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What is wrong with this reasoning in with respect to working with differential of z
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I know implicit differentiation. Just playing around with the differential of $z$ I did something like this, which I am sure is wrong somewhere. I want to know where I am wrong. Given a function $z = f(x,y)$, its differential is
beginalign
delta z &= fracpartial zpartial x cdot delta x + fracpartial zpartial y cdot delta y
endalign
Now dividing by $delta x$ throughout and taking the limit $delta x$, $delta y$, $delta z$ tending to zero, I get something like
beginalign
fracpartial zpartial x &= fracpartial zpartial x cdot 1 + fracpartial zpartial y cdot fracpartial ypartial x
endalign
Cancelling the $fracpartial zpartial x$ term, I get $fracpartial zpartial y cdot fracpartial ypartial x = 0 $, which is not an identity.
My question is, where am I wrong?
calculus derivatives partial-derivative differential-forms
edited 4 hours ago
Mattos
2,66321121
asked 5 hours ago
Abhishek_04
142
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up vote
2
down vote
favorite
I know implicit differentiation. Just playing around with the differential of $z$ I did something like this, which I am sure is wrong somewhere. I want to know where I am wrong. Given a function $z = f(x,y)$, its differential is
beginalign
delta z &= fracpartial zpartial x cdot delta x + fracpartial zpartial y cdot delta y
endalign
Now dividing by $delta x$ throughout and taking the limit $delta x$, $delta y$, $delta z$ tending to zero, I get something like
beginalign
fracpartial zpartial x &= fracpartial zpartial x cdot 1 + fracpartial zpartial y cdot fracpartial ypartial x
endalign
Cancelling the $fracpartial zpartial x$ term, I get $fracpartial zpartial y cdot fracpartial ypartial x = 0 $, which is not an identity.
My question is, where am I wrong?
calculus derivatives partial-derivative differential-forms
edited 4 hours ago
Mattos
2,66321121
asked 5 hours ago
Abhishek_04
142
New contributor
Abhishek_04 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If we were allowed to treat $partial x$ and $dx$ (which you note $delta x$) as interchangeable (and cancelable), then we'd have a more basic problem: in the original equation we could cancel $delta x$ and also $delta y$ and we'd get $delta z = 2 delta z$...
– leonbloy
4 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I know implicit differentiation. Just playing around with the differential of $z$ I did something like this, which I am sure is wrong somewhere. I want to know where I am wrong. Given a function $z = f(x,y)$, its differential is
beginalign
delta z &= fracpartial zpartial x cdot delta x + fracpartial zpartial y cdot delta y
endalign
Now dividing by $delta x$ throughout and taking the limit $delta x$, $delta y$, $delta z$ tending to zero, I get something like
beginalign
fracpartial zpartial x &= fracpartial zpartial x cdot 1 + fracpartial zpartial y cdot fracpartial ypartial x
endalign
Cancelling the $fracpartial zpartial x$ term, I get $fracpartial zpartial y cdot fracpartial ypartial x = 0 $, which is not an identity.
My question is, where am I wrong?
calculus derivatives partial-derivative differential-forms
edited 4 hours ago
Mattos
2,66321121
asked 5 hours ago
Abhishek_04
142
New contributor
Abhishek_04 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I know implicit differentiation. Just playing around with the differential of $z$ I did something like this, which I am sure is wrong somewhere. I want to know where I am wrong. Given a function $z = f(x,y)$, its differential is
beginalign
delta z &= fracpartial zpartial x cdot delta x + fracpartial zpartial y cdot delta y
endalign
Now dividing by $delta x$ throughout and taking the limit $delta x$, $delta y$, $delta z$ tending to zero, I get something like
beginalign
fracpartial zpartial x &= fracpartial zpartial x cdot 1 + fracpartial zpartial y cdot fracpartial ypartial x
endalign
Cancelling the $fracpartial zpartial x$ term, I get $fracpartial zpartial y cdot fracpartial ypartial x = 0 $, which is not an identity.
My question is, where am I wrong?
calculus derivatives partial-derivative differential-forms
calculus derivatives partial-derivative differential-forms
edited 4 hours ago
Mattos
2,66321121
asked 5 hours ago
Abhishek_04
142
New contributor
Abhishek_04 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Mattos
2,66321121
asked 5 hours ago
Abhishek_04
142
New contributor
Abhishek_04 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Mattos
2,66321121
edited 4 hours ago
Mattos
2,66321121
edited 4 hours ago
Mattos
2,66321121
2,66321121
asked 5 hours ago
Abhishek_04
142
New contributor
Abhishek_04 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Abhishek_04
142
asked 5 hours ago
Abhishek_04
142
142
New contributor
Abhishek_04 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Abhishek_04 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Abhishek_04 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If we were allowed to treat $partial x$ and $dx$ (which you note $delta x$) as interchangeable (and cancelable), then we'd have a more basic problem: in the original equation we could cancel $delta x$ and also $delta y$ and we'd get $delta z = 2 delta z$...
