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
multivariable-calculus
add a comment |
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
multivariable-calculus
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
add a comment |
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
multivariable-calculus
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
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
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 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
add a comment |
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
$$left( y frac{partial}{partial z} right)left( z frac{partial}{partial x} right) = y frac{partial}{partial x} + yz frac{partial^2}{partial z partial x}$$
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
add a comment |
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 $frac{partial}{partial z}$!
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'spartial
.
– 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
add a comment |
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
$$left( y frac{partial}{partial z} right)left( z frac{partial}{partial x} right) = y frac{partial}{partial x} + yz frac{partial^2}{partial z partial x}$$
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
add a comment |
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
$$left( y frac{partial}{partial z} right)left( z frac{partial}{partial x} right) = y frac{partial}{partial x} + yz frac{partial^2}{partial z partial x}$$
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
add a comment |
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
$$left( y frac{partial}{partial z} right)left( z frac{partial}{partial x} right) = y frac{partial}{partial x} + yz frac{partial^2}{partial z partial x}$$
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 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
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 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
add a comment |
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 $frac{partial}{partial z}$!
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'spartial
.
– 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
add a comment |
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 $frac{partial}{partial z}$!
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'spartial
.
– 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
add a comment |
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 $frac{partial}{partial z}$!
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 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'spartial
.
– 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
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 at 16:28
@AlopiJameson: it'spartial
.
– 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
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
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