Let \(f(x)=\sec^{-1}x\). Then \(\sec f(x)=x\), and differentiating both sides with respect to \(x\) gives
\((\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1\)
\(f'(x)=\dfrac1{\sec f(x)\tan f(x)}\)
Now, when \(x=\sqrt2\), you get
\((\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}\)
You have \(\sec^{-1}\sqrt2=\dfrac\pi4\), so \(\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2\) and \(\tan\left(\sec^{-1}\sqrt2\right)=1\). So \((\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}\)