– leonbloy
4 hours ago
add a comment |Â
If we were allowed to treat $partial x$ and $dx$ (which you note $delta x$) as interchangeable (and cancelable), then we'd have a more basic problem: in the original equation we could cancel $delta x$ and also $delta y$ and we'd get $delta z = 2 delta z$...
– leonbloy
4 hours ago
If we were allowed to treat $partial x$ and $dx$ (which you note $delta x$) as interchangeable (and cancelable), then we'd have a more basic problem: in the original equation we could cancel $delta x$ and also $delta y$ and we'd get $delta z = 2 delta z$...
– leonbloy
4 hours ago
If we were allowed to treat $partial x$ and $dx$ (which you note $delta x$) as interchangeable (and cancelable), then we'd have a more basic problem: in the original equation we could cancel $delta x$ and also $delta y$ and we'd get $delta z = 2 delta z$...
– leonbloy
4 hours ago
add a comment |Â
3 Answers
3
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oldest
votes
up vote
3
down vote
It's the difference between the partial derivative $fracpartial zpartial x$ and the total derivative $fracmathrmd zmathrmd x$. The total derivative accounts for potential dependency of $x$ and $y$ on each other. In our case, we would have
$$fracmathrmd zmathrmd x = fracpartial zpartial x + fracpartial zpartial y fracmathrmd ymathrmd x.$$
Note that this is precisely the (single-variable) derivative of a two variable function $z$, with $x$ and a function $y(x)$ substituted in. Rearranging this gives us the differential of $z$ as before.
$$mathrmd z = fracpartial zpartial xmathrmd x + fracpartial zpartial y mathrmd y.$$
Your mistake was treating $mathrmd z$ and $partial z$ as interchangeable, which they aren't.
answered 4 hours ago
Theo Bendit
13.9k12045
Thanks,it makes sense now.
– Abhishek_04
3 hours ago
add a comment |Â
up vote
1
down vote
You are dividing $$beginalign
delta z = fracpartial zpartial x * delta x + fracpartial zpartial y * delta y
endalign$$
by $ delta x$ to get $$ beginalign
fracpartial zpartial x = fracpartial zpartial x * 1 + fracpartial zpartial y * fracpartial ypartial x
endalign$$
How did you get $$fracpartial zpartial x$$
on the $LHS$ ?
Then you cancelled two different things from both sides.
Somehow you are confused between partial derivatives and total derivative.
answered 4 hours ago
Mohammad Riazi-Kermani
32k41853
add a comment |Â
up vote
1
down vote
In the right-hand side you don't have $z=z(x,y)$, but the composition $tildez(x)= tildez(x,y(x))$. It is another different function. The chain rule gives $$fracrm dtildezrm dx(x) = fracpartial zpartial x(x,y(x))+fracpartial zpartial y(x,y(x))fracrm dyrm dx(x). $$You were a victim of the abuse of notation $tildezequiv z$. In fact, I am commiting one more abuse myself: using $y$ both for a variable of the function $z$ and to denote a function of $x$ (but I suppose you understand that no one wants to keep writing $tildez(x)=z(x,tildey(x))$, etc.).
answered 4 hours ago
Ivo Terek
44.1k949135
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
It's the difference between the partial derivative $fracpartial zpartial x$ and the total derivative $fracmathrmd zmathrmd x$. The total derivative accounts for potential dependency of $x$ and $y$ on each other. In our case, we would have
$$fracmathrmd zmathrmd x = fracpartial zpartial x + fracpartial zpartial y fracmathrmd ymathrmd x.$$
Note that this is precisely the (single-variable) derivative of a two variable function $z$, with $x$ and a function $y(x)$ substituted in. Rearranging this gives us the differential of $z$ as before.
$$mathrmd z = fracpartial zpartial xmathrmd x + fracpartial zpartial y mathrmd y.$$
Your mistake was treating $mathrmd z$ and $partial z$ as interchangeable, which they aren't.
answered 4 hours ago
Theo Bendit
13.9k12045
Thanks,it makes sense now.
