Mixing
Where do I go wrong with the product rule with three variables?
I am not sure how to use the product rule for differentiation, when I have three (or more) variables. For example, how would I solve this?
$$(y frac{d}{dz} - z frac{d}{dy}) (z frac{d}{dx} - x frac{d}{dz})$$
Because I thought it goes like this (For example, just for the $y frac {d}{dz} * z frac{d}{dx}$ part):
You first multiply everything, so you get:
$$ yz frac{d}{dz} frac{d}{dx}$$
and then get the sum of the individual derivatives, like:
$$yz frac{d}{dz} + yz frac{d}{dx}$$
But it's wrong and I don't know why, where did I go wrong?
Edit: I didn't realise you need a function, but I am not sure how to answer, so here is the whole example (The title says: Calculate the commutator):
The whole problem
multivariable-calculus
asked Nov 20 '18 at 15:23
Alopi Jameson
83
add a comment |
I am not sure how to use the product rule for differentiation, when I have three (or more) variables. For example, how would I solve this?
$$(y frac{d}{dz} - z frac{d}{dy}) (z frac{d}{dx} - x frac{d}{dz})$$
Because I thought it goes like this (For example, just for the $y frac {d}{dz} * z frac{d}{dx}$ part):
You first multiply everything, so you get:
$$ yz frac{d}{dz} frac{d}{dx}$$
and then get the sum of the individual derivatives, like:
$$yz frac{d}{dz} + yz frac{d}{dx}$$
But it's wrong and I don't know why, where did I go wrong?
Edit: I didn't realise you need a function, but I am not sure how to answer, so here is the whole example (The title says: Calculate the commutator):
The whole problem
multivariable-calculus
asked Nov 20 '18 at 15:23
Alopi Jameson
83
1
What are you trying to find? Is there a function $w$ that involves all three variables? Is one of the variables a function of the other 2?
– Joel Pereira
Nov 20 '18 at 15:29
Yes, there is a function that involves all three variables, sorry. It's a spatial function so the variables are coordinates x, y and z. The exercise is from quantum chemistry, and you have to calculate the commutator. This is just a part of that
– Alopi Jameson
Nov 20 '18 at 15:33
And what is the function you are trying to differentiate?
– saulspatz
Nov 20 '18 at 15:35
I have added the whole problem, if it helps.
– Alopi Jameson
Nov 20 '18 at 15:41
add a comment |
I am not sure how to use the product rule for differentiation, when I have three (or more) variables. For example, how would I solve this?
$$(y frac{d}{dz} - z frac{d}{dy}) (z frac{d}{dx} - x frac{d}{dz})$$
Because I thought it goes like this (For example, just for the $y frac {d}{dz} * z frac{d}{dx}$ part):
You first multiply everything, so you get:
$$ yz frac{d}{dz} frac{d}{dx}$$
and then get the sum of the individual derivatives, like:
$$yz frac{d}{dz} + yz frac{d}{dx}$$
But it's wrong and I don't know why, where did I go wrong?
Edit: I didn't realise you need a function, but I am not sure how to answer, so here is the whole example (The title says: Calculate the commutator):
The whole problem
multivariable-calculus
asked Nov 20 '18 at 15:23
Alopi Jameson
83
I am not sure how to use the product rule for differentiation, when I have three (or more) variables. For example, how would I solve this?
$$(y frac{d}{dz} - z frac{d}{dy}) (z frac{d}{dx} - x frac{d}{dz})$$
Because I thought it goes like this (For example, just for the $y frac {d}{dz} * z frac{d}{dx}$ part):
You first multiply everything, so you get:
$$ yz frac{d}{dz} frac{d}{dx}$$
and then get the sum of the individual derivatives, like:
$$yz frac{d}{dz} + yz frac{d}{dx}$$
But it's wrong and I don't know why, where did I go wrong?
Edit: I didn't realise you need a function, but I am not sure how to answer, so here is the whole example (The title says: Calculate the commutator):
The whole problem
multivariable-calculus
multivariable-calculus
asked Nov 20 '18 at 15:23
Alopi Jameson
83
asked Nov 20 '18 at 15:23
Alopi Jameson
83
edited Nov 20 '18 at 15:40
asked Nov 20 '18 at 15:23
Alopi Jameson
83
asked Nov 20 '18 at 15:23
Alopi Jameson
83
asked Nov 20 '18 at 15:23
Alopi Jameson
83
83
1
What are you trying to find? Is there a function $w$ that involves all three variables? Is one of the variables a function of the other 2?
