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"The derivative of a sum is the sum of derivatives"

Above theorem can be mathematically expressed as:

$$h'(x)=f'(x)+g'(x)$$

where $f(x)$ and $g(x)$ are two differentiable functions. What is the right way to express the statement of this theorem in Leibniz notation? Is it $$\frac{d}{dx}h=\frac{d}{dx}(f+g)=\frac{d}{dx}f+\frac{d}{dx}g$$

OR

$$\frac{d}{dx}h(x)=\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)?$$

In other words, is it permissible to write the derivative of functions in such a way so as to express the variable(s) on which they depend?

Samama Fahim
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5 Answers5

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Honestly, you should also write $$h'=f'+g',$$ since you take derivative of functions, and not of values of functions. Anyway, Leibniz' notation emphasizes the variable (you can't write $\frac{d}{dx} \sin t = \cos t$), so you might prefer to write $\frac{d}{dx}h(x) = \frac{d}{dx} f(x) + \frac{d}{dx}g(x)$. My favorite notation remains $Dh=Df+Dg$, which is rather common in infinite-dimensional settings: it stresses the functional nature of differentiation.

Siminore
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it's common to write the variable to express that it's the variable you are deriving according to. but anyway it's not that critical

Hanna Khalil
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I have only seen $\frac{d}{dx}h(x)=\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)$.

Even with the prime notation, you don't say $h'=f'+g'$, unless it is for shorthand for scrap work. If it is being graded, I would always specify that the function is some function of a variable- in this case $x$.

Sujaan Kunalan
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$f(x)$ is a value, not a function, so technically it's wrong to say that $f(x)$ is differentiable. To be precise you should say "$h'(x) = f'(x) + g'(x)$ for any $x \in D$ for any differentiable functions $f,g$ where $h(x) = f(x) + g(x)$ for any $x \in D$", but it is more concise and equally precise to say "$(f+g)' = f' + g'$ on D". Likewise the middle line with $\frac{d}{dx} f$ is probably the 'original' Leibniz notation, but it makes no sense if $f,g,h$ are strictly functions; the only way to get the correct semantic meaning is to interpret $f,g,h$ not as functions but as values that are determined by $x$. Historically $\frac{df}{dx}$ meant the change in $f$ divided by the change in $x$, with "d" standing for "Δ" / "δ" (delta). But if you want to interpret $f,g,h$ as true (1-variable) functions, then none of them are externally bound to any variables, so $x$ has no relation to $f,g,h$.

At the same time, Wikipedia says that Leibniz's notation is indeed in the form of the last line, but again we cannot interpret $f(x)$ within the $\frac{d}{dx}$ as the value of $f$ on input $x$, but rather have to interpret $\frac{d}{dx}(f(x))$ as something more like $\frac{d}{dx}($"f(x)"$)$ where $\frac{d}{dx}$ behaves like a procedure that looks at how the value described by the input string containing some occurrences of "$x$" depends on the value of $x$. This somewhat strange abuse of notation is quite common in many areas of mathematics, which you should be aware of, and when using such notation you should be clear about which expressions denote the objects themselves, which denote their descriptions, and which denote their state as bound to the state of other objects.

user21820
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There's a distinction between function and expression.

If $f$ and $g$ are differentiable functions $f,g : I \rightarrow \mathbb{R}$ where $I$ is an open set, then the correct notation is:

$$\frac{d}{dx}(f(x)+g(x)) = \frac{d}{dx}f(x) + \frac{d}{dx}g(x)$$

If, on the other hand, $f$ and $g$ are expressions of type $\mathbb{R}$ having a free variable $x$ bound to the set $I$, and if $f$ and $g$ are furthermore differentiable in $x$, then the correct notation is:

$$\frac{d}{dx}(f+g) = \frac{d}{dx}f + \frac{d}{dx}g$$

goblin GONE
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  • In the first interpretation, $\dfrac{\mathrm d}{\mathrm dx}$ is an operator which takes functions as inputs. As you said $f(x)+g(x)$ is not a function. Correct would be $\dfrac{\mathrm d}{\mathrm dx}(f+g)(x)$. – Git Gud May 22 '14 at 14:41
  • @GitGud, actually $f(x)+g(x)$ can be viewed as a function, namely a function $I^{{x}} \rightarrow \mathbb{R}$ where $x$ is a formal symbol. This explains why $\forall xPx$ is a well-formed expression in predicate calculus; $Px$ isn't being viewed as a truthvalue, but rather as a function of type $D^{{x}} \rightarrow {\mbox{True},\mbox{False}}$ where $D$ is the domain of discourse. – goblin GONE May 22 '14 at 14:46
  • But then $\color{red}I^{{x}}$ the codomain is wrong, since you're actually thinking about a string of symbols and not an interval. – Git Gud May 22 '14 at 14:50
  • @GitGud, I'm not sure I understand what you mean. To ensure that everything makes sense, we need to think of $I$ as a set, and $x$ as a formal symbol. Of course, to be truly rigorous, we should really tell the reader that this is how we're defining things. "Dear reader, $x$ is going to be used as a formal symbol for the duration of this proof." – goblin GONE May 22 '14 at 14:53
  • Perhaps I misinterpreted what you meant as $I$. At any rate this purely syntactic interpretation completely avoids the objects at hand, namely real differentiable functions of real variable. – Git Gud May 22 '14 at 14:55
  • @GitGud, the way I have tried to explain my point of view here in the comments was not intended to be syntactic. In particular, the following sets are in a clear bijection. 0) The set of all functions $I \rightarrow \mathbb{R}$ that are differentiable 1) The set of all functions $I^{{x}} \rightarrow \mathbb{R}$ that are differentiable in $x$. Frankly, I'm surprised you're not more... "convinced." It was only once I started looking at things from the vantage point I'm describing that previously "incomprehensible" notation became comprehensible to me. – goblin GONE May 22 '14 at 15:04
  • There is a bijection between them, but they are different. We're only interested in o) here. – Git Gud May 22 '14 at 15:16
  • @GitGud, but the point is that 1) does not have to be relegated to the metatheory. It can be viewed as a genuine set, within ZFC, say. – goblin GONE May 22 '14 at 15:17
  • OK. But what you end up with in 1) is different from what you end up with in o). I really don't see where you're getting at. Are you trying to say it is possible to make sense of the notation in a non-unreasonable way? – Git Gud May 22 '14 at 16:13
  • @GitGud, yes, that is all I'm trying to say! – goblin GONE May 23 '14 at 01:35