– Abhishek_04
3 hours ago
add a comment |Â
up vote
3
down vote
It's the difference between the partial derivative $fracpartial zpartial x$ and the total derivative $fracmathrmd zmathrmd x$. The total derivative accounts for potential dependency of $x$ and $y$ on each other. In our case, we would have
$$fracmathrmd zmathrmd x = fracpartial zpartial x + fracpartial zpartial y fracmathrmd ymathrmd x.$$
Note that this is precisely the (single-variable) derivative of a two variable function $z$, with $x$ and a function $y(x)$ substituted in. Rearranging this gives us the differential of $z$ as before.
$$mathrmd z = fracpartial zpartial xmathrmd x + fracpartial zpartial y mathrmd y.$$
Your mistake was treating $mathrmd z$ and $partial z$ as interchangeable, which they aren't.
answered 4 hours ago
Theo Bendit
13.9k12045
Thanks,it makes sense now.
– Abhishek_04
3 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
It's the difference between the partial derivative $fracpartial zpartial x$ and the total derivative $fracmathrmd zmathrmd x$. The total derivative accounts for potential dependency of $x$ and $y$ on each other. In our case, we would have
$$fracmathrmd zmathrmd x = fracpartial zpartial x + fracpartial zpartial y fracmathrmd ymathrmd x.$$
Note that this is precisely the (single-variable) derivative of a two variable function $z$, with $x$ and a function $y(x)$ substituted in. Rearranging this gives us the differential of $z$ as before.
$$mathrmd z = fracpartial zpartial xmathrmd x + fracpartial zpartial y mathrmd y.$$
Your mistake was treating $mathrmd z$ and $partial z$ as interchangeable, which they aren't.
answered 4 hours ago
Theo Bendit
13.9k12045
It's the difference between the partial derivative $fracpartial zpartial x$ and the total derivative $fracmathrmd zmathrmd x$. The total derivative accounts for potential dependency of $x$ and $y$ on each other. In our case, we would have
$$fracmathrmd zmathrmd x = fracpartial zpartial x + fracpartial zpartial y fracmathrmd ymathrmd x.$$
Note that this is precisely the (single-variable) derivative of a two variable function $z$, with $x$ and a function $y(x)$ substituted in. Rearranging this gives us the differential of $z$ as before.
$$mathrmd z = fracpartial zpartial xmathrmd x + fracpartial zpartial y mathrmd y.$$
Your mistake was treating $mathrmd z$ and $partial z$ as interchangeable, which they aren't.
answered 4 hours ago
Theo Bendit
13.9k12045
answered 4 hours ago
Theo Bendit
13.9k12045
answered 4 hours ago
Theo Bendit
13.9k12045
answered 4 hours ago
Theo Bendit
13.9k12045
13.9k12045
Thanks,it makes sense now.
– Abhishek_04
3 hours ago
add a comment |Â
Thanks,it makes sense now.
– Abhishek_04
3 hours ago
Thanks,it makes sense now.
– Abhishek_04
3 hours ago
Thanks,it makes sense now.
– Abhishek_04
3 hours ago
add a comment |Â
up vote
1
down vote
You are dividing $$beginalign
delta z = fracpartial zpartial x * delta x + fracpartial zpartial y * delta y
endalign$$
by $ delta x$ to get $$ beginalign
fracpartial zpartial x = fracpartial zpartial x * 1 + fracpartial zpartial y * fracpartial ypartial x
endalign$$
How did you get $$fracpartial zpartial x$$
on the $LHS$ ?
Then you cancelled two different things from both sides.
Somehow you are confused between partial derivatives and total derivative.
answered 4 hours ago
Mohammad Riazi-Kermani
32k41853
add a comment |Â
up vote
1
down vote
You are dividing $$beginalign
delta z = fracpartial zpartial x * delta x + fracpartial zpartial y * delta y
endalign$$
by $ delta x$ to get $$ beginalign
fracpartial zpartial x = fracpartial zpartial x * 1 + fracpartial zpartial y * fracpartial ypartial x
endalign$$
How did you get $$fracpartial zpartial x$$
on the $LHS$ ?
Then you cancelled two different things from both sides.
Somehow you are confused between partial derivatives and total derivative.
answered 4 hours ago
Mohammad Riazi-Kermani
32k41853
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You are dividing $$beginalign
delta z = fracpartial zpartial x * delta x + fracpartial zpartial y * delta y
endalign$$
by $ delta x$ to get $$ beginalign
fracpartial zpartial x = fracpartial zpartial x * 1 + fracpartial zpartial y * fracpartial ypartial x
endalign$$
How did you get $$fracpartial zpartial x$$
on the $LHS$ ?