– Joel Pereira
Nov 20 '18 at 15:29
Yes, there is a function that involves all three variables, sorry. It's a spatial function so the variables are coordinates x, y and z. The exercise is from quantum chemistry, and you have to calculate the commutator. This is just a part of that
– Alopi Jameson
Nov 20 '18 at 15:33
And what is the function you are trying to differentiate?
– saulspatz
Nov 20 '18 at 15:35
I have added the whole problem, if it helps.
– Alopi Jameson
Nov 20 '18 at 15:41
add a comment |
1
What are you trying to find? Is there a function $w$ that involves all three variables? Is one of the variables a function of the other 2?
– Joel Pereira
Nov 20 '18 at 15:29
Yes, there is a function that involves all three variables, sorry. It's a spatial function so the variables are coordinates x, y and z. The exercise is from quantum chemistry, and you have to calculate the commutator. This is just a part of that
– Alopi Jameson
Nov 20 '18 at 15:33
And what is the function you are trying to differentiate?
– saulspatz
Nov 20 '18 at 15:35
I have added the whole problem, if it helps.
– Alopi Jameson
Nov 20 '18 at 15:41
1
1
What are you trying to find? Is there a function $w$ that involves all three variables? Is one of the variables a function of the other 2?
– Joel Pereira
Nov 20 '18 at 15:29
What are you trying to find? Is there a function $w$ that involves all three variables? Is one of the variables a function of the other 2?
– Joel Pereira
Nov 20 '18 at 15:29
Yes, there is a function that involves all three variables, sorry. It's a spatial function so the variables are coordinates x, y and z. The exercise is from quantum chemistry, and you have to calculate the commutator. This is just a part of that
– Alopi Jameson
Nov 20 '18 at 15:33
Yes, there is a function that involves all three variables, sorry. It's a spatial function so the variables are coordinates x, y and z. The exercise is from quantum chemistry, and you have to calculate the commutator. This is just a part of that
– Alopi Jameson
Nov 20 '18 at 15:33
And what is the function you are trying to differentiate?
– saulspatz
Nov 20 '18 at 15:35
And what is the function you are trying to differentiate?
– saulspatz
Nov 20 '18 at 15:35
I have added the whole problem, if it helps.
– Alopi Jameson
Nov 20 '18 at 15:41
I have added the whole problem, if it helps.
– Alopi Jameson
Nov 20 '18 at 15:41
add a comment |
2 Answers
2
active
oldest
votes
The differential operator $dfrac {partial}{partial x}$ acts on functions via differentiation, on other differential operators via composition, and obeys the product rule. For instance,
$$frac{partial}{partial z} left( z frac{partial}{partial x} right) = frac{partial z }{partial z} frac{partial}{partial x} + zfrac{partial}{partial z}frac{partial}{partial x} = frac{partial}{partial x} + z frac{partial^2}{partial z partial x}$$
so that
$$left( y frac{partial}{partial z} right)left( z frac{partial}{partial x} right) = y frac{partial}{partial x} + yz frac{partial^2}{partial z partial x}$$
answered Nov 20 '18 at 15:59
Umberto P.
38.5k13064
Thank you! I understand how this works now. The differential operator acts only on the things behind it, so y is just left there and doesn't do anything.
– Alopi Jameson
Nov 20 '18 at 16:29
add a comment |
You are making three mistakes.
First, you assume that $yzfrac{partial}{partial z}frac{partial}{partial x}$ is the same as $yfrac{partial}{partial z}zfrac{partial}{partial x}$. Let's put a simple function $w(x,y,z)=x$ into the mix to see why this is wrong:
$yzfrac{partial}{partial z}(frac{partial w}{partial x}) = yzfrac{partial}{partial z}(1)=0$
$yfrac{partial}{partial z}(zfrac{partial w}{partial x})=yfrac{partial}{partial z}(z)=y$
In fact, by the product rule, $yfrac{partial}{partial z}zfrac{partial}{partial x}$ is equal to $y(frac{partial z}{partial z}frac{partial}{partial x}+yzfrac{partial}{partial z}frac{partial}{partial x})=y(frac{partial}{partial x}+zfrac{partial^2}{partial zpartial x})$
Second, $frac{partial}{partial z}frac{partial}{partial x}$ is not equal to $frac{partial}{partial z}+frac{partial}{partial x}$. Why would it be?
Third: $frac{d}{dz}$ is not the same as $frac{partial}{partial z}$!
answered Nov 20 '18 at 15:59
TonyK
41.6k353132
Thank you, this helped a lot! (I was using d because I didn't find the proper symbol)
– Alopi Jameson
Nov 20 '18 at 16:28
@AlopiJameson: it'spartial.
