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












1














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




















  • 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














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




















  • 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








1







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















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






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


















0














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



















0














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









0














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
















0














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














0












0








0






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.

38.5k13064




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 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











0














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
















0














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














0












0








0






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

41.4k353132




41.4k353132












  • 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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Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
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