Then you cancelled two different things from both sides.
Somehow you are confused between partial derivatives and total derivative.
answered 4 hours ago
Mohammad Riazi-Kermani
32k41853
You are dividing $$beginalign
delta z = fracpartial zpartial x * delta x + fracpartial zpartial y * delta y
endalign$$
by $ delta x$ to get $$ beginalign
fracpartial zpartial x = fracpartial zpartial x * 1 + fracpartial zpartial y * fracpartial ypartial x
endalign$$
How did you get $$fracpartial zpartial x$$
on the $LHS$ ?
Then you cancelled two different things from both sides.
Somehow you are confused between partial derivatives and total derivative.
answered 4 hours ago
Mohammad Riazi-Kermani
32k41853
answered 4 hours ago
Mohammad Riazi-Kermani
32k41853
answered 4 hours ago
Mohammad Riazi-Kermani
32k41853
answered 4 hours ago
Mohammad Riazi-Kermani
32k41853
32k41853
add a comment |Â
add a comment |Â
up vote
1
down vote
In the right-hand side you don't have $z=z(x,y)$, but the composition $tildez(x)= tildez(x,y(x))$. It is another different function. The chain rule gives $$fracrm dtildezrm dx(x) = fracpartial zpartial x(x,y(x))+fracpartial zpartial y(x,y(x))fracrm dyrm dx(x). $$You were a victim of the abuse of notation $tildezequiv z$. In fact, I am commiting one more abuse myself: using $y$ both for a variable of the function $z$ and to denote a function of $x$ (but I suppose you understand that no one wants to keep writing $tildez(x)=z(x,tildey(x))$, etc.).
answered 4 hours ago
Ivo Terek
44.1k949135
add a comment |Â
up vote
1
down vote
In the right-hand side you don't have $z=z(x,y)$, but the composition $tildez(x)= tildez(x,y(x))$. It is another different function. The chain rule gives $$fracrm dtildezrm dx(x) = fracpartial zpartial x(x,y(x))+fracpartial zpartial y(x,y(x))fracrm dyrm dx(x). $$You were a victim of the abuse of notation $tildezequiv z$. In fact, I am commiting one more abuse myself: using $y$ both for a variable of the function $z$ and to denote a function of $x$ (but I suppose you understand that no one wants to keep writing $tildez(x)=z(x,tildey(x))$, etc.).
answered 4 hours ago
Ivo Terek
44.1k949135
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In the right-hand side you don't have $z=z(x,y)$, but the composition $tildez(x)= tildez(x,y(x))$. It is another different function. The chain rule gives $$fracrm dtildezrm dx(x) = fracpartial zpartial x(x,y(x))+fracpartial zpartial y(x,y(x))fracrm dyrm dx(x). $$You were a victim of the abuse of notation $tildezequiv z$. In fact, I am commiting one more abuse myself: using $y$ both for a variable of the function $z$ and to denote a function of $x$ (but I suppose you understand that no one wants to keep writing $tildez(x)=z(x,tildey(x))$, etc.).
answered 4 hours ago
Ivo Terek
44.1k949135
In the right-hand side you don't have $z=z(x,y)$, but the composition $tildez(x)= tildez(x,y(x))$. It is another different function. The chain rule gives $$fracrm dtildezrm dx(x) = fracpartial zpartial x(x,y(x))+fracpartial zpartial y(x,y(x))fracrm dyrm dx(x). $$You were a victim of the abuse of notation $tildezequiv z$. In fact, I am commiting one more abuse myself: using $y$ both for a variable of the function $z$ and to denote a function of $x$ (but I suppose you understand that no one wants to keep writing $tildez(x)=z(x,tildey(x))$, etc.).
answered 4 hours ago
Ivo Terek
44.1k949135
answered 4 hours ago
Ivo Terek
44.1k949135
answered 4 hours ago
Ivo Terek
44.1k949135
answered 4 hours ago
Ivo Terek
44.1k949135
44.1k949135
add a comment |Â
add a comment |Â
Abhishek_04 is a new contributor. Be nice, and check out our Code of Conduct.
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If we were allowed to treat $partial x$ and $dx$ (which you note $delta x$) as interchangeable (and cancelable), then we'd have a more basic problem: in the original equation we could cancel $delta x$ and also $delta y$ and we'd get $delta z = 2 delta z$...
– leonbloy
4 hours ago