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Why chain rule isn't working here?
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Let $(x,y,z)$ and $(x',y',z')$ be two points in space. Let the distance between them be $r$. Also let:
$x-x'=xi$
$y-y'=eta$
$z-z'=zeta$
Since $(x,y,z)$ and $(x',y',z')$ are two separate points, changing only the point $(x,y,z)$ will not change the point $(x',y',z')$. Therefore:
$dfracpartial x'partial x=dfracpartial xpartial x'=0$
Then:
$dfracpartial rpartial x=dfracdrdr^2dfracpartial r^2partial x=dfrac12rdfracpartial(xi^2 +eta^2 +zeta^2)partial xidfracpartial xipartial x=dfrac12r.2 xi.dfracpartial(x-x')partial x=dfracxir$
and
$dfracpartial rpartial x'=dfracdrdr^2dfracpartial r^2partial x'=dfrac12rdfracpartial(xi^2 +eta^2 +zeta^2)partial xidfracpartial xipartial x'=dfrac12r.2 xi.dfracpartial(x-x')partial x'=-dfracxir$
Now let's find $dfracpartial rpartial x'$ by another way using chain rule:
$dfracpartial rpartial x'=dfracpartial rpartial xdfracpartial xpartial x'=dfracxir.0=0$
I guess I am getting different answers because there is something wrong in the application of chain rule here. Can someone explain why am I getting different answers?
calculus derivatives partial-derivative chain-rule
asked Jul 31 at 14:40
Joe
373113
add a comment |Â
up vote
-2
down vote
favorite
Let $(x,y,z)$ and $(x',y',z')$ be two points in space. Let the distance between them be $r$. Also let:
$x-x'=xi$
$y-y'=eta$
$z-z'=zeta$
Since $(x,y,z)$ and $(x',y',z')$ are two separate points, changing only the point $(x,y,z)$ will not change the point $(x',y',z')$. Therefore:
$dfracpartial x'partial x=dfracpartial xpartial x'=0$
Then:
$dfracpartial rpartial x=dfracdrdr^2dfracpartial r^2partial x=dfrac12rdfracpartial(xi^2 +eta^2 +zeta^2)partial xidfracpartial xipartial x=dfrac12r.2 xi.dfracpartial(x-x')partial x=dfracxir$
and
$dfracpartial rpartial x'=dfracdrdr^2dfracpartial r^2partial x'=dfrac12rdfracpartial(xi^2 +eta^2 +zeta^2)partial xidfracpartial xipartial x'=dfrac12r.2 xi.dfracpartial(x-x')partial x'=-dfracxir$
Now let's find $dfracpartial rpartial x'$ by another way using chain rule:
$dfracpartial rpartial x'=dfracpartial rpartial xdfracpartial xpartial x'=dfracxir.0=0$
I guess I am getting different answers because there is something wrong in the application of chain rule here. Can someone explain why am I getting different answers?
calculus derivatives partial-derivative chain-rule
asked Jul 31 at 14:40
Joe
373113
$partial$ doesn't satisfy the chain rule that $mathrmd$ does.
– Hurkyl
Jul 31 at 14:51
You can't take the partial derivative on $x'$ while keeping $x'$ constant.
– Yves Daoust
Jul 31 at 14:52
1
By using the chain rule in the last step you're implying that $x=x(x')$, otherwise this will not make sense. Using you're logic we could define the function $y=x+c$ ,where $c$ is constant and then saying $fracpartial ypartial x = fracpartial ypartial c fracpartial cpartial x=0$.
