Partial Derivative and fraction

Question

In the George F Simmons Calculus with Analytical Geometry, textbook, it is mentioned that the Partial derivatives of a function cannot be treated as fractions as in case of single variable function.

In the example provided it has been proved that,

from
$pV = nRT$
where, - p - pressure
- T - Temperature
- V - Volume
- n - number of moles
- R - Universal gas constant

$\dfrac{\partial{p}}{\partial{T}} \dfrac{\partial{T}}{\partial{V}} \dfrac{\partial{V}}{\partial{p}} = -1$

which is fine, but in the explanation, it is given as, as the right hand side is -1 and not 1 we cannot treat the LHS of the above statement as fractions

If I am not wrong the 1st statement would mean, we cannot expand a partial differential as say, $\dfrac{d y}{d t} = \dfrac{dy}{dx}\dfrac{dx}{dt}$, where, $y=y(x)$ and $x = x(t)$,for a single variable function and we cannot represent the same for partial differentials the same manner if $y = y(x,t)$

What what is the significance of +1 or -1 for a differential to represented as a fraction?

Answer 1

Yes. $$\dfrac{\partial{x}}{\partial{y}} \dfrac{\partial{y}}{\partial{x}} \neq \dfrac{\partial{x}}{\partial{z}}$$ The equation above is an example to this phenomenon.

Answer 2

Although there's nothing wrong with the answer from mouthetics I would like to expand on something: ultimately what allows you to go from the lhs to rhs is the chain rule. For example, if $y = y(x, t)$ and $x = x(t)$ then

$$ \frac{{\rm d}y}{{\rm d} t} = \frac{\partial y}{\partial x}\frac{{\rm d}x}{{\rm d}t} + \frac{\partial y}{\partial t} $$

However, consider this other case. Imagine a function $f(x, y, z) = 0$. Now suppose you can invert this relation to obtain $z = z(x, y)$, if you apply the chain rule you will get

$$ {\rm d}z = \left(\frac{\partial z}{\partial x}\right)_{y}{\rm d}x + \left(\frac{\partial z}{\partial y}\right)_x{\rm d}y \tag{1} $$

where I have used the script to indicate that variable remains fixed. Suppose now that you move along a path on the surface $f(x,y,z) = 0$, where $z$ is kept constant. In this case you can also invert the function $f$ to obtain $y = y(x)$ and

$$ {\rm d}y = \left(\frac{\partial y}{\partial x}\right)_z{\rm d}x \tag{2} $$

Here it is more obvious why this notation is useful, it emphasizes the fact that we keep $z$ constant. Replace that in (1) with ${\rm d}z = 0$ (remember, $z$ is constant) and you will get

$$ 0 = \left(\frac{\partial z}{\partial x}\right)_{y}{\rm d}x + \left(\frac{\partial z}{\partial y}\right)_x \left(\frac{\partial y}{\partial x}\right)_z{\rm d}x \tag{3} $$

And from here is trivial to obtain

$$ \left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y = -1 \tag{4} $$

Note that this is a result of consistently using the chain rule, and keeping track what function depends on what variables.

In your case just set $x = P, y = V, z = T$, the function $f(x,y,z)$ is called a equation of state