How do I differentiate a differential?

Question

Consider:

$$ dU = T dS - PdV$$

Now, how would I differentiate both sides with temperature...?

Or more simply:

$$ dy = f'(x) dx$$

Can I differentiate this expression above? (I'm not talking about moving dx to the denominator on left)

More particularly:

$$ (dy)' = (f'(x) dx)'$$

does the above expression have any meaning?

Answer 1

I'm not exactly sure of your question, but you can indeed differentiate a differential. It is a slightly different notation, in that the typical notation for a second derivative doesn't allow for this, but if you adopt a different notation for the second derivative, it works perfectly fine. The typical notation for the second derivative of $y$ with respect to $x$ is $\frac{d^2y}{dx^2}$. Using a notation that allows for taking second differentials, this becomes $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$.

The process of taking a differential of a differential is straightforward. The differential of a differential (say $dx$) is simply $d^2x$. In other words, $dx$ has the meaning $d(x)$ and $d^2x$ has the meaning $d(d(x))$. The thing to keep in mind is that if you have something like $x\,dx$, this has to be differentiated with the product rule. So $d(x\,dx) = x\,d^2(x) + dx^2$.

So, if you are wanting to differentiate your equation above, you would get:

$$ dU = T\,dS - P\,dV \\ d(dU) = d(T\,dS - P\,dV) \\ d^2U = d(T\,dS) - d(P\,dV) \\ d^2U = T\,d^2S + dT\,dS - P\,d^2V - dP\,dV $$

If you wanted that in terms of a second derivative, you would have to get it in the form of the second derivative I noted above.

You can find out more about using higher order differentials algebraically from my paper, "Extending the Algebraic Manipulability of Differentials."

Answer 2

If you have the basic knowledge about differential form and exterior differentiation, then $\mathrm{d}\mathrm{d}U$ do have its mathematical meaning. $$ \mathrm{d}\mathrm{d}U=\mathrm{d}T\wedge\mathrm{d}S-\mathrm{d}P\wedge\mathrm{d}V=(\dfrac{\partial T}{\partial S}\mathrm{d}S+\dfrac{\partial T}{\partial V}\mathrm{d}V)\wedge\mathrm{d}S-(\dfrac{\partial P}{\partial S}\mathrm{d}S+\dfrac{\partial P}{\partial V}\mathrm{d}V)\wedge\mathrm{d}V\\=(\dfrac{\partial T}{\partial V}+\dfrac{\partial P}{\partial S})\mathrm{d}V\wedge\mathrm{d}S $$ But I do not know the physical meaning of this.

In thermal physics, maybe the Legendre transformation is more common. $$ \mathrm{d}(U-TS)=-P\mathrm{d}V-S\mathrm{d}T $$ move $\mathrm{d}T$ to the left and we can differentiate again.