– TonyK
Nov 20 '18 at 16:30
Thank you. Is there a way to accept more than one answer?
– Alopi Jameson
Nov 20 '18 at 16:32
@AlopiJameson: No there isn't. (But you can up-vote.)
– TonyK
Nov 20 '18 at 16:34
Not enough reputation unfortunately.
– Alopi Jameson
Nov 20 '18 at 16:35
add a comment |
Your Answer
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2 Answers
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2 Answers
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The differential operator $dfrac {partial}{partial x}$ acts on functions via differentiation, on other differential operators via composition, and obeys the product rule. For instance,
$$frac{partial}{partial z} left( z frac{partial}{partial x} right) = frac{partial z }{partial z} frac{partial}{partial x} + zfrac{partial}{partial z}frac{partial}{partial x} = frac{partial}{partial x} + z frac{partial^2}{partial z partial x}$$
so that
$$left( y frac{partial}{partial z} right)left( z frac{partial}{partial x} right) = y frac{partial}{partial x} + yz frac{partial^2}{partial z partial x}$$
answered Nov 20 '18 at 15:59
Umberto P.
38.5k13064
Thank you! I understand how this works now. The differential operator acts only on the things behind it, so y is just left there and doesn't do anything.
– Alopi Jameson
Nov 20 '18 at 16:29
add a comment |
The differential operator $dfrac {partial}{partial x}$ acts on functions via differentiation, on other differential operators via composition, and obeys the product rule. For instance,
$$frac{partial}{partial z} left( z frac{partial}{partial x} right) = frac{partial z }{partial z} frac{partial}{partial x} + zfrac{partial}{partial z}frac{partial}{partial x} = frac{partial}{partial x} + z frac{partial^2}{partial z partial x}$$
so that
$$left( y frac{partial}{partial z} right)left( z frac{partial}{partial x} right) = y frac{partial}{partial x} + yz frac{partial^2}{partial z partial x}$$
answered Nov 20 '18 at 15:59
Umberto P.
38.5k13064
Thank you! I understand how this works now. The differential operator acts only on the things behind it, so y is just left there and doesn't do anything.
– Alopi Jameson
Nov 20 '18 at 16:29
add a comment |
The differential operator $dfrac {partial}{partial x}$ acts on functions via differentiation, on other differential operators via composition, and obeys the product rule. For instance,
$$frac{partial}{partial z} left( z frac{partial}{partial x} right) = frac{partial z }{partial z} frac{partial}{partial x} + zfrac{partial}{partial z}frac{partial}{partial x} = frac{partial}{partial x} + z frac{partial^2}{partial z partial x}$$
so that
$$left( y frac{partial}{partial z} right)left( z frac{partial}{partial x} right) = y frac{partial}{partial x} + yz frac{partial^2}{partial z partial x}$$
answered Nov 20 '18 at 15:59
Umberto P.
38.5k13064
The differential operator $dfrac {partial}{partial x}$ acts on functions via differentiation, on other differential operators via composition, and obeys the product rule. For instance,
$$frac{partial}{partial z} left( z frac{partial}{partial x} right) = frac{partial z }{partial z} frac{partial}{partial x} + zfrac{partial}{partial z}frac{partial}{partial x} = frac{partial}{partial x} + z frac{partial^2}{partial z partial x}$$
so that
$$left( y frac{partial}{partial z} right)left( z frac{partial}{partial x} right) = y frac{partial}{partial x} + yz frac{partial^2}{partial z partial x}$$
answered Nov 20 '18 at 15:59
Umberto P.
38.5k13064
answered Nov 20 '18 at 15:59
Umberto P.
38.5k13064
answered Nov 20 '18 at 15:59
Umberto P.
38.5k13064
answered Nov 20 '18 at 15:59
Umberto P.
38.5k13064
38.5k13064
Thank you! I understand how this works now. The differential operator acts only on the things behind it, so y is just left there and doesn't do anything.
– Alopi Jameson
Nov 20 '18 at 16:29
add a comment |
Thank you! I understand how this works now. The differential operator acts only on the things behind it, so y is just left there and doesn't do anything.
– Alopi Jameson
Nov 20 '18 at 16:29
Thank you! I understand how this works now. The differential operator acts only on the things behind it, so y is just left there and doesn't do anything.
– Alopi Jameson
Nov 20 '18 at 16:29
Thank you! I understand how this works now. The differential operator acts only on the things behind it, so y is just left there and doesn't do anything.