– User123456789
Jul 31 at 14:53
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Let $(x,y,z)$ and $(x',y',z')$ be two points in space. Let the distance between them be $r$. Also let:
$x-x'=xi$
$y-y'=eta$
$z-z'=zeta$
Since $(x,y,z)$ and $(x',y',z')$ are two separate points, changing only the point $(x,y,z)$ will not change the point $(x',y',z')$. Therefore:
$dfracpartial x'partial x=dfracpartial xpartial x'=0$
Then:
$dfracpartial rpartial x=dfracdrdr^2dfracpartial r^2partial x=dfrac12rdfracpartial(xi^2 +eta^2 +zeta^2)partial xidfracpartial xipartial x=dfrac12r.2 xi.dfracpartial(x-x')partial x=dfracxir$
and
$dfracpartial rpartial x'=dfracdrdr^2dfracpartial r^2partial x'=dfrac12rdfracpartial(xi^2 +eta^2 +zeta^2)partial xidfracpartial xipartial x'=dfrac12r.2 xi.dfracpartial(x-x')partial x'=-dfracxir$
Now let's find $dfracpartial rpartial x'$ by another way using chain rule:
$dfracpartial rpartial x'=dfracpartial rpartial xdfracpartial xpartial x'=dfracxir.0=0$
I guess I am getting different answers because there is something wrong in the application of chain rule here. Can someone explain why am I getting different answers?
calculus derivatives partial-derivative chain-rule
asked Jul 31 at 14:40
Joe
373113
Let $(x,y,z)$ and $(x',y',z')$ be two points in space. Let the distance between them be $r$. Also let:
$x-x'=xi$
$y-y'=eta$
$z-z'=zeta$
Since $(x,y,z)$ and $(x',y',z')$ are two separate points, changing only the point $(x,y,z)$ will not change the point $(x',y',z')$. Therefore:
$dfracpartial x'partial x=dfracpartial xpartial x'=0$
Then:
$dfracpartial rpartial x=dfracdrdr^2dfracpartial r^2partial x=dfrac12rdfracpartial(xi^2 +eta^2 +zeta^2)partial xidfracpartial xipartial x=dfrac12r.2 xi.dfracpartial(x-x')partial x=dfracxir$
and
$dfracpartial rpartial x'=dfracdrdr^2dfracpartial r^2partial x'=dfrac12rdfracpartial(xi^2 +eta^2 +zeta^2)partial xidfracpartial xipartial x'=dfrac12r.2 xi.dfracpartial(x-x')partial x'=-dfracxir$
Now let's find $dfracpartial rpartial x'$ by another way using chain rule:
$dfracpartial rpartial x'=dfracpartial rpartial xdfracpartial xpartial x'=dfracxir.0=0$
I guess I am getting different answers because there is something wrong in the application of chain rule here. Can someone explain why am I getting different answers?
calculus derivatives partial-derivative chain-rule
asked Jul 31 at 14:40
Joe
373113
asked Jul 31 at 14:40
Joe
373113
asked Jul 31 at 14:40
Joe
373113
asked Jul 31 at 14:40
Joe
373113
373113
$partial$ doesn't satisfy the chain rule that $mathrmd$ does.
– Hurkyl
Jul 31 at 14:51
You can't take the partial derivative on $x'$ while keeping $x'$ constant.
– Yves Daoust
Jul 31 at 14:52
1
By using the chain rule in the last step you're implying that $x=x(x')$, otherwise this will not make sense. Using you're logic we could define the function $y=x+c$ ,where $c$ is constant and then saying $fracpartial ypartial x = fracpartial ypartial c fracpartial cpartial x=0$.
– User123456789
Jul 31 at 14:53
add a comment |Â
$partial$ doesn't satisfy the chain rule that $mathrmd$ does.
– Hurkyl
Jul 31 at 14:51
You can't take the partial derivative on $x'$ while keeping $x'$ constant.
– Yves Daoust
Jul 31 at 14:52
1
By using the chain rule in the last step you're implying that $x=x(x')$, otherwise this will not make sense. Using you're logic we could define the function $y=x+c$ ,where $c$ is constant and then saying $fracpartial ypartial x = fracpartial ypartial c fracpartial cpartial x=0$.
