I define best as the simplest most consistent notation.

Question: What is the best derivative notation?

Summary of what others have done:

Let $f$ be a function that maps any $(x,y)$ to $f(x,y)$. I have seen the following with respect to the partial derivative of $f$ with respect to $x$:

$$ \frac{\delta}{\delta x} f = \frac{\delta f}{\delta x} = \frac{\delta f(x,y)}{\delta x} = \frac{\delta f}{\delta x} (x,y) = f'_x = f^{(1)}_x $$

And the following for their 2nd derivative: $$ \frac{\delta^2}{\delta x^2} f = \frac{\delta^2 f}{\delta x^2} = \frac{\delta^2 f(x,y)}{\delta x^2} = \frac{\delta^2 f}{\delta x^2} (x,y) = f''_x = f^{(2)}_x $$

Then, if we wish to use such derivative functions later to find their value at some point $(x,y)$, we use: $$ \left(\frac{\delta}{\delta x} f\right)(x,y) = \left(\frac{\delta f}{\delta x}\right)(x,y) = \left(\frac{\delta f(x,y)}{\delta x}\right)(x,y) = \left(\frac{\delta f}{\delta x} (x)\right)(x,y) = f'_x(x,y) = f^{(1)}_x(x,y) $$

But now there is a conflict between $x$ in the name of the derivative functions, and $x$ in the point $(x,y)$. To resolve this, we need to use some other name for the input point $(x,y)$, such as $(a,b)$ as follows $$ \left(\frac{\delta}{\delta x} f\right)(a,b) = \left(\frac{\delta f}{\delta x}\right)(a,b) = \left(\frac{\delta f(x,y)}{\delta x}\right)(a,b) = \left(\frac{\delta f}{\delta x} (x)\right)(a,b) = f'_x(a,b) = f^{(1)}_x(a,b) $$

What I have done:

I think the best notation for the partial derivative function is $$ f^{(n)}_m $$ where $n$ denotes the $n^{th}$ derivative, and $m$ denotes that the derivative is with respect to the $m^{th}$ argument of function $f$.

This is because of the fact that function arguments are addressed based on their order, not based on their names. E.g. $f(\pi, e)$ means that $\pi$ is fed as the first argument, and $e$ is fed as the 2nd argument.

More specifically my equivalent notation to $f''_x$ is: $$ f^{(2)}_1 $$ since $x$ is the 1st argument.

This way, we could use $(x,y)$ as the input point without causing the confusion earlier: $$ f^{(2)}_1(x,y) $$

Another thought that I have is the following: $$ f^{\nabla 2}_1 $$ which also relates to the gradient $\nabla f$, or the second gradient $\nabla^2 f$ (gradient of gradient) as follows: $$ \nabla^2 f = f^{\nabla 2} = f^{\nabla 2}_1 \mathbf{\hat i} + f^{\nabla 2}_2 \mathbf{\hat j} $$

Update 1: After thinking a bit more on mixed partial derivatives (thanks to the discussion with @axiom_of_choice under his answer), I found this idea to handle 1) mixed partial derivatives, as well as the case when $f$ takes highly dimensional input, e.g. $f(x_1, x_2, \ldots, x_{100})$: $$ f'_{2,1}(x_1, x_2, \ldots, x_{100}) $$ to denote the value that the mixed partial derivative maps against when given input $(x_1, x_2, \ldots, x_{100})$. This is traditionally probably denoted by $$ \frac{\partial^2}{\partial x_2 \partial x_1}f(x_{1,0}, x_{2,0}, \ldots, x_{100,0}) $$

More generally, I could say: $$ f'_{2^2,1^5} $$ to denote the $2+5=7^{th}$ mixed partial derivative, which is traditionally denoted by $$ \frac{\partial^7}{\partial x_2^2 \partial x_1^5}f $$

I also think using $f^{\nabla}$ instead of $\nabla f$, to denote the gradient, is better to avoid the confusion $\nabla f(x_1,\ldots)$ against $\nabla \big(f(x_1,\ldots)\big)$: $$ \nabla \left(\frac{\partial^7}{\partial x_2^2 \partial x_1^5}f\right) = f^{\nabla}_{2^2,1^5} = f'_{1,2^2,1^5} \mathbf{\hat i_1} + f'_{2^3,1^5} \mathbf{\hat i_2} + \ldots + f'_{100,2^2,1^5} \mathbf{\hat i_{100}} $$ where $\mathbf{\hat i_1}, \mathbf{\hat i_2}, \ldots, \mathbf{\hat i_{100}}$ are perpendicular unit vectors that make up the axis of a $100$-dimensional cartesian coordinate system.

If one really wants to say $\nabla \big(f(x_1,\ldots)\big)$, then I think this to be used: $f(x_1,\ldots)^{\nabla}$


1 Answer 1


You should use $\frac{\partial}{\partial x^i}$ since it easily allows the notation of mixed partial derivatives. (By the way the partial symbol is done by "\partial").

  • $\begingroup$ Thank you. Could you please elaborate on what you mean by mixed partial derivatives? $\endgroup$
    – caveman
    Sep 9, 2017 at 22:55
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    $\begingroup$ For example take $f(x,y)=x\cdot y$, then $\frac{\partial}{\partial x}f(x,y)=y$ and $\frac{\partial^2 }{\partial x^2}f(x,y)=0$ i.e. if you see $\frac{\partial}{\partial x}f(x,y)$ as a function of $(x,y)$ and do another partial derivative in the $x$ direction it is 0. But if you do $\frac{\partial^2}{\partial y\partial x}f(x,y)=\frac{\partial}{\partial y}(\frac{\partial}{\partial x}f(x,y))=\frac{\partial}{\partial y}y=1$. You call $\frac{\partial^2}{\partial y\partial x}f(x,y)$ a mixed partial derivative since you first derive in one direction and then in another. $\endgroup$
    – F.R.
    Sep 9, 2017 at 23:04
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    $\begingroup$ @caveman, the y partial of the x partial, for instance $\endgroup$
    – Mark S.
    Sep 9, 2017 at 23:06
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    $\begingroup$ You're right, this notation might be confusing but it is common. So actually if you want to have the first partial derivative at $(x_0,y_0)$ you write $\frac{\partial}{\partial x}f(x_0,y_0)$. The first partial derivative as a function would then map $(x_0,y_0)\mapsto\frac{\partial}{\partial x}f(x_0,y_0)$. If you really want to use your other notation of course you can do so as long as you state the meaning at the beginning, but i haven't seen them in any mathematical text i have read (if you decide to use them i would use $f^{(2)}_{2,1}$ since it seems more natural to me). $\endgroup$
    – F.R.
    Sep 9, 2017 at 23:28
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    $\begingroup$ You're right, though you should start counting at 1. Then you can also write something like $\frac{\partial}{\partial x_1}f(x)$. $\endgroup$
    – F.R.
    Sep 10, 2017 at 9:10

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