$2.$ For the following truth table

$\backslash$begin$\{$tabular$\}\{|$c$|$c$|$c$|$c$|$c$|\}$

$\backslash$hline A & B & C & D & E $\backslash \backslash$

$\backslash$hline $0$ & $0$ & $0$ & $0$ & $0\backslash \backslash$

$\backslash$hline $0$ & $0$ & $0$ & $1$ & $1\backslash \backslash$

$\backslash$hline $0$ & $0$ & $1$ & $0$ & $0\backslash \backslash$

$\backslash$hline $0$ & $0$ & $1$ & $1$ & $0\backslash \backslash$

$\backslash$hline $0$ & $1$ & $0$ & $0$ & $0\backslash \backslash$

$\backslash$hline $0$ & $1$ & $0$ & $1$ & $1\backslash \backslash$

$\backslash$hline $0$ & $1$ & $1$ & $0$ & $1\backslash \backslash$

$\backslash$hline $0$ & $1$ & $1$ & $1$ & $1\backslash \backslash$

$\backslash$hline $1$ & $0$ & $0$ & $0$ & $0\backslash \backslash$

$\backslash$hline $1$ & $0$ & $0$ & $1$ & $0\backslash \backslash$

$\backslash$hline $1$ & $0$ & $1$ & $0$ & $0\backslash \backslash$

$\backslash$hline $1$ & $0$ & $1$ & $1$ & $0\backslash \backslash$

$\backslash$hline $1$ & $1$ & $0$ & $0$ & $1\backslash \backslash$

$\backslash$hline $1$ & $1$ & $0$ & $1$ & $1\backslash \backslash$

$\backslash$hline $1$ & $1$ & $1$ & $0$ & $0\backslash \backslash$

$\backslash$hline $1$ & $1$ & $1$ & $1$ & $0\backslash \backslash$

$\backslash$hline

$\backslash$end$\{$tabular$\}$

$-[3$ points$]$ Derive a Sum of Products expression and a Product of Sums expression for the output $\backslash ($ E $\backslash )$ in terms of the input signals $\backslash ($ A$,$ B$,$ C $\backslash )$ and $\backslash ($ D $\backslash ).$

$-[4$ points$]$ Simplify the Sum of Products expression that you obtained from the previous question using Karnaugh maps

$-[3$ points$]$ Draw the corresponding circuit diagram for this simplified expression.