Answer
${u_{xxx}} = - 8sec{h^2}\left( {x - t} \right){\tanh ^3}\left( {x - t} \right) + 16sec{h^4}\left( {x - t} \right)\tanh \left( {x - t} \right)$
Work Step by Step
We have $u\left( {x,t} \right) = sec{h^2}\left( {x - t} \right)$.
From Section 7.9 on page 378, we have
$\frac{d}{{dv}}sechv = - sechv\tanh v$
$\frac{d}{{dv}}\tanh v = sec{h^2}v$
So,
${u_x} = \left( {2sech\left( {x - t} \right)} \right)\left( { - sech\left( {x - t} \right)\tanh \left( {x - t} \right)} \right)$
$ = - 2sec{h^2}\left( {x - t} \right)\tanh \left( {x - t} \right)$
$ = - 2u\tanh \left( {x - t} \right)$
${u_{xx}} = - 2{u_x}\tanh \left( {x - t} \right) - 2usec{h^2}\left( {x - t} \right)$
$ = - 2{u_x}\tanh \left( {x - t} \right) - 2{u^2}$
${u_{xxx}} = - 2{u_{xx}}\tanh \left( {x - t} \right) - 2{u_x}sec{h^2}\left( {x - t} \right) - 4u{u_x}$
$ = - 2{u_{xx}}\tanh \left( {x - t} \right) - 2{u_x}u - 4u{u_x}$
$ = - 2{u_{xx}}\tanh \left( {x - t} \right) - 6u{u_x}$
From previous results we have
${u_x} = - 2sec{h^2}\left( {x - t} \right)\tanh \left( {x - t} \right)$
${u_{xx}} = - 2{u_x}\tanh \left( {x - t} \right) - 2usec{h^2}\left( {x - t} \right)$
$ = 4sec{h^2}\left( {x - t} \right){\tanh ^2}\left( {x - t} \right) - 2sec{h^4}\left( {x - t} \right)$
Substituting these in ${u_{xxx}}$
${u_{xxx}} = - 2{u_{xx}}\tanh \left( {x - t} \right) - 6u{u_x}$
gives
${u_{xxx}} = - 8sec{h^2}\left( {x - t} \right){\tanh ^3}\left( {x - t} \right) + 4sec{h^4}\left( {x - t} \right)\tanh \left( {x - t} \right)$
$ + 12sec{h^4}\left( {x - t} \right)\tanh \left( {x - t} \right)$
Finally,
${u_{xxx}} = - 8sec{h^2}\left( {x - t} \right){\tanh ^3}\left( {x - t} \right) + 16sec{h^4}\left( {x - t} \right)\tanh \left( {x - t} \right)$
