# Question about partial derivatives (it's probably based on flawed reasoning, but )

1. May 25, 2010

### 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...

2. May 25, 2010

### Cyosis

Re: Question about partial derivatives (it's probably based on flawed reasoning, but.

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

3. May 25, 2010

### Battlemage!

Re: Question about partial derivatives (it's probably based on flawed reasoning, but.

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?

4. May 25, 2010

### Cyosis

Re: Question about partial derivatives (it's probably based on flawed reasoning, but.

This is gibberish, such an expression does not exist.

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.

5. May 25, 2010

### Battlemage!

Re: Question about partial derivatives (it's probably based on flawed reasoning, but.

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

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.

6. Jun 1, 2010

### Quisquis

Re: Question about partial derivatives (it's probably based on flawed reasoning, but.

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