– User123456789
Jul 31 at 14:53
$partial$ doesn't satisfy the chain rule that $mathrmd$ does.
– Hurkyl
Jul 31 at 14:51
$partial$ doesn't satisfy the chain rule that $mathrmd$ does.
– Hurkyl
Jul 31 at 14:51
You can't take the partial derivative on $x'$ while keeping $x'$ constant.
– Yves Daoust
Jul 31 at 14:52
You can't take the partial derivative on $x'$ while keeping $x'$ constant.
– Yves Daoust
Jul 31 at 14:52
1
1
By using the chain rule in the last step you're implying that $x=x(x')$, otherwise this will not make sense. Using you're logic we could define the function $y=x+c$ ,where $c$ is constant and then saying $fracpartial ypartial x = fracpartial ypartial c fracpartial cpartial x=0$.
– User123456789
Jul 31 at 14:53
By using the chain rule in the last step you're implying that $x=x(x')$, otherwise this will not make sense. Using you're logic we could define the function $y=x+c$ ,where $c$ is constant and then saying $fracpartial ypartial x = fracpartial ypartial c fracpartial cpartial x=0$.
– User123456789
Jul 31 at 14:53
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The thing you have to bear in mind about partial derivatives is they assume something is held constant, but $partial_u vpartial_v w = partial_u w$ is only guaranteed when the partial derivatives on the left-hand side hold the same things constant. Define $alpha:=left(xi,,eta,,zetaright)$. In your first calculation, you show $(partial_x'x)_alpha=0$, where the subscript indicates what's held constant. But in your final calculation, you try $(partial_x'r)_?=(partial_x r)_x'(partial_x'x)_alpha$, about which nothing follows from the chain rule.
answered Jul 31 at 14:51
J.G.
12.8k11423
Could you strictly explain what $(partial_x' x)_alpha$ is, given that $x$ and $x'$ are assumed to be coordinates of some points and no function or relation between them has been defined?
– Adayah
Jul 31 at 15:04
Since $x=x'+xi$, that derivative is $1$.
– J.G.
Jul 31 at 15:09
@J.G.- I had a bit trouble understanding your derivative notation.... Thanks anyway
– Joe
Jul 31 at 15:31
Formally, to even begin speaking of differentiation and the chain rule, we must define one or more functions (and obviously, it must be made absolutely clear which thing is a function and which is a variable). But here we have as much as a collection of indeterminates ($x$, $xi$, $r$, etc.) and some relations between them, e.g. $xi = x - x'$. No distinction between functions and variables is made, yet we proceed to differentiate those indeterminates one with respect to another in every possible way. (cont'd)
– Adayah
Jul 31 at 15:55
Even if the original relation $xi = x - x'$ slightly suggests that $xi$ is a function of variables $x$ and $x'$, it is freely transformed into $x = x' + xi$, which supposedly justifies treating $x$ as a function of $x'$ and $xi$ and differentiating it. It feels very alarming and unclear. That's why if you could explain in possibly strict terms some rigorous theory behind such seemingly incorrect operations, I think it would be very useful to many people here (including myself).
– Adayah
Jul 31 at 15:55
 |Â
show 5 more comments
up vote
0
down vote
What you've written in the yellow box is nonsense.
An expression like $fracpartial xpartial x'$ only makes sense when $x'$ is thought of as one of several independent variables, including presumably $y'$ and $z'$, when $x$ is as a function of all three of them (in that setting, $fracpartial xpartial x'$ is defined by holding $y'$ and $z'$ constant while letting $x'$ vary, and taking the limit of a very carefully formulated difference quotient).
Similarly, and expression like $fracpartial x'partial x$ only makes sense when $x$ is thought of as one of several independent variables, including presumably $y$ and $x$, and when $x'$ is as a function of all three of them.
