Product rule OR Partial differentiation

[Jahnavi]

I have a very basic knowledge of calculus of one variable .

In the chapter on heat and thermodynamics , ideal gas law PV =nRT is given .

Then the book says, differentiating you get

PdV +VdP = nRdT .

The book doesn't explain the differentiation step .

I think , there are two ways to differentiate the gas law PV =nRT

1) Applying product rule of differentiation when a single variable is involved :

Assuming all the three P, V, T are functions of a common variable x , I can differentiate both sides of PV = nRT by x .

d[P(x)V(x)]/dx = d[nRT]dx

Applying product rule on left side I get ,

VdP/dx+PdV/dx = nRdT/dx

Eliminating dx from the denominator from both sides I get ,

VdP+PdV = nRdT

2) Taking total derivative of both sides ,

d(PV) = d(nRT)

[∂(PV)/∂P]dP + [∂(PV)/∂V]dV = [∂(nRT)/∂T]dT

This also gives PdV +VdP = nRdT

Both approaches give same result(equation) .

Is it the product rule that is applied or is it the total derivative (involving partial differentiation ) being applied here ?

Thank you

 

[fresh_42]

I have a very basic knowledge of calculus of one variable .
In the chapter on heat and thermodynamics , ideal gas law $PV =nRT$ is given .
Then the book says, differentiating you get
$$PdV +VdP = nRdT$$
The book doesn't explain the differentiation step .
I think , there are two ways to differentiate the gas law $PV =nRT$
1) Applying product rule of differentiation when a single variable is involved :
Assuming all the three $P, V, T$ are functions of a common variable $x$ , I can differentiate both sides of $PV = nRT$ by $x$ .
$$d[P(x)V(x)]/dx = d[nRT]dx$$
Applying product rule on left side I get ,
$$VdP/dx+PdV/dx = nRdT/dx$$
Eliminating $dx$ from the denominator from both sides I get ,
$$VdP+PdV = nRdT$$

2) Taking total derivative of both sides ,
$$d(PV) = d(nRT) \\
[∂(PV)/∂P]dP + [∂(PV)/∂V]dV = [∂(nRT)/∂T]dT$$
This also gives $PdV +VdP = nRdT$
Both approaches give same result(equation) .
Is it the product rule that is applied or is it the total derivative (involving partial differentiation ) being applied here ?

Thank you

These two approaches are basically the same. You applied the product rule in both cases.

The first case means the application of a differentiation and thus the product rule on functions in one variable $x$ as you've said. In the last step, eliminating $\frac{1}{dx}$ means, it is here where you passed from an equation of functions to an equation of differential forms.

The second case means to consider differential forms from the beginning and to write them as a linear combination of its coordinates, the partial derivatives. Here you apply the product rule on the coordinate forms.

 

[BvU]

Hello Jahnavi,

The answer is 'yes'

I think your path 2 is somewhat safer. (basically you derive the product rule there...)

 

[Jahnavi]

The second case means to consider differential forms from the beginning and to write them as a linear combination of its coordinates, the partial derivatives. Here you apply the product rule on the coordinate forms.

If possible , could you explain how the second approach (use of partial derivatives ) is same as the product rule of differentiation of a single variable .

 

[fresh_42]

The product rule is part of the definition of derivatives, so it is applied in both cases: first on $P(x)V(x)=nRT(x)$ as single variable functions and second in $\frac{\partial PV}{\partial P}\; , \;\frac{\partial PV}{\partial V}\, , \,\frac{\partial nRT}{\partial T}$ as functions in $P,V,T$.

The first case is an equation of dependencies of $x$:
$$
(P(x)\cdot V(x))' = P(x)' \cdot V(x) + P(x) \cdot V(x)'
$$
The second case is an equation of dependencies of functions $P,V,T$.
Differentiation is always basically Linearity plus Leibniz Rule per definition, resp. construction.
$$
d(P\cdot V) = d(P) \cdot V + P\cdot d(V)
$$
and with a function $F(P,V) := P\cdot V$ of functions, i.e. in a subspace spanned by the functions $P,V,T$. So we get in coordinates
$$
dF(P,V) = \dfrac{\partial F}{\partial P} dP + \dfrac{\partial F}{\partial V} dV = VdP +PdV
$$
The major difference is, that in the first case we more or less implicitly carry the point of evaluation with us, because it is actually
$$
\left. \dfrac{d}{dx}\right|_{x=x_0} (P(x)V(x)) = \left( \left. \dfrac{d}{dx}\right|_{x=x_0} P(x)\right) \cdot V(x_0) + P(x_0) \cdot \left( \left. \dfrac{d}{dx}\right|_{x=x_0} V(x)\right)
$$
what we have written, whereas in the second case the evaluation at $x=x_0$ means
$$
d_{x_0}(P\cdot V) = d_{x_0}(P) \cdot V(x_0) + P(x_0) \cdot d_{x_0}(V)
$$
It is simply a different point of view.