Official solution
Let \[f(x) = {\log _{\cos x}}\cos ec\,x\]
\[ = {{\log \cos ec\,x} \over {\log \cos x}}\]
\[ \Rightarrow f'(x) = {{\log \cos x\,.\,\sin x\,.\,\left( { - \cos ec\,x\cot x - \log \cos ec\,x\,.\,{1 \over {\cos x}}\,.\, - \sin x} \right)} \over {{{(\log \cos x)}^2}}}\]
at \[x = {\pi \over 4}\]
\[f'\left( {{\pi \over 4}} \right) = {{ - \log \left( {{1 \over {\sqrt 2 }}} \right) + \log \sqrt 2 } \over {{{\left( {\log {1 \over {\sqrt 2 }}} \right)}^2}}} = {2 \over {\log \sqrt 2 }}\]
\[\therefore\] \[{\log _e}2f'(x)\] at \[x = {\pi \over 4} = 4\]