ここで勉強すれば数学検定１級の壁は超えられるか。

$f\left( x\right) =\tan ^{-1}\left( \sec x\right)$

$f'\left( x\right) =\dfrac {1}{1+\sec ^{2}x}\cdot \left( \sec x\right) '$

$f'\left( x\right) =\dfrac {1}{1+\sec ^{2}x}\cdot \dfrac {\sin x}{\cos ^{2}x}$

$f'\left( \dfrac {\pi }{4}\right) =\dfrac {1}{1+2}\cdot \dfrac {\dfrac {1}{\sqrt {2}}}{\dfrac {1}{2}}=\dfrac {2}{3\sqrt {2}}=\dfrac {\sqrt {2}}{3}$・・・答え