Can Quotient Rule Be Applied to Partial Derivatives?

[Battlemage!]

My question revolves around the following derivative:

for z(x,y)​

*sorry I can't seem to get the latex right.

∂/∂x (∂z/∂y)
​

What I thought about doing was using the quotient rule to see what I would get (as if these were regular differentials). So, I "factored out" the 1/∂x, then did the quotient rule with the "differential," then divided by ∂x.

∂/∂x (∂z/∂y) = 1/∂x (∂ (∂z/∂y))​

Doing the quotient rule with the bold:

1/∂x "(low d high - high d low )/low squared" ​

which gave:

1/∂x (∂y∂²z - ∂z∂²y)/(∂y²)
​

Now divide by ∂x:

(∂y∂²z - ∂z∂²y)/(∂x∂y²)​

Now, if I assume the bold above is somehow zero, suddenly I have the right answer:

(∂²z)/(∂x∂y)​

Now, I know this is probably horrid math(I can't emphasize this enough. Battlemage! ≠ crank), but if only that second term in the top of the fraction is zero then it works.

So, my question is, is there any legitimacy whatsoever to this?




Oddly, if I do it with this:

∂/∂x (∂z/∂x)

I get again the same result, with the second term in the top of the fraction = 0, then it's the right answer:

1/∂x ("(low d high - high d low)/low squared" )

1/∂x ((∂x ∂²z - ∂z∂²x)/(∂x²) )

((∂x ∂²z - ∂z∂²x)/(∂x³)

assume right term in numerator = 0

(∂x ∂²z)/(∂x³) = ∂²z/(∂x²)



Just what is going on here...

 

[Cyosis]

What I thought about doing was using the quotient rule to see what I would get (as if these were regular differentials). So, I "factored out" the 1/∂x, then did the quotient rule with the "differential," then divided by ∂x.

This would make no sense with regular differentials let alone partial derivatives.

 

[Battlemage!]

This would make no sense with regular differentials let alone partial derivatives.

So you mean that it is not the case that:

d(dz/dy) = (dyd²z - dzd²y)/(dy²)

Because obviously d(dz/dy) is just d²z/dy


but again even in this case if dzd²y = 0 then it's the correct answer (this is what I am wondering about)

(dyd²z - dzd²y)/(dy²)

(dyd²z - 0)/(dy²)

(dyd²z)/(dy²) = d²z/dy


And I guess my real question is, why is it that using the quotient rule for d(dz/dy) gives something that is almost the right answer? Just a coincidence?

 

[Cyosis]

Because obviously d(dz/dy) is just d²z/dy

This is gibberish, such an expression does not exist.

And I guess my real question is, why is it that using the quotient rule for d(dz/dy) gives something that is almost the right answer? Just a coincidence?

If you allow yourself to set things equal to zero or one or both at random without a proper argument then you can basically transform any expression into the 'correct' one.

 

[Battlemage!]

This is gibberish, such an expression does not exist.

So you have to keep the d/dx operator together, right?

If you allow yourself to set things equal to zero or one or both at random without a proper argument then you can basically transform any expression into the 'correct' one.

Ah, so coincidence. Thanks. I was just trying to feel my way through this. I've never really seen a good source explaining differentials to where I could understand it.

 

[Quisquis]

From what I understand, you can't treat the differentials in partial differentiation the same as you do a regular differential.