jaydfox Wrote:By the way, the recurrence relation should be obvious:
\(
\begin{eqnarray}
T'(b, x)
& = & T(b, x)T'(b, x-1) \\
& = & T(b, x)T(b, x-1)T'(b, x-2) \\
& = & T(b, x)T(b, x-1)T(b, x-2)T'(b, x-2) \\
& = & T(b, x)T(b, x-1)T(b, x-2)T(b, x-3)T'(b, x-3)
\end{eqnarray}
\)
This process can be continued, so that you can find the derivative of the tetration over the entire domain, so long as you know the value of the tetration function and its derivative on the critical interval.
And for completeness, so we can go to the left of the critical interval:
\(
\begin{eqnarray}
T'(b, x) & = & T(b, x)T'(b, x-1)\\
T'(b, x+1) & = & T(b, x+1)T'(b, x)\\
T'(b, x) & = & \frac{T'(b, x+1)}{T(b, x+1)}
\end{eqnarray}
\)
Now we use the recurrence relation:
\(
\begin{eqnarray}
T'(b, x) & = & \frac{T'(b, x+1)}{T(b, x+1)}\\
\\[10pt]
\\
& = & \frac{\frac{T'(b, x+2)}{T(b, x+2)}}{T(b, x+1)}\\
\\[5pt]
\\
& = & \frac{T'(b, x+2)}{T(b, x+2)T(b, x+1)} \\
\\[5pt]
\\
& = & \frac{T'(b, x+3)}{T(b, x+3)T(b, x+2)T(b, x+1)}
\end{eqnarray}
\)
~ Jay Daniel Fox
