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?

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⋅δ(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)=δ(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)⋅δ′(x)=−f(x)′⋅δ(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)