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In a book I saw this expression: $$dv = v(p + dt, t + dt) - v(p,t) = \frac{\partial v}{\partial t}dt + \frac{\partial p}{ \partial t} dp$$

I'm wondering what is the correct way to get this result.

I'm guessing I have to multiply $$v(p + dt, t + dt) - v(p,t)\quad \text{by}\quad \frac{dt}{dt} \frac{dp}{dp},$$ since $$\frac{dt}{dt} \frac{dp}{dp} = 1$$ that doesn't change the expression.

I know that $$\frac{v(p + dt, t + dt) - v(p,t)}{dt}$$ is the definition of a partial derivative, however I'm not sure how to calculate it correctly, especially the $+$ sign.

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    $\begingroup$ See Appendix: Theorems on Partial Differentiation, specifically the discussion starting after Eq. A.8. $\endgroup$ Commented Sep 20, 2022 at 15:13
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    $\begingroup$ This is also called the multivariable chain rule and is derived in various places under that name, including here. $\endgroup$ Commented Sep 20, 2022 at 17:17
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    $\begingroup$ Are you certain the expression is not a typographical error? The correct expression would be $dv = \frac{\partial v}{\partial t}dt + \frac{\partial v}{ \partial p} dp$. The one you mention in your first equation does not have correct dimensions – final term is $\sim p^2/t$ which cannot equal $\sim v$ $\endgroup$
    – RC_23
    Commented Sep 20, 2022 at 17:22
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    $\begingroup$ I actually just answered a very similar question physics.stackexchange.com/a/728371/313823 $\endgroup$
    – RC_23
    Commented Sep 20, 2022 at 17:24

1 Answer 1

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I believe the expression you're referring to is

$$dv = \left(\frac{\partial v}{\partial t}\right)_pdt + \left(\frac{\partial v}{ \partial p}\right)_t dp,$$

where I've correct some apparent typos and added subscripts to indicate variables that are kept constant during certain differentiation steps.

This is called the multivariable chain rule. It's also referred to as expanding a differential (over its variables).

I've linked a couple proofs in the comments; the strategy familiar to physicists is to generalize the 1D expression

$$dy(x)=\frac{dy(x)}{dx}dx,$$

which states that the rise of a curve over an infinitesimal run equals the slope times the run. Put another way, if we zoom in enough (i.e., consider infinitesimal displacements), any curve looks like a straight line. This is equivalent to saying that we'll ignore $(dx)^2$ terms and higher.

We now generalize to the multivariable case. For a function of two variables, for instance, one can imagine a curved sheet $z(x,y)$ (with orthogonal directions $x$ and $y$) that for infinitesimal displacements looks like a tilted plane. The rise in this case is

$$dz(x,y)=\left(\frac{dz(x,y)}{dx}\right)_ydx+\left(\frac{dz(x,y)}{dy}\right)_xdy;$$

again, any curvature (corresponding to $(dx)^2$, $(dy)^2$, and $dx\,dy$ terms) is negligible in the infinitesimal limit. One can continue the series with additional dimensions/relevant variables, although the geometric visualization becomes more challenging.

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