# Derivative of the delta function at some point

The derivative of the delta function can be treated similar to the actual delta function. Suppose I have an expression like

$$\frac{\mathrm{d}}{\mathrm{d}x}\delta\left(x-x_0\right)$$

what does this mean for the integral

$$\int\mathrm{d}x\ f\left(x\right)\frac{\mathrm{d}}{\mathrm{d}x}\delta\left(x-x_0\right)$$

Furthermore, does it matter whether or not i derive with respect to the variable before or after the minus sign, i.e. does

$$\int\mathrm{d}x\ f\left(x\right)\frac{\mathrm{d}}{\mathrm{d}x}\delta\left(x_0-x\right)$$

give a different result?

Finally, does it matter whether or not I derive the delta function with the argument $$x_0-x$$ or the delta function alone and then plug in the argument:

$$\int\mathrm{d}x\ f\left(x\right)\delta'\left(x_0-x\right)$$

where $$\delta'\left(x\right)$$ is first "differentiated" with respect to $$x$$ and then evaluated at $$x_0-x$$ and/or $$x-x_0$$.

The derivative is usually defined via integration by parts, because we can't really apply the usual definition. The basic idea is the following: If we take 2 compactly supported, infinitely differentiable function $$u$$ and $$v$$, we get that $$\int u'v=-\int uv'$$ So if $$f$$ is compactly supported and infinitely differentiable, we "can" do the same (in the sense that it's the defining property of the derivative): $$\int \mathrm{d}x f(x)\delta'(x-x_0)=-\int \mathrm{d}x f'(x)\delta(x-x_0)=-f'(x_0)$$ This concept is called distributional derivative: https://en.m.wikipedia.org/wiki/Distribution_(mathematics)#Derivatives_of_distributions
Note: the weak derivative of an $$L^1$$ function is usually defined the same way.
• Just as the delta function $\delta(x-x_0)$ can be considered to be a certain measure (point mass at the point $x_0$), so its derivative can be considered to be a more complicated object at the point $x_0$. This object is known in physics as a "dipole". Commented Jul 3, 2020 at 14:43
Regarding your last question, the expression $$\delta'$$ as a function, i.e. $$\delta': x \mapsto \delta'(x-x_0)$$, the latter being some value, is not defined and hence not meaningful. This is due to the fact (which was well explained in the above answer) that we cannot make sense of $$\delta'$$ as a pointwise object; instead, the only means to understand its action is by pairing it with a function.
More generell, given some distribution $$\psi$$, i.e. a continuous linear functional acting on the classical test function space $$C_c^{\infty}$$, the derivative $$d\backslash dx \psi$$ is (in general) NOT something you can evaluate pointwise, but itself only a distribution, the acting on $$f \in C_c^{\infty}$$ being given as $$d\backslash dx \psi (f) = -\psi (d\backslash dx f).$$ Note that the RHS is meaningful, since $$\psi$$ is a continuous linear functional on $$C_c^{\infty}$$ and $$f$$ is, by choice, an element of that space. Also, this definition (!) is consistent with the smooth case, i.e. if $$\psi$$ and $$f$$ are smooth functions with compact support (think: integration by parts!) .You may think: Distributions are in a sense "too rough" to be evaluated pointwise (unlike classical functions $$u: \mathbb{R}^d \to \mathbb{R}$$), you can only make sense of them through a "coarser lense", through their action on very good functions. Of course, any reasonable function (in this sense, "reasonable" $$:= L_{loc}^1$$) is also a distribution - in rough words: An object which can be studied through a fine lense (i.e. pointwise) can also be studied through a coarser lense.
Now the shifted delta distribution $$\delta(x-x_0)$$ is a distribution - and nothing better (i.e. not a proper function), so everything said above applies to it and answers your question concerning $$\delta'$$ - I hope :)