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