*need not itself be a measure*. Let $P={\mathbb{R}}$ and $\mu_t=\delta_t$, the Dirac delta at $t\in{\mathbb{R}}$. Then for all $f\in C^\infty(P)$ we have $$\left.\frac{d}{dt}\right|_{t=0}\int f\,d\mu_t=\left.\frac{d}{dt}\right|_{t=0}f(t)=f'(0),$$ so that the derivative $\eta=d\mu_t/dt|_{t=0}$ is the distribution $\eta=-\partial_x \delta_0$, where $\partial_x$ indicates the derivative in the domain ${\mathbb{R}}=P$ of the measures, and is taken here in the sense of distributions.

\[rmk:notdistributional\] The derivative $d\mu_t/dt|_{t=0+}$ that we are interested in *does not coincide with the usual distributional derivative*. In order to take the derivative in the sense of distributions, we would need to consider $(\mu_t)_{t{\geqslant}0}$ as a distribution on the set $P\times [0,+\infty)$, and we do not do that.

Let us consider the situation in Example \[ex:deltas\] as a purely distributional derivative: we take $\mu_t=\delta_t$ on $P={\mathbb{R}}$ and, since it is well defined and smooth for all $t$ in ${\mathbb{R}}$, we think of the entire family $\mu_t$ as defining a distribution $\nu$ on ${\mathbb{R}}^2$, acting on functions $f\in C^\infty({\mathbb{R}}^2)$ by integration: $$\begin{gathered}
  \langle\nu,f\rangle=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(x,t)d\mu(x)dt= \\
  \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(x,t)d\delta_t(x)dt=\int_{-\infty}^{+\infty}f(t,t)dt.
 \end{gathered}$$ Then the distribution $\partial_t\nu$ (where the derivative is taken in the sense of distributions) acts like this on $f\in C^\infty_c({\mathbb{R}}^2)$: $$\langle\partial_t\nu,f\rangle=-\langle\nu,\partial_t f\rangle=-\int_{-\infty}^{+\infty}\frac{\partial f}{\partial t}(s,s)\,ds.$$ Clearly, there is no reason for this to coincide with the result we obtain in Example 