{partial v} = frac{partial f}{partial x} + u frac{partial f}{partial y}$$
Thus we have :
$$frac{partial}{partial v} frac{partial g}{partial u} = left ( frac{partial }{partial x} + u frac{partial }{partial y} right) left ( frac{partial f}{partial x} + v frac{partial f}{partial y} right)$$
But here I have a problem when developing this expression. Indeed we have : $frac{partial v}{partial v} = 1$ so I should expect that :
$$left ( frac{partial }{partial x} + u frac{partial }{partial y} right) v = 1$$
Yet for me we have :
$frac{partial v}{partial x} = 0$ since there isn't any occurence of $x$ in $v$. So I get that : $(frac{partial }{partial x} + u frac{partial }{partial y} )v = 0 ne 1. $
So what is the problem here ? I think it comes from the fact that $x$ depends on $u$, but normally this shouldn't be the case right ? I mean $x$ and $y$ are defined at the beginning there are just the composant of the function $f$, and they are independant. Moreover if $x$ depends an $u$ it would mean that $f$ is a function of $u$, but it doesn't mean anything to differentiate at a function, moreover as said earlier $x$ and $y$ doesn't depend on anything.
Hence what is really going on here ?
Thank you !
real-analysis calculus multivariable-calculus chain-rule
edited Jan 27 at 23:22
Bernard
123k741117
asked Jan 27 at 22:58
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
add a comment |
$begingroup$
I have a function $f:mathbb{R}^2 to mathbb{R}$. So we can write $f(x,y)$ where $x$ and $y$ are independant variables.
Now we define the function $g(u,v) = f(u+v, uv)$.
Using the chain rule we have :
$$frac{partial g}{partial u} = frac{partial f}{partial x} + v frac{partial f}{partial y}$$
$$frac{partial g}{partial v} = frac{partial f}{partial x} + u frac{partial f}{partial y}$$
Thus we have :
$$frac{partial}{partial v} frac{partial g}{partial u} = left ( frac{partial }{partial x} + u frac{partial }{partial y} right) left ( frac{partial f}{partial x} + v frac{partial f}{partial y} right)$$
But here I have a problem when developing this expression. Indeed we have : $frac{partial v}{partial v} = 1$ so I should expect that :
$$left ( frac{partial }{partial x} + u frac{partial }{partial y} right) v = 1$$
Yet for me we have :
$frac{partial v}{partial x} = 0$ since there isn't any occurence of $x$ in $v$. So I get that : $(frac{partial }{partial x} + u frac{partial }{partial y} )v = 0 ne 1. $
So what is the problem here ? I think it comes from the fact that $x$ depends on $u$, but normally this shouldn't be the case right ? I mean $x$ and $y$ are defined at the beginning there are just the composant of the function $f$, and they are independant. Moreover if $x$ depends an $u$ it would mean that $f$ is a function of $u$, but it doesn't mean anything to differentiate at a function, moreover as said earlier $x$ and $y$ doesn't depend on anything.
Hence what is really going on here ?
Thank you !
real-analysis calculus multivariable-calculus chain-rule
edited Jan 27 at 23:22
Bernard
123k741117
asked Jan 27 at 22:58
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
3
$begingroup$
Why does $frac{partial g}{partial v} = frac{partial f}{partial x} + u frac{partial f}{partial y}$ imply $frac{partial}{partial v} = frac{partial}{partial x} + u frac{partial}{partial y}$?
$endgroup$
– angryavian
Jan 27 at 23:04
$begingroup$
$v$ isn't a function so how are you taking the partial derivative of it with respect to itself??
$endgroup$
– Displayname
Jan 27 at 23:10
$begingroup$
@angryavian Thank you. I don't really know actually... But then I don't see what $partial / partial v$ is equal to ?
$endgroup$
– dghkgfzyukz
Jan 27 at 23:10
$begingroup$
@Displayname it implicitily defined the function $ v mapsto v$
$endgroup$
– dghkgfzyukz
Jan 27 at 23:10
add a comment |
$begingroup$
I have a function $f:mathbb{R}^2 to mathbb{R}$. So we can write $f(x,y)$ where $x$ and $y$ are independant variables.