Obviously these two situations are incompatible, so any kind of relation or equation between $fracpartial xpartial x'$ and $fracpartial x'partial x$ makes no sense, and any assumption of an equation like $fracpartial xpartial x' = 0$ makes no sense.
answered Jul 31 at 14:52
Lee Mosher
45.4k33478
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The thing you have to bear in mind about partial derivatives is they assume something is held constant, but $partial_u vpartial_v w = partial_u w$ is only guaranteed when the partial derivatives on the left-hand side hold the same things constant. Define $alpha:=left(xi,,eta,,zetaright)$. In your first calculation, you show $(partial_x'x)_alpha=0$, where the subscript indicates what's held constant. But in your final calculation, you try $(partial_x'r)_?=(partial_x r)_x'(partial_x'x)_alpha$, about which nothing follows from the chain rule.
answered Jul 31 at 14:51
J.G.
12.8k11423
Could you strictly explain what $(partial_x' x)_alpha$ is, given that $x$ and $x'$ are assumed to be coordinates of some points and no function or relation between them has been defined?
– Adayah
Jul 31 at 15:04
Since $x=x'+xi$, that derivative is $1$.
– J.G.
Jul 31 at 15:09
@J.G.- I had a bit trouble understanding your derivative notation.... Thanks anyway
– Joe
Jul 31 at 15:31
Formally, to even begin speaking of differentiation and the chain rule, we must define one or more functions (and obviously, it must be made absolutely clear which thing is a function and which is a variable). But here we have as much as a collection of indeterminates ($x$, $xi$, $r$, etc.) and some relations between them, e.g. $xi = x - x'$. No distinction between functions and variables is made, yet we proceed to differentiate those indeterminates one with respect to another in every possible way. (cont'd)
– Adayah
Jul 31 at 15:55
Even if the original relation $xi = x - x'$ slightly suggests that $xi$ is a function of variables $x$ and $x'$, it is freely transformed into $x = x' + xi$, which supposedly justifies treating $x$ as a function of $x'$ and $xi$ and differentiating it. It feels very alarming and unclear. That's why if you could explain in possibly strict terms some rigorous theory behind such seemingly incorrect operations, I think it would be very useful to many people here (including myself).
– Adayah
Jul 31 at 15:55
 |Â
show 5 more comments
up vote
1
down vote
accepted
The thing you have to bear in mind about partial derivatives is they assume something is held constant, but $partial_u vpartial_v w = partial_u w$ is only guaranteed when the partial derivatives on the left-hand side hold the same things constant. Define $alpha:=left(xi,,eta,,zetaright)$. In your first calculation, you show $(partial_x'x)_alpha=0$, where the subscript indicates what's held constant. But in your final calculation, you try $(partial_x'r)_?=(partial_x r)_x'(partial_x'x)_alpha$, about which nothing follows from the chain rule.
answered Jul 31 at 14:51
J.G.
12.8k11423
Could you strictly explain what $(partial_x' x)_alpha$ is, given that $x$ and $x'$ are assumed to be coordinates of some points and no function or relation between them has been defined?
– Adayah
Jul 31 at 15:04
Since $x=x'+xi$, that derivative is $1$.
– J.G.
Jul 31 at 15:09
@J.G.- I had a bit trouble understanding your derivative notation.... Thanks anyway
– Joe
Jul 31 at 15:31
Formally, to even begin speaking of differentiation and the chain rule, we must define one or more functions (and obviously, it must be made absolutely clear which thing is a function and which is a variable). But here we have as much as a collection of indeterminates ($x$, $xi$, $r$, etc.) and some relations between them, e.g. $xi = x - x'$. No distinction between functions and variables is made, yet we proceed to differentiate those indeterminates one with respect to another in every possible way. (cont'd)
– Adayah
Jul 31 at 15:55
Even if the original relation $xi = x - x'$ slightly suggests that $xi$ is a function of variables $x$ and $x'$, it is freely transformed into $x = x' + xi$, which supposedly justifies treating $x$ as a function of $x'$ and $xi$ and differentiating it. It feels very alarming and unclear. That's why if you could explain in possibly strict terms some rigorous theory behind such seemingly incorrect operations, I think it would be very useful to many people here (including myself).
