[tex]~~~~y= \sec^{-1} (14x)\\\\\\\implies \dfrac{dy}{dx} = \dfrac{1}{14x \sqrt{(14x)^2 -1}} \cdot 14\\\\\\\implies \dfrac{dy}{dx}=\dfrac{1}{x\sqrt{196x^2 -1}}\\\\\\\text{At point,}~~ (x_1,y_1) = \left(\dfrac{\sqrt2}{14}, \dfrac{\pi}4} \right), \text{slope of the tangent line,}\\\\\\m = \dfrac{dy}{dx} = \dfrac{1}{\dfrac{\sqrt 2}{14} \sqrt{196 \left(\dfrac{\sqrt 2}{14} \right)^2-1}} \\\\\\~~~~~~~~~~~=\dfrac{14}{\sqrt 2 }\\\\\\~~~~~~~~~~~=\dfrac{7\sqrt 2 \sqrt 2}{\sqrt 2}\\\\\\[/tex]
[tex]=7\sqrt 2\\\\\\\text{Equation of tangent line,}\\\\\\~~~~~y-y_1 = m(x -x_1)\\\\\\\implies y- \dfrac{\pi}4 = 7\sqrt 2 \left(x - \dfrac{\sqrt 2}{14} \right)\\ \\\\\implies y= 7x\sqrt 2 - 1+ \dfrac{\pi}4[/tex]
