Where do I go wrong with the product rule with three variables?











up vote
1
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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










share|cite|improve this question




















  • 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 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 at 15:33












  • And what is the function you are trying to differentiate?
    – saulspatz
    Nov 20 at 15:35










  • I have added the whole problem, if it helps.
    – Alopi Jameson
    Nov 20 at 15:41















up vote
1
down vote

favorite












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










share|cite|improve this question




















  • 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 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 at 15:33












  • And what is the function you are trying to differentiate?
    – saulspatz
    Nov 20 at 15:35










  • I have added the whole problem, if it helps.
    – Alopi Jameson
    Nov 20 at 15:41













up vote
1
down vote

favorite









up vote
1
down vote

favorite











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










share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 at 15:40

























asked Nov 20 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 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 at 15:33












  • And what is the function you are trying to differentiate?
    – saulspatz
    Nov 20 at 15:35










  • I have added the whole problem, if it helps.
    – Alopi Jameson
    Nov 20 at 15:41














  • 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 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 at 15:33












  • And what is the function you are trying to differentiate?
    – saulspatz
    Nov 20 at 15:35










  • I have added the whole problem, if it helps.
    – Alopi Jameson
    Nov 20 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 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 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 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 at 15:33














And what is the function you are trying to differentiate?
– saulspatz
Nov 20 at 15:35




And what is the function you are trying to differentiate?
– saulspatz
Nov 20 at 15:35












I have added the whole problem, if it helps.
– Alopi Jameson
Nov 20 at 15:41




I have added the whole problem, if it helps.
– Alopi Jameson
Nov 20 at 15:41










2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










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
$$left( y frac{partial}{partial z} right)left( z frac{partial}{partial x} right) = y frac{partial}{partial x} + yz frac{partial^2}{partial z partial x}$$






share|cite|improve this answer





















  • 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 at 16:29


















up vote
0
down vote













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 $frac{partial}{partial z}$!






share|cite|improve this answer





















  • Thank you, this helped a lot! (I was using d because I didn't find the proper symbol)
    – Alopi Jameson
    Nov 20 at 16:28










  • @AlopiJameson: it's partial.
    – TonyK
    Nov 20 at 16:30










  • Thank you. Is there a way to accept more than one answer?
    – Alopi Jameson
    Nov 20 at 16:32










  • @AlopiJameson: No there isn't. (But you can up-vote.)
    – TonyK
    Nov 20 at 16:34












  • Not enough reputation unfortunately.
    – Alopi Jameson
    Nov 20 at 16:35











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










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
$$left( y frac{partial}{partial z} right)left( z frac{partial}{partial x} right) = y frac{partial}{partial x} + yz frac{partial^2}{partial z partial x}$$






share|cite|improve this answer





















  • 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 at 16:29















up vote
0
down vote



accepted










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
$$left( y frac{partial}{partial z} right)left( z frac{partial}{partial x} right) = y frac{partial}{partial x} + yz frac{partial^2}{partial z partial x}$$






share|cite|improve this answer





















  • 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 at 16:29













up vote
0
down vote



accepted







up vote
0
down vote



accepted






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
$$left( y frac{partial}{partial z} right)left( z frac{partial}{partial x} right) = y frac{partial}{partial x} + yz frac{partial^2}{partial z partial x}$$






share|cite|improve this answer












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
$$left( y frac{partial}{partial z} right)left( z frac{partial}{partial x} right) = y frac{partial}{partial x} + yz frac{partial^2}{partial z partial x}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 20 at 15:59









Umberto P.

38.3k13063




38.3k13063












  • 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 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 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 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 at 16:29










up vote
0
down vote













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 $frac{partial}{partial z}$!






share|cite|improve this answer





















  • Thank you, this helped a lot! (I was using d because I didn't find the proper symbol)
    – Alopi Jameson
    Nov 20 at 16:28










  • @AlopiJameson: it's partial.
    – TonyK
    Nov 20 at 16:30










  • Thank you. Is there a way to accept more than one answer?
    – Alopi Jameson
    Nov 20 at 16:32










  • @AlopiJameson: No there isn't. (But you can up-vote.)
    – TonyK
    Nov 20 at 16:34












  • Not enough reputation unfortunately.
    – Alopi Jameson
    Nov 20 at 16:35















up vote
0
down vote













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 $frac{partial}{partial z}$!






share|cite|improve this answer





















  • Thank you, this helped a lot! (I was using d because I didn't find the proper symbol)
    – Alopi Jameson
    Nov 20 at 16:28










  • @AlopiJameson: it's partial.
    – TonyK
    Nov 20 at 16:30










  • Thank you. Is there a way to accept more than one answer?
    – Alopi Jameson
    Nov 20 at 16:32










  • @AlopiJameson: No there isn't. (But you can up-vote.)
    – TonyK
    Nov 20 at 16:34












  • Not enough reputation unfortunately.
    – Alopi Jameson
    Nov 20 at 16:35













up vote
0
down vote










up vote
0
down vote









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 $frac{partial}{partial z}$!






share|cite|improve this answer












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 $frac{partial}{partial z}$!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 20 at 15:59









TonyK

41.2k353131




41.2k353131












  • Thank you, this helped a lot! (I was using d because I didn't find the proper symbol)
    – Alopi Jameson
    Nov 20 at 16:28










  • @AlopiJameson: it's partial.
    – TonyK
    Nov 20 at 16:30










  • Thank you. Is there a way to accept more than one answer?
    – Alopi Jameson
    Nov 20 at 16:32










  • @AlopiJameson: No there isn't. (But you can up-vote.)
    – TonyK
    Nov 20 at 16:34












  • Not enough reputation unfortunately.
    – Alopi Jameson
    Nov 20 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 at 16:28










  • @AlopiJameson: it's partial.
    – TonyK
    Nov 20 at 16:30










  • Thank you. Is there a way to accept more than one answer?
    – Alopi Jameson
    Nov 20 at 16:32










  • @AlopiJameson: No there isn't. (But you can up-vote.)
    – TonyK
    Nov 20 at 16:34












  • Not enough reputation unfortunately.
    – Alopi Jameson
    Nov 20 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 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 at 16:28












@AlopiJameson: it's partial.
– TonyK
Nov 20 at 16:30




@AlopiJameson: it's partial.
– TonyK
Nov 20 at 16:30












Thank you. Is there a way to accept more than one answer?
– Alopi Jameson
Nov 20 at 16:32




Thank you. Is there a way to accept more than one answer?
– Alopi Jameson
Nov 20 at 16:32












@AlopiJameson: No there isn't. (But you can up-vote.)
– TonyK
Nov 20 at 16:34






@AlopiJameson: No there isn't. (But you can up-vote.)
– TonyK
Nov 20 at 16:34














Not enough reputation unfortunately.
– Alopi Jameson
Nov 20 at 16:35




Not enough reputation unfortunately.
– Alopi Jameson
Nov 20 at 16:35


















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