Now we define the function $g(u,v) = f(u+v, uv)$.
Using the chain rule we have :
$$frac{partial g}{partial u} = frac{partial f}{partial x} + v frac{partial f}{partial y}$$
$$frac{partial g}{partial v} = frac{partial f}{partial x} + u frac{partial f}{partial y}$$
Thus we have :
$$frac{partial}{partial v} frac{partial g}{partial u} = left ( frac{partial }{partial x} + u frac{partial }{partial y} right) left ( frac{partial f}{partial x} + v frac{partial f}{partial y} right)$$
But here I have a problem when developing this expression. Indeed we have : $frac{partial v}{partial v} = 1$ so I should expect that :
$$left ( frac{partial }{partial x} + u frac{partial }{partial y} right) v = 1$$
Yet for me we have :
$frac{partial v}{partial x} = 0$ since there isn't any occurence of $x$ in $v$. So I get that : $(frac{partial }{partial x} + u frac{partial }{partial y} )v = 0 ne 1. $
So what is the problem here ? I think it comes from the fact that $x$ depends on $u$, but normally this shouldn't be the case right ? I mean $x$ and $y$ are defined at the beginning there are just the composant of the function $f$, and they are independant. Moreover if $x$ depends an $u$ it would mean that $f$ is a function of $u$, but it doesn't mean anything to differentiate at a function, moreover as said earlier $x$ and $y$ doesn't depend on anything.
Hence what is really going on here ?
Thank you !
real-analysis calculus multivariable-calculus chain-rule
edited Jan 27 at 23:22
Bernard
123k741117
asked Jan 27 at 22:58
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
I have a function $f:mathbb{R}^2 to mathbb{R}$. So we can write $f(x,y)$ where $x$ and $y$ are independant variables.
Now we define the function $g(u,v) = f(u+v, uv)$.
Using the chain rule we have :
$$frac{partial g}{partial u} = frac{partial f}{partial x} + v frac{partial f}{partial y}$$
$$frac{partial g}{partial v} = frac{partial f}{partial x} + u frac{partial f}{partial y}$$
Thus we have :
$$frac{partial}{partial v} frac{partial g}{partial u} = left ( frac{partial }{partial x} + u frac{partial }{partial y} right) left ( frac{partial f}{partial x} + v frac{partial f}{partial y} right)$$
But here I have a problem when developing this expression. Indeed we have : $frac{partial v}{partial v} = 1$ so I should expect that :
$$left ( frac{partial }{partial x} + u frac{partial }{partial y} right) v = 1$$
Yet for me we have :
$frac{partial v}{partial x} = 0$ since there isn't any occurence of $x$ in $v$. So I get that : $(frac{partial }{partial x} + u frac{partial }{partial y} )v = 0 ne 1. $
So what is the problem here ? I think it comes from the fact that $x$ depends on $u$, but normally this shouldn't be the case right ? I mean $x$ and $y$ are defined at the beginning there are just the composant of the function $f$, and they are independant. Moreover if $x$ depends an $u$ it would mean that $f$ is a function of $u$, but it doesn't mean anything to differentiate at a function, moreover as said earlier $x$ and $y$ doesn't depend on anything.
Hence what is really going on here ?
Thank you !
real-analysis calculus multivariable-calculus chain-rule
real-analysis calculus multivariable-calculus chain-rule
edited Jan 27 at 23:22
Bernard
123k741117
asked Jan 27 at 22:58
dghkgfzyukzdghkgfzyukz
16612
edited Jan 27 at 23:22
Bernard
123k741117
asked Jan 27 at 22:58
dghkgfzyukzdghkgfzyukz
16612
edited Jan 27 at 23:22
Bernard
123k741117
edited Jan 27 at 23:22
Bernard
123k741117
edited Jan 27 at 23:22
Bernard
123k741117
123k741117
asked Jan 27 at 22:58
dghkgfzyukzdghkgfzyukz
16612
asked Jan 27 at 22:58
dghkgfzyukzdghkgfzyukz
16612
asked Jan 27 at 22:58
dghkgfzyukzdghkgfzyukz
16612
16612
3
$begingroup$
Why does $frac{partial g}{partial v} = frac{partial f}{partial x} + u frac{partial f}{partial y}$ imply $frac{partial}{partial v} = frac{partial}{partial x} + u frac{partial}{partial y}$?
