After reading many answers on the subject I feel like I am close to finally understanding why the derivative is a linear map. I think that if someone helps me understand the following example I might "get it"

So I have $f(x)= e^x$ then $f(x)' = e^x$ is not linear, but instead the function defined by $y \rightarrow e^x y$ is linear, but what is $y$? is it introduced for the sole purpose to have a linear map? Say that I fix x, then for different values of $y$ what am I computing?

So I should not call $f(x)'$ the derivative any more if the derivative is a linear map, then what should I call $f(x)'$?

  • 2
    $\begingroup$ No, I think in this case the "linear" refers not to the actual function identified as the derivative, but to the fact that (a) $\frac{d}{dx} 0 = 0$; (b) $\frac{d}{dx} a \, f(x) = a \, \frac{d}{dx} f(x)$; and (c) $\frac{d}{dx} [f(x)+g(x)] = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)$. $\endgroup$ – Brian Tung Feb 11 '16 at 21:35
  • 1
    $\begingroup$ @Monolite: what you are looking for is clearly spelled in this answer. $\endgroup$ – Martin Argerami Feb 12 '16 at 11:36

Think of the derivative (more properly the differential) as a rule that assigns a linear map to each point of the domain. This linear map approximates the change in the function’s value for a small displacement from that point.

In your example, you have the rule $f'(x)=e^x$. This rule assigns the map “multiply by $e^{x_0}$” to the point $x_0$. For any given value of $x_0$, this is multiplication by a constant, which is obviously linear. This linear map can then be used to approximate $f(x_0+h)$ by $f(x_0)+e^{x_0}h$. (I used $h$ instead of $y$ to avoid confusing a displacement from a point in the domain with a value in the function’s range.)

  • $\begingroup$ Thank you, I could not decide who to give the check-mark to, I feel like your answer was very important to my understanding also. $\endgroup$ – Monolite Feb 11 '16 at 22:19
  • 2
    $\begingroup$ I think this is the only appropriate answer. It would be more complete if it included this explanation. $\endgroup$ – Martin Argerami Feb 12 '16 at 5:56

You are confusing the derivative of a function with the differential operator.

The derivative of a function is general is not linear, as you show: $f'(x) $ can be equal to $x^2$, $x^{29}$, $e^x$, etc, all non - linear function.

What is linear is the differential operator: this is NOT a function, is an operator that goes from a space of functions to another space of functions, very different^[1] than something like $x \mapsto e^x$ which is a function and maps numbers to other numbers.

So what it means in general for an operator (or a function) to be linear? It means that

$$f(ax + by) = af(x) + bf(y)$$

where $f$ is the operator, $a,b$ are numbers and $x,y$ elements in the domain of the opeator.

So if for a function $f:\mathbb R \mapsto \mathbb R$ being linear means that

$$f(ax + by) = af(x) + bf(y)$$ with $a,b,x,y \in \mathbb R$ (and this then implies that $f(x) = ax + b$ for some $a,b$) for an operator $L$ it means that

$$L(af + bg) = aL(f) + bL(g)$$ where $L$ is our operator, $a,b \in \mathbb R$ and $f,g$ are functions in the domain of the operator.

Now if $L = \frac d{dx}$ is the differential operator that associates to a function its derivative, that is $L(f) = f'(x)$, then we say $\frac d{dx}$ is linear because

$$L(af + bg) = (af + bg)' = af' + bg' = aL(f) + bL(g)$$

Hope it's more clear now.

[1] Okay, it's not very different. Technically they are the same thing, just their domain are different (reals vs space of functions). It's good to keep in mind their distinctions though :)

  • 1
    $\begingroup$ @leftaroundabout Indeed, you're right :) I wanted to remark the difference though! I tried to sacrifice a little formalism for the sake of clarity. (another point is that $a,b$ don't have to be reals but scalar in the field of the vector space where the functions live) Anyhow I edited $\endgroup$ – Ant Feb 12 '16 at 10:54
  • 2
    $\begingroup$ Actually, there are three different things that might be being confused: The derivative of a function, the differential operator, and the derivative of a function at a point viewed as a linear operator. This answer refers to the differential operator, which operates on functions, while amd's refers to the derivative at a point viewed as a linear map, which operates on displacements. $\endgroup$ – filipos Feb 12 '16 at 11:12
  • 1
    $\begingroup$ @MartinArgerami See also my previous comment. I totally disagree, when we say "derivatives are linear" we refer precisely to the fact that they operate in the way I explained above. Again, for every function $f$ we can associate the linear map $y \mapsto f(x_0)y$. So what? Why is that significant? The explanation is Taylor's theorem that tells basically that it's a good approximation if we are close to $x_0$, but again, so what? How can this possibly be related to the phrase "derivatives are linear"? $\endgroup$ – Ant Feb 12 '16 at 11:29
  • 1
    $\begingroup$ I just wanted to make it clear that these two answers refer to different things, lest the OP be confused. $\endgroup$ – filipos Feb 12 '16 at 11:31
  • 2
    $\begingroup$ @Ant: it is very "possibly related" since it is the right generalization to higher dimensions. $\endgroup$ – Martin Argerami Feb 12 '16 at 11:32

I think you may be misunderstanding what is being called linear.

It isn't the function $f$ that is linear or not, it is the action of taking a derivative -- the differential operator $\frac d{dx}$ is linear in the sense that

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

where $a$ and $b$ are constants and $f$ and $g$ are differentiable functions. Linearity here means that the operator "distributes" across a sum of terms, and the multiplicative constants "pull through".

  • 1
    $\begingroup$ I'm afraid it's you who misunderstands : math.stackexchange.com/questions/1260050/… $\endgroup$ – Martin Argerami Feb 12 '16 at 11:30
  • 1
    $\begingroup$ @MartinArgerami: I may have misinterpreted OP's question, but no context was given. My response, however, is completely correct. If OP was really asking about local linear approximations, rather than linearity of differential operators, then I indeed didn't address that. In that case, I would say the question is unclear. Moreover, the notion that I did address is commonly misunderstood by students first encountering the notion of a linear operator. $\endgroup$ – MPW Feb 12 '16 at 13:44
  • $\begingroup$ the second paragraph of the question clearly talks about the linear map $y\longmapsto e^xy$. What's unclear about that? $\endgroup$ – Martin Argerami Feb 12 '16 at 18:56
  • $\begingroup$ I chalked that up to a random stab, as it seemed to be reaching. Three of the four answers were essentially the same as mine, so I'm evidently not alone in thinking this. Moreover, one of those was the accepted answer. I do think that the comment by @filipos hits the nail on the head, though. $\endgroup$ – MPW Feb 12 '16 at 19:20

The linearity of a system, map, function, etc. is tested by the superposition principle. For example, consider a function $f(x)$. It is linear if

$f(\alpha x_1 + \beta x_2) = \alpha f(x_1) + \beta f(x_2)$.

In this case, the derivative is also linear operator because

$D[\alpha y_1(x) + \beta y_2(x)] = \alpha D[y_1(x)] + \beta D[y_2(x)]$.

  • $\begingroup$ Nice answer. It might work better for the OP if you didn't use $f$ in both examples, and if you wrote the differentiation as an operator rather than with the prime. $\endgroup$ – Ethan Bolker Feb 11 '16 at 21:55
  • $\begingroup$ This answer is not relevant to the very concrete question asked by the OP. $\endgroup$ – Martin Argerami Feb 12 '16 at 11:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.