Could you please help me in a simple way, what is the first derivative of a Dirac delta function? I found this answer:

The informal answer is a positive Delta function immediately followed by a negative-going Delta function.

Could you please explain this?


6 Answers 6


If you imagine a Dirac delta impulse as the limit of a very narrow very high rectangular impulse with unit area centered at $t=0$, then it's clear that its derivative must be a positive impulse at $0^-$ (because that's where the original impulse goes from zero to a very large value), and a negative impulse at $0^+$ (where the impulse goes from a very large value back to zero).

Actually, the generalized derivative of a Dirac delta impulse $\delta(t)$, denoted by $\delta'(t)$, is a generalized function (distribution) with the following properties:


Property $(1)$ is basically the definition of the derivative of a distribution. The right-most equality is of course a consequence of considering the special distribution $\delta(t)$. Property $(3)$ means that convolution with the derivative of a Dirac impulse results in the derivative of the convolved function. I.e., the distribution $\delta'(t)$ is the impulse response of an ideal differentiator. From property $(1)$ (with $f(t)=1$) it follows that


Furthermore, from $(1)$ it also follows that $\delta'(t)$ is odd, because for every even function $f(t)$ whose derivative exists at $t=0$ we have


since for even (and differentiable) $f(t)$, $f'(0)=0$ holds.


First of all the dirac delta is NOT a function, it's a distribution. See for example http://web.mit.edu/8.323/spring08/notes/ft1ln04-08-2up.pdf

Treating it as a conventional function can lead to misunderstandings. Example: "informally" the dirac delta is often defined as "infinity at x=0 and zero everywhere else". Now let's look at a function

$$y(t) = 2 \cdot \delta (t)$$

How would you describe that. Using the informal approach you would say "twice infinity at x=0 and 0 everywhere else". But there is no such thing as "twice infinity". Any number (other than 0) times infinity is just infinity again. That means that $y(t)$ ALSO matches the informal definition of the delta dirac so we would get $y(t) = \delta (t)$ which is non-sense.

Instead, we define the dirac delta by what it does:

  • any integration interval over the dirac delta that includes x=0 is 1. If it doesn't include x=0, it's 0
  • An integral over a function multiplied with a dirac delta will return the value of the function at x= 0 (or wherever the argument into the dirac vanishes)

So something strange happens in the dirac delta at x=0. We can only describe what it does, but we don't know how exactly it's doing it.

Once we get over that hurdle, the derivative question becomes easier. The derivative is NOT a function, it's a distribution. For the first derivative we can derive (see https://physicspages.com/pdf/Mathematics/Derivatives%20of%20delta%20function.pdf)

$$f(x) \cdot \delta ^{'} (x) = -f(x)^{'} \cdot \delta (x) $$

So we can describe the derivative similar to the original

  • An integral over a function multiplied with 1st derivative of a dirac delta will return the negative value of the first derivative of the function at x= 0 (or wherever the argument into the dirac vanishes)
  • 2
    $\begingroup$ Could some moderator please change the 9 above to a zero, so that the post makes sense? This is a mathematical stackexchange, where just one character can make a big difference, yet apparently edits must be "substantial". $\endgroup$
    – TimWescott
    Commented Jun 30, 2020 at 15:00
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    $\begingroup$ That last equation is not true in general. It is true only if $f(0)=0$. What is true is $\int_{-\infty}^{\infty}f(x)\delta'(x)dx=-\int_{-\infty}^{\infty}f'(x)\delta(x)dx$, but the conclusion that the equality must also hold for the integrands is wrong. $\endgroup$
    – Matt L.
    Commented Jul 1, 2020 at 6:46

Maybe a picture is worth a thousand words? Here's how a Gaussian pulse of variable width and its derivatives look like:


As others have said, Dirac is a distribution, hence the Gaussian pulse, and its width gets narrower and narrower. The derivative of


Which says that the derivative is the same as the function, multiplied by $-2x$, or a ramp going from positive to negative, centered on 0. Since the impulse is positive and centered around 0, the result comes out as two impulses of opposite signs.

Here, both the impulses and the derivatives have normalized amplitude for better viewing. As you can see, as the pulse gets narrower and narrower, the derivatives follow, so for a fixed amplitude, when the width of the input impulse becomes zero, the resulting derivatives will have zero widths and two, opposing signs peaks.

  • 2
    $\begingroup$ To match the properties of the delta "function" you should keep the area under the gaussian constant as you reduce its width.(i.e. you should normalize by area under the curve rather than by peak value) $\endgroup$
    – The Photon
    Commented Jul 1, 2020 at 18:16
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    $\begingroup$ @ThePhoton I know, but if I kept the area constant as it should be then it wouldn't have plotted nicely, particularly the derivatives, that's why I normalized the amplitudes of both (while making sure I let OP know about it). $\endgroup$ Commented Jul 1, 2020 at 22:03
  • $\begingroup$ If you don't mind the discontinuities, an isosceles triangle model would work just as well, maybe better conceptually. $\endgroup$ Commented Jul 5, 2020 at 23:33
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    $\begingroup$ @CedronDawg Believe it or not, initially I actually prepared three examples: a half sine pulse, a triangle (as you say), and a slow rise/fall pulse, and their derivatives. I started to write the answer progressively, starting with the triangle and how the derivatives are, then moving to the pulse, and then to the smooth sine, but then I got to the part where I said "as it gets narrower and narrower...", and none of them looked good plotted. Then I switched to the Gaussian pulse, also because the others made it plain clear that it's a distribution, so it would have matched the theme better. $\endgroup$ Commented Jul 6, 2020 at 6:38
  • $\begingroup$ I believe you. I didn't see that Andy Walls had covered it either. I do like your intuitive pictorial approach. You got my only upvote in this set of answers because your answer is the only one that attempts to give meaning of the derivative of delta itself, not in the context of being applied to a function. $\endgroup$ Commented Jul 6, 2020 at 11:41

Dirac's $\delta$ is a distribution. Distributions can be interpreted as limits of smooth functions under an integral or as operators acting on functions in ways which are defined by integrals. Both approaches have in common that basic properties of integrals are expected to work, partial integration in particular. Other answers have showed you the limiting processes, so in this answer I want to work a little bit with the abstract mathematical objects.