– Adayah
Jul 31 at 15:55
 |Â
show 5 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The thing you have to bear in mind about partial derivatives is they assume something is held constant, but $partial_u vpartial_v w = partial_u w$ is only guaranteed when the partial derivatives on the left-hand side hold the same things constant. Define $alpha:=left(xi,,eta,,zetaright)$. In your first calculation, you show $(partial_x'x)_alpha=0$, where the subscript indicates what's held constant. But in your final calculation, you try $(partial_x'r)_?=(partial_x r)_x'(partial_x'x)_alpha$, about which nothing follows from the chain rule.
answered Jul 31 at 14:51
J.G.
12.8k11423
The thing you have to bear in mind about partial derivatives is they assume something is held constant, but $partial_u vpartial_v w = partial_u w$ is only guaranteed when the partial derivatives on the left-hand side hold the same things constant. Define $alpha:=left(xi,,eta,,zetaright)$. In your first calculation, you show $(partial_x'x)_alpha=0$, where the subscript indicates what's held constant. But in your final calculation, you try $(partial_x'r)_?=(partial_x r)_x'(partial_x'x)_alpha$, about which nothing follows from the chain rule.
answered Jul 31 at 14:51
J.G.
12.8k11423
answered Jul 31 at 14:51
J.G.
12.8k11423
answered Jul 31 at 14:51
J.G.
12.8k11423
answered Jul 31 at 14:51
J.G.
12.8k11423
12.8k11423
Could you strictly explain what $(partial_x' x)_alpha$ is, given that $x$ and $x'$ are assumed to be coordinates of some points and no function or relation between them has been defined?
– Adayah
Jul 31 at 15:04
Since $x=x'+xi$, that derivative is $1$.
– J.G.
Jul 31 at 15:09
@J.G.- I had a bit trouble understanding your derivative notation.... Thanks anyway
– Joe
Jul 31 at 15:31
Formally, to even begin speaking of differentiation and the chain rule, we must define one or more functions (and obviously, it must be made absolutely clear which thing is a function and which is a variable). But here we have as much as a collection of indeterminates ($x$, $xi$, $r$, etc.) and some relations between them, e.g. $xi = x - x'$. No distinction between functions and variables is made, yet we proceed to differentiate those indeterminates one with respect to another in every possible way. (cont'd)
– Adayah
Jul 31 at 15:55
Even if the original relation $xi = x - x'$ slightly suggests that $xi$ is a function of variables $x$ and $x'$, it is freely transformed into $x = x' + xi$, which supposedly justifies treating $x$ as a function of $x'$ and $xi$ and differentiating it. It feels very alarming and unclear. That's why if you could explain in possibly strict terms some rigorous theory behind such seemingly incorrect operations, I think it would be very useful to many people here (including myself).
– Adayah
Jul 31 at 15:55
 |Â
show 5 more comments
Could you strictly explain what $(partial_x' x)_alpha$ is, given that $x$ and $x'$ are assumed to be coordinates of some points and no function or relation between them has been defined?
– Adayah
Jul 31 at 15:04
Since $x=x'+xi$, that derivative is $1$.
– J.G.
Jul 31 at 15:09
@J.G.- I had a bit trouble understanding your derivative notation.... Thanks anyway
– Joe
Jul 31 at 15:31
Formally, to even begin speaking of differentiation and the chain rule, we must define one or more functions (and obviously, it must be made absolutely clear which thing is a function and which is a variable). But here we have as much as a collection of indeterminates ($x$, $xi$, $r$, etc.) and some relations between them, e.g. $xi = x - x'$. No distinction between functions and variables is made, yet we proceed to differentiate those indeterminates one with respect to another in every possible way. (cont'd)
– Adayah
Jul 31 at 15:55
Even if the original relation $xi = x - x'$ slightly suggests that $xi$ is a function of variables $x$ and $x'$, it is freely transformed into $x = x' + xi$, which supposedly justifies treating $x$ as a function of $x'$ and $xi$ and differentiating it. It feels very alarming and unclear. That's why if you could explain in possibly strict terms some rigorous theory behind such seemingly incorrect operations, I think it would be very useful to many people here (including myself).