$endgroup$
– angryavian
Jan 27 at 23:04
$begingroup$
$v$ isn't a function so how are you taking the partial derivative of it with respect to itself??
$endgroup$
– Displayname
Jan 27 at 23:10
$begingroup$
@angryavian Thank you. I don't really know actually... But then I don't see what $partial / partial v$ is equal to ?
$endgroup$
– dghkgfzyukz
Jan 27 at 23:10
$begingroup$
@Displayname it implicitily defined the function $ v mapsto v$
$endgroup$
– dghkgfzyukz
Jan 27 at 23:10
add a comment |
3
$begingroup$
Why does $frac{partial g}{partial v} = frac{partial f}{partial x} + u frac{partial f}{partial y}$ imply $frac{partial}{partial v} = frac{partial}{partial x} + u frac{partial}{partial y}$?
$endgroup$
– angryavian
Jan 27 at 23:04
$begingroup$
$v$ isn't a function so how are you taking the partial derivative of it with respect to itself??
$endgroup$
– Displayname
Jan 27 at 23:10
$begingroup$
@angryavian Thank you. I don't really know actually... But then I don't see what $partial / partial v$ is equal to ?
$endgroup$
– dghkgfzyukz
Jan 27 at 23:10
$begingroup$
@Displayname it implicitily defined the function $ v mapsto v$
$endgroup$
– dghkgfzyukz
Jan 27 at 23:10
3
3
$begingroup$
Why does $frac{partial g}{partial v} = frac{partial f}{partial x} + u frac{partial f}{partial y}$ imply $frac{partial}{partial v} = frac{partial}{partial x} + u frac{partial}{partial y}$?
$endgroup$
– angryavian
Jan 27 at 23:04
$begingroup$
Why does $frac{partial g}{partial v} = frac{partial f}{partial x} + u frac{partial f}{partial y}$ imply $frac{partial}{partial v} = frac{partial}{partial x} + u frac{partial}{partial y}$?
$endgroup$
– angryavian
Jan 27 at 23:04
$begingroup$
$v$ isn't a function so how are you taking the partial derivative of it with respect to itself??
$endgroup$
– Displayname
Jan 27 at 23:10
$begingroup$
$v$ isn't a function so how are you taking the partial derivative of it with respect to itself??
$endgroup$
– Displayname
Jan 27 at 23:10
$begingroup$
@angryavian Thank you. I don't really know actually... But then I don't see what $partial / partial v$ is equal to ?
$endgroup$
– dghkgfzyukz
Jan 27 at 23:10
$begingroup$
@angryavian Thank you. I don't really know actually... But then I don't see what $partial / partial v$ is equal to ?
$endgroup$
– dghkgfzyukz
Jan 27 at 23:10
$begingroup$
@Displayname it implicitily defined the function $ v mapsto v$
$endgroup$
– dghkgfzyukz
Jan 27 at 23:10
$begingroup$
@Displayname it implicitily defined the function $ v mapsto v$
$endgroup$
– dghkgfzyukz
Jan 27 at 23:10
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You miss the term $frac{partial f}{partial y}$ obtained from deriving $vtimes frac{partial f}{partial y}$ by $v$.
$begin{align}displaystyle frac{partial}{partial v}frac{partial g}{partial u}
&=frac{partial}{partial v}left(frac{partial f}{partial x} + v frac{partial f}{partial y}right)
\\displaystyle &=frac{partial}{partial v}left(frac{partial f}{partial x}right)+vfrac{partial}{partial v}left(frac{partial f}{partial y}right)+frac{partial f}{partial y}
\\displaystyle &=left(frac{partial^2 f}{partial x^2} + ufrac{partial^2 f}{partial xpartial y}right)+vleft(frac{partial^2 f}{partial xpartial y} + ufrac{partial^2 f}{partial y^2}right)+frac{partial f}{partial y}end{align}$
answered Jan 27 at 23:26
zwimzwim
12.6k831
$endgroup$
$begingroup$
Thank you. But to go from the second equality to the third you use the fact that : $frac{partial}{partial v} = frac{partial}{partial x} + ufrac{partial}{partial y}$ right ?
