# For this Partial Derivative -- Why are different results obtained?

• Silvia2023
In summary: No, that's not correct. If you insist on writing ##F((1/x), y)## instead of doing the much more convenient change of variable you did already, then you will have$$F((1/x), y) = A \frac{1}{(1/x)} \frac{1}{(1/x)} y$$Applying the chain rule to this will work the same as the change of variable version you already... explained.
Silvia2023
Homework Statement
dF(x,y)/d(1/x)
Relevant Equations
dF(x,y)/d(1/x)
Given a function F(x,y)=A*x*x*y, calculate dF(x,y)/d(1/x), to calculate this derivative I make a change of variable, let u=1/x, then the function becomes F(u,y)=A*(1/u*u)*y, calculating the derivative with respect to u, we have dF/du=-2*A*y*(1/(u*u *u)) replacing we have dF/d(1/x)=-2*A*x*x*x*y
On the other hand, if we create the same original function, but as a function of (1/x), F(x,y)=A*(1/x)*(1/x)*x*x*(x*x) *y we derive with respect to (1/x) we have dF/d(1/x)= A *x*x*x*y Why are different results obtained?. Maybe I am wrong in something simple, but fundamental. Could you help me?
Best regards
Silvia

Silvia2023 said:
Homework Statement: dF(x,y)/d(1/x)
Relevant Equations: dF(x,y)/d(1/x)

Given a function F(x,y)=A*x*x*y, calculate dF(x,y)/d(1/x), to calculate this derivative I make a change of variable, let u=1/x, then the function becomes F(u,y)=A*(1/u*u)*y, calculating the derivative with respect to u, we have dF/du=-2*A*y*(1/(u*u *u)) replacing we have dF/d(1/x)=-2*A*x*x*x*y
On the other hand, if we create the same original function, but as a function of (1/x), F(x,y)=A*(1/x)*(1/x)*x*x*(x*x) *y we derive with respect to (1/x) we have dF/d(1/x)= A *x*x*x*y Why are different results obtained?. Maybe I am wrong in something simple, but fundamental. Could you help me?
Best regards
Silvia
Think about the chain rule. ##\frac{\partial F}{\partial x} = \frac{\partial F}{\partial u} \cdot ?##

Thanks for your answer; I'm thinking about the chain rule, but that doesn't answer my question about why the difference between these 2 ways, one with a change of variable and the other replacing the variable with respect to which I want to derive the function in the original function

I cannot figure out what you are trying to do in the second part.

berkeman
Silvia2023 said:
Given a function F(x,y)=A*x*x*y, calculate dF(x,y)/d(1/x), to calculate this derivative I make a change of variable, let u=1/x, then the function becomes F(u,y)=A*(1/u*u)*y, calculating the derivative with respect to u, we have dF/du=-2*A*y*(1/(u*u *u)) replacing we have dF/d(1/x)=-2*A*x*x*x*y
The function definition says to form the product of A, the square of the first argument, and the second argument. So wouldn't F(u, y) be ##Au^2y##?

In any case, what you want is this:
$$\frac{\partial F(x, y)}{\partial (1/x)}$$
Per the chain rule
$$\frac{\partial F(x, y)}{\partial (1/x)} = \frac{\partial F(x, y)}{\partial x} \cdot \frac{dx}{d(1/x)}$$
The 2nd fraction on the right side is the reciprocal of ##\frac d {dx}\left(1/x\right)##.
Using this logic I get exactly what you got.
Silvia2023 said:
On the other hand, if we create the same original function, but as a function of (1/x), F(x,y)=A*(1/x)*(1/x)*x*x*(x*x) *y we derive with respect to (1/x) we have dF/d(1/x)= A *x*x*x*y Why are different results obtained?.
I can't make sense of this, either.

PeroK
Mark44 said:
So wouldn't F(u, y) be Au2y?
No, the OP was correct. F(u,y)=Ay/u2

Frabjous said:
No, the OP was correct. F(u,y)=Ay/u2
I don't see why -- The function is given as ##F(x, y) = Ax^2y##, so ##F(3, y) = A(3)^2y## and ##F(u, y) = Au^2y##. You can't just willy-nilly change the role of the function parameters.

Mark44 said:
I don't see why -- The function is given as ##F(x, y) = Ax^2y##, so ##F(3, y) = A(3)^2y## and ##F(u, y) = Au^2y##. You can't just willy-nilly change the role of the function parameters.
You are overthinking it. The OP defined u=1/x.

