2 added 4 characters in body edited Aug 24 '15 at 12:25 glS 6,85844 gold badges2020 silver badges6262 bronze badges For (most) physical purpuses, you can safely think of the Dirac delta function $$\delta(x-x_0)$$ as some function not vanishing only around $$x_0$$, and with the property that its integral is normalized to one: $$\tag{1} \int dx \,\delta(x) = 1.$$ With this I mean that you can think of the delta as being a proper function, satisfying (1) and being not vanishing only in a very narrow$$^\dagger$$ interval around $$x_0$$. So given this view, what is $$\delta'(x)$$? Nothing but the "usual" derivative of whatever function $$\delta(x)$$ is. And this is the crucial point: we don't know what $$\delta$$ really looks like apart from the localization and the integral property, so while there is no problem in defining its derivative, we don't know what it looks like. So how can we use it? Well it turns out (as shown in the other answer) that when $$\delta'(x)$$ appears in an integral with another function we can "transfer" (integrating by parts) the derivative to the other function, and safely carry on with the calculation. $$^\dagger$$meaning smaller that all the other physical quantities involved in the given calculation For (most) physical purpuses, you can safely think of the Dirac delta function $$\delta(x-x_0)$$ as some function not vanishing only around $$x_0$$, and with the property that its integral is normalized to one: $$\tag{1} \int dx \,\delta(x) = 1.$$ With this I mean that you can think of the delta as being a proper function, satisfying (1) and being not vanishing only in a very narrow$$^\dagger$$ interval around $$x_0$$. So given this view, what is $$\delta'(x)$$? Nothing but the "usual" derivative of whatever function $$\delta(x)$$ is. And this is the crucial point: we don't know what $$\delta$$ really looks like apart from the localization and the integral property, so while there is no problem in defining its derivative, we don't know what it looks like. So how can we use it? Well it turns (as shown in the other answer) that when $$\delta'(x)$$ appears in an integral with another function we can "transfer" (integrating by parts) the derivative to the other function, and safely carry on with the calculation. $$^\dagger$$meaning smaller that all the other physical quantities involved in the given calculation For (most) physical purpuses, you can safely think of the Dirac delta function $$\delta(x-x_0)$$ as some function not vanishing only around $$x_0$$, and with the property that its integral is normalized to one: $$\tag{1} \int dx \,\delta(x) = 1.$$ With this I mean that you can think of the delta as being a proper function, satisfying (1) and being not vanishing only in a very narrow$$^\dagger$$ interval around $$x_0$$. So given this view, what is $$\delta'(x)$$? Nothing but the "usual" derivative of whatever function $$\delta(x)$$ is. And this is the crucial point: we don't know what $$\delta$$ really looks like apart from the localization and the integral property, so while there is no problem in defining its derivative, we don't know what it looks like. So how can we use it? Well it turns out (as shown in the other answer) that when $$\delta'(x)$$ appears in an integral with another function we can "transfer" (integrating by parts) the derivative to the other function, and safely carry on with the calculation. $$^\dagger$$meaning smaller that all the other physical quantities involved in the given calculation 1 answered Jan 24 '15 at 20:51 glS 6,85844 gold badges2020 silver badges6262 bronze badges For (most) physical purpuses, you can safely think of the Dirac delta function $$\delta(x-x_0)$$ as some function not vanishing only around $$x_0$$, and with the property that its integral is normalized to one: $$\tag{1} \int dx \,\delta(x) = 1.$$ With this I mean that you can think of the delta as being a proper function, satisfying (1) and being not vanishing only in a very narrow$$^\dagger$$ interval around $$x_0$$. So given this view, what is $$\delta'(x)$$? Nothing but the "usual" derivative of whatever function $$\delta(x)$$ is. And this is the crucial point: we don't know what $$\delta$$ really looks like apart from the localization and the integral property, so while there is no problem in defining its derivative, we don't know what it looks like. So how can we use it? Well it turns (as shown in the other answer) that when $$\delta'(x)$$ appears in an integral with another function we can "transfer" (integrating by parts) the derivative to the other function, and safely carry on with the calculation. $$^\dagger$$meaning smaller that all the other physical quantities involved in the given calculation