$endgroup$
– dghkgfzyukz
Jan 27 at 23:29
$begingroup$
also at the beginning it should be $g$ instead of $f$ if I am not mistaken.
$endgroup$
– dghkgfzyukz
Jan 27 at 23:29
$begingroup$
I do not really use the operator equality as you write. I'm just considering the chain rule. Think of it as $frac{partial}{partial v}(f_x'(u+v,uv))$ when the variables appears in $f_x'$ it is easier to see what's going on, and why the operator equality won't work for $frac{partial}{partial v}(vf_y'(u+v,uv))$
$endgroup$
– zwim
Jan 27 at 23:36
$begingroup$
Oh I see. It makes a lot more sens now, and I guess there is no other simple form for : $frac{partial}{partial v}$ so you need to use the chain rule. Thank you !
$endgroup$
– dghkgfzyukz
Jan 28 at 7:54
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You miss the term $frac{partial f}{partial y}$ obtained from deriving $vtimes frac{partial f}{partial y}$ by $v$.
$begin{align}displaystyle frac{partial}{partial v}frac{partial g}{partial u}
&=frac{partial}{partial v}left(frac{partial f}{partial x} + v frac{partial f}{partial y}right)
\\displaystyle &=frac{partial}{partial v}left(frac{partial f}{partial x}right)+vfrac{partial}{partial v}left(frac{partial f}{partial y}right)+frac{partial f}{partial y}
\\displaystyle &=left(frac{partial^2 f}{partial x^2} + ufrac{partial^2 f}{partial xpartial y}right)+vleft(frac{partial^2 f}{partial xpartial y} + ufrac{partial^2 f}{partial y^2}right)+frac{partial f}{partial y}end{align}$
answered Jan 27 at 23:26
zwimzwim
12.6k831
$endgroup$
$begingroup$
Thank you. But to go from the second equality to the third you use the fact that : $frac{partial}{partial v} = frac{partial}{partial x} + ufrac{partial}{partial y}$ right ?
$endgroup$
– dghkgfzyukz
Jan 27 at 23:29
$begingroup$
also at the beginning it should be $g$ instead of $f$ if I am not mistaken.
$endgroup$
– dghkgfzyukz
Jan 27 at 23:29
$begingroup$
I do not really use the operator equality as you write. I'm just considering the chain rule. Think of it as $frac{partial}{partial v}(f_x'(u+v,uv))$ when the variables appears in $f_x'$ it is easier to see what's going on, and why the operator equality won't work for $frac{partial}{partial v}(vf_y'(u+v,uv))$
$endgroup$
– zwim
Jan 27 at 23:36
$begingroup$
Oh I see. It makes a lot more sens now, and I guess there is no other simple form for : $frac{partial}{partial v}$ so you need to use the chain rule. Thank you !
$endgroup$
– dghkgfzyukz
Jan 28 at 7:54
add a comment |
$begingroup$
You miss the term $frac{partial f}{partial y}$ obtained from deriving $vtimes frac{partial f}{partial y}$ by $v$.
$begin{align}displaystyle frac{partial}{partial v}frac{partial g}{partial u}
&=frac{partial}{partial v}left(frac{partial f}{partial x} + v frac{partial f}{partial y}right)
\\displaystyle &=frac{partial}{partial v}left(frac{partial f}{partial x}right)+vfrac{partial}{partial v}left(frac{partial f}{partial y}right)+frac{partial f}{partial y}
\\displaystyle &=left(frac{partial^2 f}{partial x^2} + ufrac{partial^2 f}{partial xpartial y}right)+vleft(frac{partial^2 f}{partial xpartial y} + ufrac{partial^2 f}{partial y^2}right)+frac{partial f}{partial y}end{align}$
answered Jan 27 at 23:26
zwimzwim
12.6k831
$endgroup$
$begingroup$
Thank you. But to go from the second equality to the third you use the fact that : $frac{partial}{partial v} = frac{partial}{partial x} + ufrac{partial}{partial y}$ right ?