Let's develop the necessary bits of theory: for any well-behaved functions $f, g$ one has $$ \int_a^b f'(x)g(x) dx = \left.f(x)g(x)\right|_a^b-\int_a^b f(x)g'(x)dx, $$ i.e. partial integration. In the context of distributions in particular, we consider integrals over the whole real line and smooth functions that vanish (sufficiently fast) at infinity ("test functions"). Thus the first term on the RHS is zero and we have $$ \int_{-\infty}^{\infty} f'(x)g(x)dx = -\int_{-\infty}^{\infty} f(x)g'(x) dx. $$ We define distributions in a way that respects this fundamental property. Let's consider a distribution $g$ instead of a function, and let's misuse notation in the usual way (two different uses of $g$) so that it's clear what happens when we move to $\delta$. Namely, for any test function $g$ define a distribution $g[\cdot]$ which operates on test functions $f$ as follows $$ g[f]:=\int_{-\infty}^{\infty} f(x)g(x)dx, $$ giving a number for any test function $f$. Now consider the distribution $g'[\cdot]$ defined in the same way for the test function $g'$. Working from the definition we have $$ g'[f]\equiv \int_{-\infty}^{\infty} f(x)g'(x)dx = -\int_{-\infty}^{\infty} f'(x)g(x)dx = -g[f'].$$

This is a property that applies to the distributions we defined so far which are derived from (test) functions via integration. We generalize by defining distributions as linear operators on (test) functions that respect this identity even if they are not derived from test functions via an integral. In other words, we call objects $\psi$ distributions only if they respect the identity $\psi'[f] = -\psi[f']$.

With this in hand, let's move to Dirac's delta function. Using the same notation, the delta distribution is defined as $$ \delta[f] := f(0). $$ Note that unlike the usual jargon no integral appears in the definition. Since we want this to be a distribution, and again without reference to an integral, we find given our partial-integration rule that $$ \delta'[f] = -\delta[f'] \equiv -f'(0), $$ which now is not a result but a definition of the derivative of the delta distribution.

Finally, we can recover the more common notation by writing this in terms of now purely formal integrals.


$\delta(t)$ is a distribution, which means it is represented by a limitng set of functions.

To find $\delta'(t)$, start with a limiting set of functions for $\delta(t)$ that at least have a first derivative. The triangle function of unit area is the simplest function to chose:

$$\delta(t) = \lim_{\epsilon \to 0} \dfrac{\Lambda\left(\frac{t}{\epsilon }\right)}{\epsilon}$$

The derivative of $\Lambda(t)$ is two, offset, rectangle functions of opposite sign. That derivative can serve as the function for the limiting set of functions for $\delta'(t)$.

More explicitly

$$\delta'(t) = \lim_{\epsilon \to 0} \dfrac{d}{dt}\dfrac{\Lambda\left(\frac{t}{\epsilon}\right)}{\epsilon} = \lim_{\epsilon \to 0} \dfrac{\Pi\left(\frac{t}{\epsilon} +\frac{\epsilon}{2}\right)-\Pi\left(\frac{t}{\epsilon} -\frac{\epsilon}{2}\right)}{\epsilon^2} $$

Those two $\Pi()$ functions, in the limit, are what was informally stated as "a positive Delta function immediately followed by a negative-going Delta function."

Note that other functions with a first derivative could have been used for $\delta(t)$, such as a Gaussian, which is infinitely differentiable. Using the triangular function was a choice made for convenience.


Simply put, $\delta'$ picks the opposite of the derivative of $f$ at the origin. Let us imagine that I can forget for a moment about that $\delta$ is not a function, that it should be defined in a strict mathematical sense (over compactly supported smooth test functions), etc.

It can be simpler to consider that $\delta$ acts as an operator on (nice enough functions) $f$. This operator acts as it is picking a value in its argument. In the above (simplified) context, $\delta[f]$ picks the value of $f(x)$ at $x=0$. This can be expressed as:

$$\delta[f] = f(0)\,.$$

Within this (awful) intuition, I consider that:

$$\delta'[f] = -f'(0)\,,$$

and more generally: $$\delta^{(n)}[f] = (-1)^nf^{(n)}(0)\,.$$

For the original $\delta'[f] = -f'(0)\,$, an interpretation is the following. The discrete impulse $$\ldots,0,0,1,0,0\ldots$$ gets a discrete derivative as:

$$\ldots,0,0,1,-1,0,0\ldots$$ which is just the opposite of the discretized differential operator: $$\ldots,0,0,-1,1,0,0\ldots$$ better understood as the opposite of the 2-point classical discrete derivative of discrete signal $x[n]$:

$$ x[n]-x[n-1]\,. $$

Now, imagine that the discrete pulse compresses in time while growing (the classical image of the Dirac distribution), and the same for the $1$ and $-1$ of the derivative, that is my mnemonic to remember the formula.


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