– Adayah
Jul 31 at 15:55
Could you strictly explain what $(partial_x' x)_alpha$ is, given that $x$ and $x'$ are assumed to be coordinates of some points and no function or relation between them has been defined?
– Adayah
Jul 31 at 15:04
Could you strictly explain what $(partial_x' x)_alpha$ is, given that $x$ and $x'$ are assumed to be coordinates of some points and no function or relation between them has been defined?
– Adayah
Jul 31 at 15:04
Since $x=x'+xi$, that derivative is $1$.
– J.G.
Jul 31 at 15:09
Since $x=x'+xi$, that derivative is $1$.
– J.G.
Jul 31 at 15:09
@J.G.- I had a bit trouble understanding your derivative notation.... Thanks anyway
– Joe
Jul 31 at 15:31
@J.G.- I had a bit trouble understanding your derivative notation.... Thanks anyway
– Joe
Jul 31 at 15:31
Formally, to even begin speaking of differentiation and the chain rule, we must define one or more functions (and obviously, it must be made absolutely clear which thing is a function and which is a variable). But here we have as much as a collection of indeterminates ($x$, $xi$, $r$, etc.) and some relations between them, e.g. $xi = x - x'$. No distinction between functions and variables is made, yet we proceed to differentiate those indeterminates one with respect to another in every possible way. (cont'd)
– Adayah
Jul 31 at 15:55
Formally, to even begin speaking of differentiation and the chain rule, we must define one or more functions (and obviously, it must be made absolutely clear which thing is a function and which is a variable). But here we have as much as a collection of indeterminates ($x$, $xi$, $r$, etc.) and some relations between them, e.g. $xi = x - x'$. No distinction between functions and variables is made, yet we proceed to differentiate those indeterminates one with respect to another in every possible way. (cont'd)
– Adayah
Jul 31 at 15:55
Even if the original relation $xi = x - x'$ slightly suggests that $xi$ is a function of variables $x$ and $x'$, it is freely transformed into $x = x' + xi$, which supposedly justifies treating $x$ as a function of $x'$ and $xi$ and differentiating it. It feels very alarming and unclear. That's why if you could explain in possibly strict terms some rigorous theory behind such seemingly incorrect operations, I think it would be very useful to many people here (including myself).
– Adayah
Jul 31 at 15:55
Even if the original relation $xi = x - x'$ slightly suggests that $xi$ is a function of variables $x$ and $x'$, it is freely transformed into $x = x' + xi$, which supposedly justifies treating $x$ as a function of $x'$ and $xi$ and differentiating it. It feels very alarming and unclear. That's why if you could explain in possibly strict terms some rigorous theory behind such seemingly incorrect operations, I think it would be very useful to many people here (including myself).
– Adayah
Jul 31 at 15:55
 |Â
show 5 more comments
up vote
0
down vote
What you've written in the yellow box is nonsense.
An expression like $fracpartial xpartial x'$ only makes sense when $x'$ is thought of as one of several independent variables, including presumably $y'$ and $z'$, when $x$ is as a function of all three of them (in that setting, $fracpartial xpartial x'$ is defined by holding $y'$ and $z'$ constant while letting $x'$ vary, and taking the limit of a very carefully formulated difference quotient).
Similarly, and expression like $fracpartial x'partial x$ only makes sense when $x$ is thought of as one of several independent variables, including presumably $y$ and $x$, and when $x'$ is as a function of all three of them.