$endgroup$
– dghkgfzyukz
Jan 27 at 23:29
$begingroup$
also at the beginning it should be $g$ instead of $f$ if I am not mistaken.
$endgroup$
– dghkgfzyukz
Jan 27 at 23:29
$begingroup$
I do not really use the operator equality as you write. I'm just considering the chain rule. Think of it as $frac{partial}{partial v}(f_x'(u+v,uv))$ when the variables appears in $f_x'$ it is easier to see what's going on, and why the operator equality won't work for $frac{partial}{partial v}(vf_y'(u+v,uv))$
$endgroup$
– zwim
Jan 27 at 23:36
$begingroup$
Oh I see. It makes a lot more sens now, and I guess there is no other simple form for : $frac{partial}{partial v}$ so you need to use the chain rule. Thank you !
$endgroup$
– dghkgfzyukz
Jan 28 at 7:54
add a comment |
$begingroup$
You miss the term $frac{partial f}{partial y}$ obtained from deriving $vtimes frac{partial f}{partial y}$ by $v$.
$begin{align}displaystyle frac{partial}{partial v}frac{partial g}{partial u}
&=frac{partial}{partial v}left(frac{partial f}{partial x} + v frac{partial f}{partial y}right)
\\displaystyle &=frac{partial}{partial v}left(frac{partial f}{partial x}right)+vfrac{partial}{partial v}left(frac{partial f}{partial y}right)+frac{partial f}{partial y}
\\displaystyle &=left(frac{partial^2 f}{partial x^2} + ufrac{partial^2 f}{partial xpartial y}right)+vleft(frac{partial^2 f}{partial xpartial y} + ufrac{partial^2 f}{partial y^2}right)+frac{partial f}{partial y}end{align}$
answered Jan 27 at 23:26
zwimzwim
12.6k831
$endgroup$
You miss the term $frac{partial f}{partial y}$ obtained from deriving $vtimes frac{partial f}{partial y}$ by $v$.
$begin{align}displaystyle frac{partial}{partial v}frac{partial g}{partial u}
&=frac{partial}{partial v}left(frac{partial f}{partial x} + v frac{partial f}{partial y}right)
\\displaystyle &=frac{partial}{partial v}left(frac{partial f}{partial x}right)+vfrac{partial}{partial v}left(frac{partial f}{partial y}right)+frac{partial f}{partial y}
\\displaystyle &=left(frac{partial^2 f}{partial x^2} + ufrac{partial^2 f}{partial xpartial y}right)+vleft(frac{partial^2 f}{partial xpartial y} + ufrac{partial^2 f}{partial y^2}right)+frac{partial f}{partial y}end{align}$
answered Jan 27 at 23:26
zwimzwim
12.6k831
edited Jan 27 at 23:31
answered Jan 27 at 23:26
zwimzwim
12.6k831
answered Jan 27 at 23:26
zwimzwim
12.6k831
answered Jan 27 at 23:26
zwimzwim
12.6k831
12.6k831
$begingroup$
Thank you. But to go from the second equality to the third you use the fact that : $frac{partial}{partial v} = frac{partial}{partial x} + ufrac{partial}{partial y}$ right ?
$endgroup$
– dghkgfzyukz
Jan 27 at 23:29
$begingroup$
also at the beginning it should be $g$ instead of $f$ if I am not mistaken.
$endgroup$
– dghkgfzyukz
Jan 27 at 23:29
$begingroup$
I do not really use the operator equality as you write. I'm just considering the chain rule. Think of it as $frac{partial}{partial v}(f_x'(u+v,uv))$ when the variables appears in $f_x'$ it is easier to see what's going on, and why the operator equality won't work for $frac{partial}{partial v}(vf_y'(u+v,uv))$
$endgroup$
– zwim
Jan 27 at 23:36
$begingroup$
Oh I see. It makes a lot more sens now, and I guess there is no other simple form for : $frac{partial}{partial v}$ so you need to use the chain rule. Thank you !
