My understanding of partial $\left( \frac{\partial}{\partial} \right)$ and total $\left( \frac{d}{d} \right)$ differentiation/derivative is that assuming $f(x_1, x_2, ...,x_n )$ where $x_i$s are not necessarily independent:

$$\frac{d f}{ dx_i}=\sum^n_1 \left(\frac{\partial f}{\partial x_j}\frac{d x_j}{dx_i} \right)$$

Where $\frac{\partial f}{\partial x_i}$ is the symbolic derivative of the equation $f(x_1, x_2, ...,x_n )$ assuming all $x_j$s except $x_i$ are constants. Of course when $x_i$s are independent:

$$\frac{\partial f}{\partial x_i}=\frac{d f}{ dx_i}$$

But in thermodynamics I see that they have this exact differential

$$\left(\frac{\partial f}{\partial x_i} \right)_{x_j}$$

which to me looks exactly the same as partial differential. For example see these videos of thermodynamic lectures from MIT. I find this concept/notation redundant and confusing. I would appreciate if you could explain the difference between partial and exact differentials and give me a tangible example when they are not the same.

P.S.1. This post also approves my point:

In fact, the constancy of the other variables is implicit in the partial differential notation (∂/∂x) but it is customary to write the variables that are constant under the derivative when discussing thermodynamics, just to keep track of what other variables we were considering in that particular case.

Which if true, is an awful idea. Partial differential equations are already long and confusing enough without these redundant notations. Why on earth should we make it even more difficult?


2 Answers 2


If you consider $f$ simply as a function of $n$ variables, there is no such definition of "exact" differentiation. The fact is that, sometimes, the variables depend on each other on "external" variables. Let me clarify with an example. Consider a functions $f:\mathbb{R}^2\to\mathbb{R}$ and $u:\mathbb{R}\to \mathbb{R}$. $f$ has two partial derivatives $f_x$ and $f_y$. Suppose that we have a composition $$ \Big( f\circ (\cdot,u(\cdot))\Big)(z) = f(z,u(z)). $$ We can still talk about partial derivatives for $f$, and these are $$ f_x(z,u(z)) \quad \text{and} \quad f_y(z,u(z)). $$ We talk about "total derivative" for $f$ when we differentiate with respect to the parameter to which all the variables depend, namely $z$. This is $$ (f(z,u(z)))' = f_x(z,u(z))+f_y(z,u(z))u'(z). $$ A nice example where both concepts come in action is Euler-Lagrange equation.

The connection with your linked page (differentiable forms) is the following. Suppose you have a function $F:\mathbb{R}^2\to\mathbb{R}^2$. For all $(x,y)\in \mathbb{R}^2$, $F$ is a form of the dual of $\mathbb{R}^2$. The theory developed to understand whether this is also a gradient of a function $f:\mathbb{R}^2\to \mathbb{R}$ requires you to perform integration over curves on the space $\mathbb{R}^2$. Curves are functions $(x(t),y(t))\in \mathbb{R}^2$, $t\in [t_0,t_1]$. If the potential $f$ exists, it's partial derivatives are $$ f_x = F_1 \quad f_y = F_2. $$ On the other side, the derivative of $f$ along a curve $(x(t),y(t))$ is $$ f(x(t),y(t))' = f_x(x(t),y(t))x'(t)+f_y(x(t),y(t))y'(t) = F(x(t),y(t))\cdot (x'(t),y'(t)). $$ Therefore you can define the total derivative of $f$ to be $$ df = f_xdx + f_ydy, $$ where $dx$ and $dy$ must be understood as "small" variation of space variables "in time".