Frabjous said:
You are overthinking it. The OP defined u=1/x.
So ##F(u, y) = Au^2y = A(1/x)^2y = F(1/x, y)##
Am I missing something?

SammyS
Mark44 said:
So ##F(u, y) = Au^2y = A(1/x)^2y = F(1/x, y)##
Am I missing something?
The OP is doing a change of variables, not a substitution.

Silvia2023 said:
if we create the same original function, but as a function of (1/x), F(x,y)=A*(1/x)*(1/x)*x*x*(x*x) *y
No, that's not correct. If you insist on writing ##F((1/x), y)## instead of doing the much more convenient change of variable you did already, then you will have

$$F((1/x), y) = A \frac{1}{(1/x)} \frac{1}{(1/x)} y$$

Applying the chain rule to this will work the same as the change of variable version you already did.

Frabjous
Frabjous said:
I cannot figure out what you are trying to do in the second part.
In the second part I am building the same original function, but with (1/x), if you look closely it is the same function as the original, the 2 elements of (1/x) are canceled with x*x and the original function remains, this is F(x,y)= A*((1/x)*(1/x)*x*x)*x*x*y , and then I differentiate with respect to (1/x)...............dF(x,y)d(1/x)=A*x*x*x*y

Silvia2023 said:
In the second part I am building the same original function, but with (1/x), if you look closely it is the same function as the original, the 2 elements of (1/x) are canceled with x*x and the original function remains, this is F(x,y)= A*((1/x)*(1/x)*x*x)*x*x*y , and then I differentiate with respect to (1/x)...............dF(x,y)d(1/x)=A*x*x*x*y
It is not obvious how you got that answer. See post 11 for the proper way to go along that path.

Silvia2023 said:
In the second part I am building the same original function, but with (1/x)
And you did so incorrectly, as I pointed out in post #11.

Silvia2023 said:
F(x,y)= A*((1/x)*(1/x)*x*x)*x*x*y
If you are going to treat this as a function of ##(1 / x)## and differentiate it, then you need to write it as:

$$F((1/x), y) = A (1 / x) (1 / x) \frac{1}{(1 / x)} \frac{1}{(1 / x)} \frac{1}{(1 / x)} \frac{1}{(1 / x)} y$$

(And of course the obvious thing to do here before differentiating would be to cancel out factors of ##(1 / x)## to get the simpler expression I gave in post #11.)

In other words, you cannot treat ##x## as a variable independent of ##(1 / x)##. To do things properly you have to either write ##F## entirely in terms of ##x##, or entirely in terms of ##(1 / x)##. You can't mix the two and pretend that ##F## is a function of ##x## and ##(1 / x)## separately.

Silvia2023
Frabjous said:
I cannot figure out what you are trying to do in the second part.
In other words, I give another example:
F(x,y)=B*x*x*y*y*y=B*(x*y)*(x*y)*y
dF(x,y)/d(xy)= B*x*y*y I handle the product of x by y as a single variable, it is the same as I want to do in the previous case, but with the factor (1/ x)

Silvia2023 said:
I give another example
Please take a look at post #11 (and #15) before you post again.

Frabjous
PeterDonis said:
If you are going to treat this as a function of ##(1 / x)## and differentiate it, then you need to write it as:

$$F((1/x), y) = A (1 / x) (1 / x) \frac{1}{(1 / x)} \frac{1}{(1 / x)} \frac{1}{(1 / x)} \frac{1}{(1 / x)} y$$

(And of course the obvious thing to do here before differentiating would be to cancel out factors of ##(1 / x)## to get the simpler expression I gave in post #11.)

In other words, you cannot treat ##x## as a variable independent of ##(1 / x)##. To do things properly you have to either write ##F## entirely in terms of ##x##, or entirely in terms of ##(1 / x)##. You can't mix the two and pretend that ##F## is a function of ##x## and ##(1 / x)## separately.
Ok, I understand what you mean, thanks

Post 11:
PeterDonis said:
$$F((1/x), y) = A \frac{1}{(1/x)} \frac{1}{(1/x)} y$$
This may seem obvious to you, but you lost me here. Since ##F(x, y) = Ax^2y## why wouldn't ##F(1/x, y)## be ##A(1/x)^2y## or ##A\frac y {x^2}##?

As I wrote before, with the substitution u = 1/x, ##F(u, y) = Au^2y = A(1/x)^2y## which equals ##A\frac y {x^2}## if we undo the substitution.

Again, is there something I'm missing here?