– Alopi Jameson
Nov 20 '18 at 16:29
add a comment |
You are making three mistakes.
First, you assume that $yzfrac{partial}{partial z}frac{partial}{partial x}$ is the same as $yfrac{partial}{partial z}zfrac{partial}{partial x}$. Let's put a simple function $w(x,y,z)=x$ into the mix to see why this is wrong:
$yzfrac{partial}{partial z}(frac{partial w}{partial x}) = yzfrac{partial}{partial z}(1)=0$
$yfrac{partial}{partial z}(zfrac{partial w}{partial x})=yfrac{partial}{partial z}(z)=y$
In fact, by the product rule, $yfrac{partial}{partial z}zfrac{partial}{partial x}$ is equal to $y(frac{partial z}{partial z}frac{partial}{partial x}+yzfrac{partial}{partial z}frac{partial}{partial x})=y(frac{partial}{partial x}+zfrac{partial^2}{partial zpartial x})$
Second, $frac{partial}{partial z}frac{partial}{partial x}$ is not equal to $frac{partial}{partial z}+frac{partial}{partial x}$. Why would it be?
Third: $frac{d}{dz}$ is not the same as $frac{partial}{partial z}$!
answered Nov 20 '18 at 15:59
TonyK
41.6k353132
Thank you, this helped a lot! (I was using d because I didn't find the proper symbol)
– Alopi Jameson
Nov 20 '18 at 16:28
@AlopiJameson: it'spartial.
– TonyK
Nov 20 '18 at 16:30
Thank you. Is there a way to accept more than one answer?
– Alopi Jameson
Nov 20 '18 at 16:32
@AlopiJameson: No there isn't. (But you can up-vote.)
– TonyK
Nov 20 '18 at 16:34
Not enough reputation unfortunately.
– Alopi Jameson
Nov 20 '18 at 16:35
add a comment |
You are making three mistakes.
First, you assume that $yzfrac{partial}{partial z}frac{partial}{partial x}$ is the same as $yfrac{partial}{partial z}zfrac{partial}{partial x}$. Let's put a simple function $w(x,y,z)=x$ into the mix to see why this is wrong:
$yzfrac{partial}{partial z}(frac{partial w}{partial x}) = yzfrac{partial}{partial z}(1)=0$
$yfrac{partial}{partial z}(zfrac{partial w}{partial x})=yfrac{partial}{partial z}(z)=y$
In fact, by the product rule, $yfrac{partial}{partial z}zfrac{partial}{partial x}$ is equal to $y(frac{partial z}{partial z}frac{partial}{partial x}+yzfrac{partial}{partial z}frac{partial}{partial x})=y(frac{partial}{partial x}+zfrac{partial^2}{partial zpartial x})$
Second, $frac{partial}{partial z}frac{partial}{partial x}$ is not equal to $frac{partial}{partial z}+frac{partial}{partial x}$. Why would it be?
Third: $frac{d}{dz}$ is not the same as $frac{partial}{partial z}$!
answered Nov 20 '18 at 15:59
TonyK
41.6k353132
Thank you, this helped a lot! (I was using d because I didn't find the proper symbol)
– Alopi Jameson
Nov 20 '18 at 16:28
@AlopiJameson: it'spartial.
– TonyK
Nov 20 '18 at 16:30
Thank you. Is there a way to accept more than one answer?
– Alopi Jameson
Nov 20 '18 at 16:32
@AlopiJameson: No there isn't. (But you can up-vote.)
– TonyK
Nov 20 '18 at 16:34
Not enough reputation unfortunately.
– Alopi Jameson
Nov 20 '18 at 16:35
add a comment |
You are making three mistakes.
First, you assume that $yzfrac{partial}{partial z}frac{partial}{partial x}$ is the same as $yfrac{partial}{partial z}zfrac{partial}{partial x}$. Let's put a simple function $w(x,y,z)=x$ into the mix to see why this is wrong:
$yzfrac{partial}{partial z}(frac{partial w}{partial x}) = yzfrac{partial}{partial z}(1)=0$
$yfrac{partial}{partial z}(zfrac{partial w}{partial x})=yfrac{partial}{partial z}(z)=y$
In fact, by the product rule, $yfrac{partial}{partial z}zfrac{partial}{partial x}$ is equal to $y(frac{partial z}{partial z}frac{partial}{partial x}+yzfrac{partial}{partial z}frac{partial}{partial x})=y(frac{partial}{partial x}+zfrac{partial^2}{partial zpartial x})$
Second, $frac{partial}{partial z}frac{partial}{partial x}$ is not equal to $frac{partial}{partial z}+frac{partial}{partial x}$. Why would it be?
