# How can I best improve my understanding on linear operator interactions: $df/dx$ vs $df(x)/dx$, and $df/dx$ vs $dy/dx$, etc?

For years now, I had always assumed that $$\frac{df}{dx}$$ was notational shorthand for $$\frac{df(x)}{dx}$$, because many teachers and authors of papers treat them interchangably. However upon getting studying deeper into vector/matrix calculus, and the general general theory of operators, I've realized that $$\frac{df}{dx} \ne \frac{df(x)}{dx}$$, likewise $$\frac{df}{dx} \ne \frac{dy}{dx}$$, and sometimes $$\frac{df(x)}{dx} \ne \frac{dy}{dx}$$. IE, the difference between $$df$$ and $$df(x)$$ is that the latter essentially involves a composition operator, which technically changes the logical/mathematical meaning of the expression. This is likewise further complicated by the fact that some authors use $$d$$ and $$\partial$$ interchangably, whereas others use them to indicate different properties of the differential.

While I able to observe a general pattern to the identities and operator interactions in terms of associative/commutative order of operations, like with integration by substitution and differentiation rules, they don't seem to be completely consistent, which means there's something I am not understanding.

TL;DR: I am basically trying to get a clear intuition on the interaction/relationship between the composition operator and differential operator, but cannot really find any sources that elucidate this - and such is further complicated by the fact that many authors seem to abuse notation using forms interchangably.

Note: I definitely don't know everything about operator theory, as there is still a lot I am learning, but I can't find many good learning resources about the topic.

• Use MathJax to improve readability Jun 23, 2020 at 0:09
• The operator $\frac{d}{dx}$ sends a function $f$ to another function - its derivative $\frac{df}{dx}$, which can be evaluated at the point $x$, which is what we mean by the term $\frac{df}{dx}(x)$ or $\frac{df(x)}{dx}$. Jun 23, 2020 at 0:33

Generally we don't distinguish much between $$\frac{\mathrm{d}f(x)}{\mathrm{d}x}$$ and $$\frac{\mathrm{d}f}{\mathrm{d}x}$$, but formally speaking we can view $$\frac{\mathrm{d}f(x)}{\mathrm{d}x}$$ as the value obtained when evaluating the derivative of $$f$$ at a point $$x$$, i.e $$\frac{\mathrm{d}f(x)}{\mathrm{d}x}=f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ Whereas we can view $$\frac{\mathrm{d}f}{\mathrm{d}x}$$ as the function defined as the derivative of $$f$$, i.e, $$\frac{\mathrm{d}f}{\mathrm{d}x}=f' = \mathcal{D}(f): x \mapsto \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ Here $$\mathcal{D}$$ denotes the differential operator.

• +1 for the suggestion that the best way to clear up this confusion is to think of differentiation as an operation on functions, ans write as much as possible without a variable "$x$". Jun 23, 2020 at 0:45
• This is absolutely correct. So many young mathematicians confuse exactly what is meant when we say $f$ and $f(x)$. Even though the value of $x$ is arbitrary, $f(x)$ is still a value, whereas $f$ is a list of rules that tell you how to map a value $x$ to its image, $f(x)$. And yes, we can formally view the derivative as a map $\mathcal{D}: \mathscr{D} \to \mathscr{I}$ from the set of differentiable functions to the set of integrable functions. Jun 23, 2020 at 0:50
• @K.defaoite Can you show an example? If I have $f:\mathbb R\to\mathbb R$, defined by $f(x)=ax^2+b$, we usually write $\frac{df(x)}{dx}=\frac{d(ax^2+b)}{dx} =a2x$. But if we consider $f(x)$ as the value, we have (pick $x=2$), $\frac{df(2)}{dx}=\frac{d(a4+b)}{dx} =0$?! But what is $\frac{df}{dx}$? Jun 23, 2020 at 14:31
• @JDoeDoe "defined by $f(x) = ax^2 +b$" is already technically wrong. It would be more correct to say $f: x \mapsto ax^2 + b$. It is true in this case that $\frac{\mathrm{d}f(x)}{\mathrm{d}x}=f'(x) = 2ax$ - this is the derivative of $f$ evaluated at a point $x$, but the function that is the derivative of $f$, which we call $\frac{\mathrm{d}f}{\mathrm{d}x}$ is the following mapping - $$\frac{\mathrm{d}f}{\mathrm{d}x}=f':x\mapsto 2ax.$$ Jun 23, 2020 at 15:08
• Perhaps instead of $\frac{\mathrm{d}f(x)}{\mathrm{d}x}$ it is clearer to write $\frac{\mathrm{d}f}{\mathrm{d}x}(x)$, though this is a bit cumbersome. Jun 23, 2020 at 15:13

There is no unique standard for the notation employed in calculus, let alone operator theory. Different authors use different notation to refer to the same operations. This is partly due to the historical debate regarding the "correct" notation for calculus (look up the Analytical Society) - which resulted in different notations for differentiation. There is also regional variation in commonly used notation by different education systems. As maddening as it is, the only guaranteed way to know what the notation means is from context.

Still, there are some generally accepted notations which are commonly used for specific operators. The ones that immediately come to mind are $$\mathbf{J}$$ for the Jacobian matrix, $$\mathbf{H}$$ for the Hessian matrix, $$\nabla$$ for the gradient, and $$\nabla\times$$ and $$\nabla\cdot$$ for curl and divergence, respectively.

Additionally, $$\partial$$ is most commonly used to indicate the partial derivative of a multivariable function - though it can be used to denote the derivative of a function of a single variable - while $$d$$ is typically used for the derivative of a function of a single variable, or the total derivative of a function of multiple variables.

As others have stated, the inclusion or exclusion of a variable (e.g. $$\frac{df}{dx}$$ vs $$\frac{df(x)}{dx}$$), if significant, is used to indicate whether the derivative of a function (which is another function) or the value of the derivative at a point is being considered.

Lastly, it is common in vector calculus to treat differentiation operators as scalars and vectors of a sort. This primarily a matter of convenience, as it allow, for instance, concise definitions like...

$$\text{div}\ \mathbf{f} = \nabla\cdot \mathbf{f}$$

...to replace larger expression like...

$$\text{div}\ \mathbf{f} = \sum_{i=1}^n \frac{\partial f_i}{\partial x_i}$$

• As vectors of a sort is what I think you meant to say? Jun 23, 2020 at 2:23
• Sure, but $\frac{\partial}{\partial x}$ is also a differentiation operator, so... Jun 23, 2020 at 2:37