First derivative analog filter
I will answer this with the understanding that $p$ is the Laplace variable (which I will call $s$). 

As you stated, differentiation in the time domain corresponds to multiplication by $s$ in the Laplace domain. So it seems that we could get what we want by just choosing $F(s)=s$; then the output is the derivative of the input, i.e., 

$$y(t) = \frac{du}{dt}$$.

So why the $\tau s+1$ in the denominator? 

This corresponds to a lowpass filter with bandwidth $1/\tau$. The output of $F(s) = s/(\tau s+1)$ is the derivative of the input signal after it has been smoothed by a lowpass filter with the desired bandwidth (or alternatively, this is the same as taking the derivative of the input signal and then smoothing it with the lowpass filter). This prevents the derivative operator from picking up high-frequency components of the input signal and amplifying noise. 

I don't know much about fault detection, but I can give a simple example. Say you're trying to detect when $u'(t)$ exceeds a fixed threshold. If you had no lowpass filter, small-amplitude (but high-frequency) noise in your measurement $u$ could trigger your detector even when it really shouldn't. But by setting $\tau$ you can filter out components of $u$ above a certain frequency ($1/\tau$) so they don't affect your derivative estimate.