Same questions with regard to post 15:
PeterDonis said:
$$F((1/x), y) = A (1 / x) (1 / x) \frac{1}{(1 / x)} \frac{1}{(1 / x)} \frac{1}{(1 / x)} \frac{1}{(1 / x)} y$$

Mark44 said:
Again, is there something I'm missing here?
You're right. The others are using physicists' math and abusing notation.

SammyS and Mark44
Mark44 said:
This may seem obvious to you, but you lost me here
It's a simple change of variable, but using the notation ##(1 / x)## for the new variable instead of ##u## to make clear the functional relationship to the old variable.

Mark44 said:
As I wrote before, with the substitution u = 1/x, ##F(u, y) = Au^2y = A(1/x)^2y## which equals ##A\frac y {x^2}## if we undo the substitution.
Now you're the one that's lost me. Undoing the change of variable should recover the original form of the function. If ##F(x, y) = A x^2 y##, and ##u = 1/x##, then ##F(u, y) = A (1 / u^2) y##, not ##A u^2 y##. Then undoing the variable change means setting ##x = 1 / u## and going back from ##F(u, y) = A (1 / u^2) y## to ##F(x, y) = A x^2 y##.

Last edited:
Mark44 said:
Since ##F(x, y) = Ax^2y## why wouldn't ##F(1/x, y)## be ##A(1/x)^2y## or ##A\frac y {x^2}##?
Because, as I put it in post #15, ##(1 / x)## is not independent of ##x##, so you can't just plug ##(1 / x)## into the formula for ##F(x, y)## as if it were a separate, independent variable.

Mark44 said:
In any case, what you want is this:
$$\frac{\partial F(x, y)}{\partial (1/x)}$$
Per the chain rule
$$\frac{\partial F(x, y)}{\partial (1/x)} = \frac{\partial F(x, y)}{\partial x} \cdot \frac{dx}{d(1/x)}$$
The 2nd fraction on the right side is the reciprocal of ##\frac d {dx}\left(1/x\right)##.
Using this logic I get exactly what you got.
Note that here you are, correctly, treating ##(1 / x)## as a function of ##x##, not an independent variable.

Randomly bumped into this thread, but I think what Mark is trying to point out is basically, that if you have a function ##f(x)##, and you want to perform a substitution ##x = \tfrac{1}{u}##, what you get is not a function ##f(u)##, but rather a function ##g(u) = f(\tfrac{1}{u})##. Meaning that with ##f## we define a certain functional relation, which changes after substitution, and so if we just write ##f(u)##, we don't account for this change(or we do maybe, but we abuse the notation and so we gotta be aware of what we're doing).
This is effectively the same thing that's pointed out later in Peter's post, where he comments on ##x## and ##\tfrac{1}{x}## being dependent.

So in your scenario: ##F(x,y) = Ax^2y##, you perform the substitution, and you find ##F(\tfrac{1}{u},y) = A\tfrac{1}{u^2}y##, you have to substitute both inside ##F## and outside, not just on one side, because that's incorrect. If you write it this way, you see that you can't evade using chain rule, which you evaded in your first post and that was the mistake.

vela, Silvia2023, PeterDonis and 1 other person

## Why do different results occur when computing partial derivatives?

Different results can occur due to various reasons such as errors in the differentiation process, incorrect application of differentiation rules, or misinterpretation of the function's variables. Ensuring that each variable is treated correctly and that the function is properly understood is crucial for accurate results.

## How does the order of differentiation affect partial derivatives?

The order of differentiation can affect the result if the function is not continuous or does not have continuous partial derivatives. For functions that are well-behaved (i.e., continuous and with continuous partial derivatives), the mixed partial derivatives should be equal regardless of the order in which they are taken, according to Clairaut's theorem.

## Can different coordinate systems cause different partial derivative results?

Yes, different coordinate systems can lead to different forms of partial derivatives. When changing coordinate systems, the transformation rules must be applied correctly to ensure that the partial derivatives are computed accurately in the new system. Misapplication of these transformations can lead to discrepancies.

## Why do symbolic computation tools sometimes give different partial derivatives?

Symbolic computation tools may use different algorithms or simplification techniques, which can lead to different forms of the same partial derivative. Additionally, these tools might have different default assumptions about the variables and functions involved. Ensuring consistent settings and understanding the tool's behavior can help mitigate discrepancies.

## What role does the function's domain play in partial derivatives?

The domain of the function can significantly impact the partial derivatives. If the function is not defined or not differentiable at certain points within its domain, this can lead to different results or undefined derivatives. Properly understanding the domain and ensuring the function is differentiable within that domain is essential for accurate partial derivatives.

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