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How do you derive a cross factor to decouple differentials into independent differentials? For example: $$ d(PV)= PdV+VdP $$ $$ PV=\int{PdV}+\int{VdP} $$

Obviously dP and dV are related. Do you simply write dP in terms of dV? What other methods are there where you can leave the differential as is but with a multiplicative cross factor term?

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  • $\begingroup$ This mse post shows under what condition $dP$ can be written in terms of $dV\,.$ It is a pure math question. Regarding the physics: The pressure $P$ is a function of volume $V$ only under certain conditions. $\endgroup$
    – Kurt G.
    Commented Sep 29, 2023 at 1:05

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The formula $d(pV)=pdV+Vdp$ is just Leibniz's rule on the differential quotient of the product of two function, $(fg)'=gf'+fg'$, written in the language of infinitesimals (differentials).

For one variable functions, $\frac{d(f(x)g(x))}{dx}=g(x)\frac{df(x)}{dx}+f(x)\frac{dg(x)}{dx}$.

For a function of several variables, $x_1, x_2, ..x_n$, along a curve $x_1(t), x_2(t),...x_n(t)$, the total derivative of $f^*(t)=f(x_1(t),x_2(t),..,x_n(t))$ is defined by $$\frac{df^*(t)}{dt} = \frac{\partial f}{\partial x_1}\frac{\partial x_1}{dt}+\frac{\partial f}{\partial x_2}\frac{dx_2}{dt}+...+\frac{\partial f}{\partial x_n}\frac{dx_n}{dt},$$ and again the identity holds for $\frac{d(f^*(t)g^*(t))}{dt}=g^*(t)\frac{df^*(t)}{dt}+f^*(t)\frac{dg^*(t)}{dt}$.

Now "simplify" your notation and with Leibniz (or Euler) drop $dt$ without losing any sleep over it and write $${d(f^*(t)g^*(t))}=g^*(t){df^*(t)}+f^*(t){dg^*(t)}$$

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