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











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












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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006452%2fwhere-do-i-go-wrong-with-the-product-rule-with-three-variables%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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.

38.2k13063




38.2k13063












  • 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

40.7k352130




40.7k352130












  • 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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006452%2fwhere-do-i-go-wrong-with-the-product-rule-with-three-variables%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Quarter-circle Tiles

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Mont Emei