Exchange Integral and Derivative respect to a parameter of a Dirac delta-function I'm trying to solve the 6.2 problem of Jackson's Classical Electrodynamics textbook. At some point, to get the desired solution, I have to exchange a derivative applied to a Dirac delta-function with the integral operator:
$$\int_{\mathbb{R^3}} \frac{\partial \delta(\mathbf{x}-\mathbf{x_0}(t))}{\partial t}f(\mathbf{x})\,d^3x=\frac{\partial}{\partial t}\int_{\mathbb{R^3}} \delta(\mathbf{x}-\mathbf{x_0}(t))f(\mathbf{x})\,d^3x=\frac{\partial}{\partial t}f(\mathbf{x_0}(t))$$
Under which hypothesis can I do something like that (i.e. exchange the order of differentiation and integration)? I expect that known theorems of real analysis do not apply in this case, since the $\delta$ is not even a proper function.
 A: By definition, if $T$ is a distribution, then $$\langle \partial_x T, f \rangle := - \langle T, \partial_x f \rangle\tag{1}$$ for every test function $f=f(x)$ in $C_0^\infty(\mathbb R^n)$ (or also $C^\infty(\mathbb R^n)$ if $T$ has compact support as the delta function). Here the derivative is just a bit more complicated. However, since the chain rule for taking the derivative of distributions composed with smooth functions is valid also for distributions (it is a general theorem) we have
$$\partial_t \delta(x-x_0(t)) = \frac{dx_0}{dt}|_{x_0}\cdot \nabla_{x_0} \delta(x-x_0(t)) = -\frac{dx_0}{dt}|_{x}\cdot \nabla_x \delta(x-x_0(t))\:. \tag{2}$$
As a consequence, for every  function $f \in C^\infty(\mathbb R^n)$, applying (2),
$$\int \partial_t \delta(x-x_0(t)) f(x) d^nx = -\int \frac{dx_0}{dt}|_{x_0}\cdot \nabla_x \delta(x-x_0(t)) f(x) d^nx$$ $$=  -\frac{dx_0}{dt}|_{x_0}\cdot\int \nabla_x\delta(x-x_0(t)) f(x) d^nx  \:.$$
Applying (1)
$$\int \partial_t \delta(x-x_0(t)) f(x) d^nx = + \frac{dx_0}{dt}|_{x_0}\cdot  \int  \delta(x-x_0(t)) \nabla_x f(x) d^nx$$ $$=  \frac{dx_0}{dt}|_{x_0}\cdot \nabla_x f(x)|_{x_0(t)} = \frac{d}{dt}f(x_0(t))\:.$$
