Differentiating Shannon's entropy Can somebody please show the steps of how differentiation of Shannon's entropy yields the following result? 
$H = -\sum_{l=0}^{L-1} p(l)\log_2[p(l)]$
The result of differentiating is
$H_m = -\sum_{l=0}^{L-1} \frac{1}{L}\log_2[\frac{1}{L}]$ = $log_2 L $
 A: Please excuse me if I just modify your notation slightly so that I can type it quickly:
1) Replace l (small letter "el") with x so as to avoid any confusion with 1=number one. 
2) p = p(x) is implied. 
3)Type the summation sign as "Sigma", with the summation from x=0 to x=L-1 implied, and 4) log=log2 or, without loss of generality, any other base simply with units conversion.
Explanation:
Step 1) Definition of Shannon entropy: $H = H(x) = -\Sigma[p*log(p)]$
This corresponds to what you have written correctly as the first equation.
Step 2) Differentiating is done so as to locate the maximum entropy, $H_m$, which occurs when the derivative is zero.
Step 3) Maximum entropy occurs when all states are equi-probable, i.e. $p = p(x) = \frac1{L}$ for all values of x in the range 0 to L-1.
Step 4) Substituting this into the original Shannon entropy equation:
$H_m = H.maximum = -\Sigma[(\frac1{L})*log(\frac1{L})]$ 
Because $log(\frac1{L})$ is a constant, this can be taken outside the summation, and so
$H_m = -log(\frac1{L})*\Sigma[\frac1{L}]$ and, as the summation is over the range 0 to L-1, i.e. consists L elements each of width $x = \frac1{L}$, therefore $\Sigma[\frac1{L}] = 1$.
Step 5) Substituting from 4): $H_m = -1*log(\frac1{L}) = log(L)$
Q.E.D, as required.