  • $\begingroup$ Thanks for the reply. But I do not understand how this answers my question "the difference between exact and partial differential". I already know what total differential is and it is not the same thing as exact differential. Please see this wikipedia page. $\endgroup$
    – Foad
    Mar 1, 2018 at 13:37
  • $\begingroup$ Well, you were asking for the difference, and in the first part, I showed you that they are not the same. Your reference links to differential forms and I talked about that in the second part. Note that in physics and engineering there is some confusion between total and partial, because sometimes they introduce the concept of time and go from one to the other "carelessly". $\endgroup$ Mar 1, 2018 at 13:46
  • $\begingroup$ I'm not sure but in the first part you are explaining the difference between total and partial differential. I have no issues or confusions between these two. The "exact" differential is the one I do not understand. If I may clarify the question is not "the difference between total and partial differential" but "the difference between exact and partial differential". $\endgroup$
    – Foad
    Mar 1, 2018 at 13:51
  • $\begingroup$ see this video for example. $\endgroup$
    – Foad
    Mar 1, 2018 at 13:54
  • 2
    $\begingroup$ I think you are confusing things. A form is exact, or it is an exact differential, when it is total derivative of a given function. Therefore, there is not such thing as exact derivate in mathematics. Maybe your question fits more a forum on physics! $\endgroup$ Mar 1, 2018 at 14:00

Derivatives and differentials (total and partial)

If $f$ is a function of one variable with $y=f(x)$:

  • derivative of $f$ (or $y$) at $x$:

$$\frac{dy}{dx}=\frac{df}{dx}=f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$

If $f$ is a function of more than one variable with $z=f(x,y)$:

  • partial derivatives of $f$ (or $z$) at $(x,y)$:

$$\frac{\partial z}{\partial x}=\frac{\partial f}{\partial x}=f'_x(x,y)=\lim_{h\rightarrow 0}\frac{f(x+h,y)-f(x,y)}{h}$$

$$\frac{\partial z}{\partial y}=\frac{\partial f}{\partial y}=f'_y(x,y)=\lim_{h\rightarrow 0}\frac{f(x,y+h)-f(x,y)}{h}$$

  • total (exact) differential of $f$ (or $z$):

    $$dz=df(x,y,dx,dy)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$$

    where $x$, $y$, $dx$ and $dy$ are independent variables, $df$ or $dz$ are dependent variables, and $dx$, $dy$, $dz$ and $df$ are infinitesimals. (For an introduction to the rigorous use of infinitesimals in calculus, see the book by H. Jerome Keisler: Elementary Calculus: An Infinitesimal Approach.)

  • inexact differential:

    When $f$ (or $z$) is a quantity defined by a relation of the type:


    we are not sure that the expression is the total (exact) differential of a function $f(x,y)$. It is only the case if one can identify $M(x,y)$ with $\partial f/\partial x$ and $N(x,y)$ with $\partial f/\partial y$, which is only ensured if:

    $$\frac{\partial^2 f}{\partial y \partial x}=\frac{\partial^2 f}{\partial x \partial y}$$

    that is, if:

    $$\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}$$

    If $M(x,y)$ and $N(x,y)$ do not verify this relation, the quantity

    $$\Delta z=\Delta f(x,y,x_0,y_0)=\int_{(x_0,y_0)}^{(x,y)}df$$

    depends on the path taken for the integration to go from $(x_0,y_0)$ to $(x,y)$. In thermodynamics, it is said that $z$ or $f$ are not state functions but instead process functions, or process quantities or path functions.

    Example of total (exact) differential: $$dz=(9x^2+6xy+y^2)dx+(3x^2+2xy)dy$$

    $dz$ is the total (exact) differential of a function of the form $$f(x,y)=3x^3+3yx^2+xy^2+c$$ because it verifies: $$\begin{align}\frac{\partial f}{\partial x}(x, y) & = (9x^2+6xy+y^2) \\ \\ \frac{\partial f}{\partial y}(x, y) & =(3x^2+2xy)\end{align}$$

    Example of inexact differential: $$dz=(2x+y)dx+(x+y)dy$$

    There is no function $f(x,y)$ of the variables $x$ and $y$ such that: $$\begin{align}\frac{\partial f}{\partial x}(x, y) & = (2x+y) \\ \\ \frac{\partial f}{\partial y}(x, y) & = (x+y)\end{align}$$