Third: $frac{d}{dz}$ is not the same as $frac{partial}{partial z}$!
answered Nov 20 '18 at 15:59
TonyK
41.6k353132
You are making three mistakes.
First, you assume that $yzfrac{partial}{partial z}frac{partial}{partial x}$ is the same as $yfrac{partial}{partial z}zfrac{partial}{partial x}$. Let's put a simple function $w(x,y,z)=x$ into the mix to see why this is wrong:
$yzfrac{partial}{partial z}(frac{partial w}{partial x}) = yzfrac{partial}{partial z}(1)=0$
$yfrac{partial}{partial z}(zfrac{partial w}{partial x})=yfrac{partial}{partial z}(z)=y$
In fact, by the product rule, $yfrac{partial}{partial z}zfrac{partial}{partial x}$ is equal to $y(frac{partial z}{partial z}frac{partial}{partial x}+yzfrac{partial}{partial z}frac{partial}{partial x})=y(frac{partial}{partial x}+zfrac{partial^2}{partial zpartial x})$
Second, $frac{partial}{partial z}frac{partial}{partial x}$ is not equal to $frac{partial}{partial z}+frac{partial}{partial x}$. Why would it be?
Third: $frac{d}{dz}$ is not the same as $frac{partial}{partial z}$!
answered Nov 20 '18 at 15:59
TonyK
41.6k353132
answered Nov 20 '18 at 15:59
TonyK
41.6k353132
answered Nov 20 '18 at 15:59
TonyK
41.6k353132
answered Nov 20 '18 at 15:59
TonyK
41.6k353132
41.6k353132
Thank you, this helped a lot! (I was using d because I didn't find the proper symbol)
– Alopi Jameson
Nov 20 '18 at 16:28
@AlopiJameson: it'spartial.
– TonyK
Nov 20 '18 at 16:30
Thank you. Is there a way to accept more than one answer?
– Alopi Jameson
Nov 20 '18 at 16:32
@AlopiJameson: No there isn't. (But you can up-vote.)
– TonyK
Nov 20 '18 at 16:34
Not enough reputation unfortunately.
– Alopi Jameson
Nov 20 '18 at 16:35
add a comment |
Thank you, this helped a lot! (I was using d because I didn't find the proper symbol)
– Alopi Jameson
Nov 20 '18 at 16:28
@AlopiJameson: it'spartial.
– TonyK
Nov 20 '18 at 16:30
Thank you. Is there a way to accept more than one answer?
– Alopi Jameson
Nov 20 '18 at 16:32
@AlopiJameson: No there isn't. (But you can up-vote.)
– TonyK
Nov 20 '18 at 16:34
Not enough reputation unfortunately.
– Alopi Jameson
Nov 20 '18 at 16:35
Thank you, this helped a lot! (I was using d because I didn't find the proper symbol)
– Alopi Jameson
Nov 20 '18 at 16:28
Thank you, this helped a lot! (I was using d because I didn't find the proper symbol)
– Alopi Jameson
Nov 20 '18 at 16:28
@AlopiJameson: it's
partial.– TonyK
Nov 20 '18 at 16:30
@AlopiJameson: it's
partial.– TonyK
Nov 20 '18 at 16:30
Thank you. Is there a way to accept more than one answer?
– Alopi Jameson
Nov 20 '18 at 16:32
Thank you. Is there a way to accept more than one answer?
– Alopi Jameson
Nov 20 '18 at 16:32
@AlopiJameson: No there isn't. (But you can up-vote.)
– TonyK
Nov 20 '18 at 16:34
@AlopiJameson: No there isn't. (But you can up-vote.)
– TonyK
Nov 20 '18 at 16:34
Not enough reputation unfortunately.
– Alopi Jameson
Nov 20 '18 at 16:35
Not enough reputation unfortunately.
– Alopi Jameson
Nov 20 '18 at 16:35
add a comment |
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1
What are you trying to find? Is there a function $w$ that involves all three variables? Is one of the variables a function of the other 2?
– Joel Pereira
Nov 20 '18 at 15:29
Yes, there is a function that involves all three variables, sorry. It's a spatial function so the variables are coordinates x, y and z. The exercise is from quantum chemistry, and you have to calculate the commutator. This is just a part of that
– Alopi Jameson
Nov 20 '18 at 15:33
And what is the function you are trying to differentiate?
– saulspatz
Nov 20 '18 at 15:35
I have added the whole problem, if it helps.
– Alopi Jameson
Nov 20 '18 at 15:41