Answer
$$y = 2\sqrt 2 x - 1 + \frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& y = \operatorname{arcsec} 4x,{\text{ }}\left( {\frac{{\sqrt 2 }}{4},\frac{\pi }{4}} \right) \cr
& {\text{Differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\operatorname{arcsec} 4x} \right] \cr
& {\text{Use }}\frac{d}{{dx}}\left[ {\operatorname{arcsec} u} \right] = \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left| {4x} \right|\sqrt {{{\left( {4x} \right)}^2} - 1} }}\frac{d}{{dx}}\left( {4x} \right) \cr
& \frac{{dy}}{{dx}} = \frac{4}{{\left| {4x} \right|\sqrt {16{x^2} - 1} }} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left| x \right|\sqrt {16{x^2} - 1} }} \cr
& {\text{Calculate the slope at the given point }}\left( {\frac{{\sqrt 2 }}{4},\frac{\pi }{4}} \right) \cr
& {\left. {m = \frac{{dy}}{{dx}}} \right|_{x = \frac{{\sqrt 2 }}{4}}} = \frac{1}{{\left| {\sqrt 2 /4} \right|\sqrt {16{{\left( {\sqrt 2 /4} \right)}^2} - 1} }} \cr
& {\left. {m = \frac{{dy}}{{dx}}} \right|_{x = \frac{{\sqrt 2 }}{4}}} = \frac{4}{{\sqrt 2 \sqrt 1 }} \cr
& {\left. {m = \frac{{dy}}{{dx}}} \right|_{x = \frac{{\sqrt 2 }}{4}}} = 2\sqrt 2 \cr
& {\text{The equation of the tangent line at the point }}\left( {\frac{{\sqrt 2 }}{4},\frac{\pi }{4}} \right){\text{ is}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - \frac{\pi }{4} = 2\sqrt 2 \left( {x - \frac{{\sqrt 2 }}{4}} \right) \cr
& y - \frac{\pi }{4} = 2\sqrt 2 x - 1 \cr
& y = 2\sqrt 2 x - 1 + \frac{\pi }{4} \cr} $$