  • partial (inexact) differentials of $f$ (or $z$):

    Here, I am not sure if the term "partial differential" is really used that way (see below for a different use in a different context), but in the context of a relation $y=f(x_1,x_2,x_3)$, by analogy with the partial derivatives, one could use the term "partial differential" to designate either of the forms (see for example this page):

    $$d_{x_1} f=\frac{\partial f}{\partial x_1}dx_1$$

    $$d_{x_2} f=\frac{\partial f}{\partial x_2}dx_2$$

    $$d_{x_3} f=\frac{\partial f}{\partial x_3}dx_3$$

    (One could possibly extend the term to any incomplete sum of such components:

    $$d_{x_1 x_2}f=\frac{\partial f}{\partial x_1}dx_1 + \frac{\partial f}{\partial x_2}dx_2$$

    $$d_{x_1 x_3}f=\frac{\partial f}{\partial x_1}dx_1 + \frac{\partial f}{\partial x_3}dx_3$$

    $$d_{x_2 x_3}f=\frac{\partial f}{\partial x_2}dx_2 + \frac{\partial f}{\partial x_3}dx_3$$

    but I have no reference for that.)

If $y$ can be expressed as a composed function $y=f(x(t))=(f\circ g)(t)=h(t)$:

(Here is where the language gets a bit tricky...)

  • total derivative of $y$:

    $$\frac{dy}{dt}=\frac{dh}{dt}=h'(t)=\frac{d(f\circ g)}{dt}=(f\circ g)'(t)=g'(t)\cdot (f'\circ g)(t)$$

    Note that I did not write "total derivative of the function $f$" but "the total derivative of $y$" (the depedent variable). This is on purpose, because here "total derivative" means that we take $y$ as a function of the ultimate variable $t$ taken as the independent variable—that is, the function considered is not $f$ but really $(f\circ g)=h$.

  • partial derivative of $y$:

    $$\frac{\partial y}{\partial x}=\frac{\partial f(x(t)) }{\partial x}=\frac{df}{dx}=f'(x)=f'(x(t))=(f'\circ g)(t)$$

    Note that here we take $x$ as the independent variable, ignoring any dependence of $x$ on $t$ for the computation of the derivative. So in this very specific context, $\partial f/\partial x=df/dx$, and the function considered is not $h = (f\circ g)$ but $f$.

The above way of defining "total" and "partial" might seem strange in the context of a relation $y=f(x(t))=(f\circ g)(t)$, but it makes more sens in the context of a relations of the type:

$$z = f(t,x(t),y(t))=(f\circ g)(t) = h(t).$$

In such a case, $f$ is a single-valued function of three variables, $f(t,x,y)$, whereas $g$ is a three-valued function of one variable, $g(t) = (g_1(t),g_2(t),g_3(t))$.

For example, if you have a function $f(x_1,x_2,\ldots,x_n)$ and then wish to take into account some dependencies between some of the variables $x_1$, $x_2$, ..., $x_n$, for example $x_n=u(x_1, x_2,\ldots,x_{n-1})$, then you are really considering a new function: $$h(x_1, x_2,\ldots, x_{n-1}) = f(x_1,x_2,\ldots,x_{n-1},u(x_1,x_2,\ldots,x_{n-1})).$$

  • total derivative of $z$:

    In this new set-up:

    $$\begin{align}\frac{dz}{dt} & = \frac{dh}{dt} = h'(t)\\ & = \frac{d(f\circ g)}{dt} = (f\circ g)'(t)\\ & = \frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}\\ & = \frac{\partial f}{\partial t}\frac{dg_1}{dt}+\frac{\partial f}{\partial x}\frac{dg_2}{dt}+\frac{\partial f}{\partial y}\frac{dg_3}{dt}\\ & = g'(t)\cdot (f'\circ g)(t)\end{align}$$