$endgroup$
– dghkgfzyukz
Jan 28 at 7:54
add a comment |
$begingroup$
Thank you. But to go from the second equality to the third you use the fact that : $frac{partial}{partial v} = frac{partial}{partial x} + ufrac{partial}{partial y}$ right ?
$endgroup$
– dghkgfzyukz
Jan 27 at 23:29
$begingroup$
also at the beginning it should be $g$ instead of $f$ if I am not mistaken.
$endgroup$
– dghkgfzyukz
Jan 27 at 23:29
$begingroup$
I do not really use the operator equality as you write. I'm just considering the chain rule. Think of it as $frac{partial}{partial v}(f_x'(u+v,uv))$ when the variables appears in $f_x'$ it is easier to see what's going on, and why the operator equality won't work for $frac{partial}{partial v}(vf_y'(u+v,uv))$
$endgroup$
– zwim
Jan 27 at 23:36
$begingroup$
Oh I see. It makes a lot more sens now, and I guess there is no other simple form for : $frac{partial}{partial v}$ so you need to use the chain rule. Thank you !
$endgroup$
– dghkgfzyukz
Jan 28 at 7:54
$begingroup$
Thank you. But to go from the second equality to the third you use the fact that : $frac{partial}{partial v} = frac{partial}{partial x} + ufrac{partial}{partial y}$ right ?
$endgroup$
– dghkgfzyukz
Jan 27 at 23:29
$begingroup$
Thank you. But to go from the second equality to the third you use the fact that : $frac{partial}{partial v} = frac{partial}{partial x} + ufrac{partial}{partial y}$ right ?
$endgroup$
– dghkgfzyukz
Jan 27 at 23:29
$begingroup$
also at the beginning it should be $g$ instead of $f$ if I am not mistaken.
$endgroup$
– dghkgfzyukz
Jan 27 at 23:29
$begingroup$
also at the beginning it should be $g$ instead of $f$ if I am not mistaken.
$endgroup$
– dghkgfzyukz
Jan 27 at 23:29
$begingroup$
I do not really use the operator equality as you write. I'm just considering the chain rule. Think of it as $frac{partial}{partial v}(f_x'(u+v,uv))$ when the variables appears in $f_x'$ it is easier to see what's going on, and why the operator equality won't work for $frac{partial}{partial v}(vf_y'(u+v,uv))$
$endgroup$
– zwim
Jan 27 at 23:36
$begingroup$
I do not really use the operator equality as you write. I'm just considering the chain rule. Think of it as $frac{partial}{partial v}(f_x'(u+v,uv))$ when the variables appears in $f_x'$ it is easier to see what's going on, and why the operator equality won't work for $frac{partial}{partial v}(vf_y'(u+v,uv))$
$endgroup$
– zwim
Jan 27 at 23:36
$begingroup$
Oh I see. It makes a lot more sens now, and I guess there is no other simple form for : $frac{partial}{partial v}$ so you need to use the chain rule. Thank you !
$endgroup$
– dghkgfzyukz
Jan 28 at 7:54
$begingroup$
Oh I see. It makes a lot more sens now, and I guess there is no other simple form for : $frac{partial}{partial v}$ so you need to use the chain rule. Thank you !
$endgroup$
– dghkgfzyukz
Jan 28 at 7:54
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3
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Why does $frac{partial g}{partial v} = frac{partial f}{partial x} + u frac{partial f}{partial y}$ imply $frac{partial}{partial v} = frac{partial}{partial x} + u frac{partial}{partial y}$?
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– angryavian
Jan 27 at 23:04
$begingroup$
$v$ isn't a function so how are you taking the partial derivative of it with respect to itself??
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– Displayname
Jan 27 at 23:10
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@angryavian Thank you. I don't really know actually... But then I don't see what $partial / partial v$ is equal to ?
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– dghkgfzyukz
Jan 27 at 23:10
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@Displayname it implicitily defined the function $ v mapsto v$
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– dghkgfzyukz
Jan 27 at 23:10