Link between energy per unit frequency and derivative of the energy in regard of frequency I take a simple example to illustrate my question. I consider $\frac{du}{d \omega}$.
It physically represents the volumic energy per unit interval of frequencies (imagine a system with electromagnetic fields inside).
If I take $\frac{du}{d \omega}(\omega_0)$, it will physically give me the volumic energy of all the waves at the frequency $\omega_0$. If I had a filter that would select only $\omega_0$ and delete all other frequencies, it would be the total energy of my system.
Well ok. But what is the link with the derivative ? Indeed this quantity is written in the same way than the derivative of the volumic energy. But I don't see the link between those quantities...
Could you help me ?
Thank you
 A: They are the same thing.  That is the reason "energy per unit frequency" is written the same was as the "derivative of energy with respect to frequency."
When we ask how energy changes with respect to frequency (the derivative) it is the same as asking "for every unit change in frequency, how does the energy change".
To put it more generally, these word expressions mean the same thing:
(1) "derivative of A with respect to B"
(2) "quantity A per unit of B"
In general people tend to think of (1) as a function, and (2) as that function evaluated at a specific point, or as an average.  However both can and should be considered as functions.  Hope that helps.
A: The derivative is meant to denote the fraction (energy of waves in the interval $[\omega,\omega+\Delta \omega]$)/($\Delta \omega$). Usually $\Delta \omega$ is considered to small enough that $u$ is almost constant when $\omega$ changes by $\Delta \omega$ (usually $\Delta \omega$ is much smaller than $\omega$).
Thus $u$ is actually function of two things; the frequency $\omega$ and the frequency bandwidth $\Delta \omega$. The derivative is really taken with respect to this bandwidth:
$$
\rho(\omega) = \frac{du(\omega, \Delta \omega)}{d\Delta \omega}
$$.
Unfortunately sometimes this is sloppily written as
$$
\rho(\omega) = \frac{du(\omega)}{d\omega}
$$
as if $u$ was a function of frequency $\omega$ only. What is really meant by this fraction is derivative with respect to bandwidth $\Delta \omega$, not frequency $\omega$.