  • partial derivatives of $z$:

    $$\frac{\partial z}{\partial t} = \frac{\partial f}{\partial t} = \lim_{h\rightarrow 0}\frac{f(t+h,x(t),y(t))-f(t,x(t),y(t))}{h}$$

    $$\frac{\partial z}{\partial x} = \frac{\partial f}{\partial x} = \lim_{h\rightarrow 0}\frac{f(t,x(t) + h,y(t))-f(t,x(t),y(t))}{h}$$

    $$\frac{\partial z}{\partial y} = \frac{\partial f}{\partial y} = \lim_{h\rightarrow 0}\frac{f(t,x(t),y(t)+h)-f(t,x(t),y(t))}{h}$$

  • total (exact) differential of $z$:

$$dz = dh(t,dt) = d(f\circ g)(t, dt) = \frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial x}\frac{dx}{dt}dt + \frac{\partial f}{\partial y}\frac{dy}{dt}dt$$

  • partial (exact) differential of $z$:

$$dz = df(t,x,y,dt,dx,dy) = \frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$$

So overall, in the context of a composed function, it is important to understand the various levels of dependence of the variables through the composition to make sense of what is meant by "total" or "partial" derivatives or differentials.


In thermodynamics, because at equilibrium there exist relations between different thermodynamic variables (for example $PV=nRT$ for a perfect gas), quantities such as the internal energy $U$ or the entropy $S$ can be expressed as functions of different sets of independent thermodynamic variables:

  • $U(S,V,n)$
    $U(T, P, n)$ ...

and similarly:

  • $S(U,V,n)$
    $S(T,P,n)$ ...

Because of these alternative choices, it is important to indicate the variables that are kept constant in the partial derivatives as a reminder of the set of independent variables actually considered.


$$\begin{align}C_V & = T\left(\frac{\partial S}{\partial T}\right)_V \\ \\ C_P & = T\left(\frac{\partial S}{\partial T}\right)_P\end{align}$$

In the first case, $S$ is expressed as a function of $T$ and $V$. In the second case it is expressed as a function of $T$ and $P$.

As for $$\left(\frac{\partial f}{\partial x_i}\right)_{x_j}$$ it is not an exact differential but a partial derivative.

  • $\begingroup$ I am sorry but your definition of partial derivative is not correct. As far as I understand partial differentiation is just a symbolic derivative and nothing more. you can't define it with the limit and the h, etc. $\endgroup$
    – Foad
    Jul 20, 2023 at 9:05
  • $\begingroup$ also the term "process functions" seems made up to me. By definition, a function is a map that relates each element of the domain with exactly one element in the range. If the value of something depends on the path, then it is not a function of the domain variables. $\endgroup$
    – Foad
    Jul 20, 2023 at 9:15
  • $\begingroup$ @Foad Partial derivative is the derivative according to one variable, the other variables staying fixed. I don't see how the definition is not correct here. $\endgroup$
    – The Quark
    Jul 20, 2023 at 10:07
  • $\begingroup$ @Foad Exactly, process functions are functions of... processes (that is of paths $(x(s),y(s),...)=(u_1(s),u_2(s),...)$ where $s$ is some parametric variable and $u_1$, $u_2$,... are some functions that define a path), so they are not just functions of values (here initial and final values) of the $x$, $y$,... variables that are used to define the (then inexact) differential. That is to say, the differential expression is not the (total) differential of a function of $x$, $y$,... as it does not define a proper function of the variables $x$, $y$,... $\endgroup$
    – The Quark
    Jul 20, 2023 at 10:43
  • $\begingroup$ the partial derivative is a shallow symbolic/analytic differentiation. It doesn't have a physical meaning. If your domain variables are not fully independent then the constancy assumption is tautological. For example, if you have z = f(x, y) while y = g(x) then how do you calculate ∂z/∂y assuming x is constant? It will be zero anyway. $\endgroup$
    – Foad
    Jul 20, 2023 at 10:44

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