Obviously these two situations are incompatible, so any kind of relation or equation between $fracpartial xpartial x'$ and $fracpartial x'partial x$ makes no sense, and any assumption of an equation like $fracpartial xpartial x' = 0$ makes no sense.
answered Jul 31 at 14:52
Lee Mosher
45.4k33478
add a comment |Â
up vote
0
down vote
What you've written in the yellow box is nonsense.
An expression like $fracpartial xpartial x'$ only makes sense when $x'$ is thought of as one of several independent variables, including presumably $y'$ and $z'$, when $x$ is as a function of all three of them (in that setting, $fracpartial xpartial x'$ is defined by holding $y'$ and $z'$ constant while letting $x'$ vary, and taking the limit of a very carefully formulated difference quotient).
Similarly, and expression like $fracpartial x'partial x$ only makes sense when $x$ is thought of as one of several independent variables, including presumably $y$ and $x$, and when $x'$ is as a function of all three of them.
Obviously these two situations are incompatible, so any kind of relation or equation between $fracpartial xpartial x'$ and $fracpartial x'partial x$ makes no sense, and any assumption of an equation like $fracpartial xpartial x' = 0$ makes no sense.
answered Jul 31 at 14:52
Lee Mosher
45.4k33478
add a comment |Â
up vote
0
down vote
up vote
0
down vote
What you've written in the yellow box is nonsense.
An expression like $fracpartial xpartial x'$ only makes sense when $x'$ is thought of as one of several independent variables, including presumably $y'$ and $z'$, when $x$ is as a function of all three of them (in that setting, $fracpartial xpartial x'$ is defined by holding $y'$ and $z'$ constant while letting $x'$ vary, and taking the limit of a very carefully formulated difference quotient).
Similarly, and expression like $fracpartial x'partial x$ only makes sense when $x$ is thought of as one of several independent variables, including presumably $y$ and $x$, and when $x'$ is as a function of all three of them.
Obviously these two situations are incompatible, so any kind of relation or equation between $fracpartial xpartial x'$ and $fracpartial x'partial x$ makes no sense, and any assumption of an equation like $fracpartial xpartial x' = 0$ makes no sense.
answered Jul 31 at 14:52
Lee Mosher
45.4k33478
What you've written in the yellow box is nonsense.
An expression like $fracpartial xpartial x'$ only makes sense when $x'$ is thought of as one of several independent variables, including presumably $y'$ and $z'$, when $x$ is as a function of all three of them (in that setting, $fracpartial xpartial x'$ is defined by holding $y'$ and $z'$ constant while letting $x'$ vary, and taking the limit of a very carefully formulated difference quotient).
Similarly, and expression like $fracpartial x'partial x$ only makes sense when $x$ is thought of as one of several independent variables, including presumably $y$ and $x$, and when $x'$ is as a function of all three of them.
Obviously these two situations are incompatible, so any kind of relation or equation between $fracpartial xpartial x'$ and $fracpartial x'partial x$ makes no sense, and any assumption of an equation like $fracpartial xpartial x' = 0$ makes no sense.
answered Jul 31 at 14:52
Lee Mosher
45.4k33478
answered Jul 31 at 14:52
Lee Mosher
45.4k33478
answered Jul 31 at 14:52
Lee Mosher
45.4k33478
answered Jul 31 at 14:52
Lee Mosher
45.4k33478
45.4k33478
add a comment |Â
add a comment |Â
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$partial$ doesn't satisfy the chain rule that $mathrmd$ does.
– Hurkyl
Jul 31 at 14:51
You can't take the partial derivative on $x'$ while keeping $x'$ constant.
– Yves Daoust
Jul 31 at 14:52
1
By using the chain rule in the last step you're implying that $x=x(x')$, otherwise this will not make sense. Using you're logic we could define the function $y=x+c$ ,where $c$ is constant and then saying $fracpartial ypartial x = fracpartial ypartial c fracpartial cpartial x=0$.
– User123456789
Jul 31 at 14:53