Answer
$$\frac{{dy}}{{dx}} = \frac{1}{{\left| {2x + 1} \right|\sqrt {{x^2} + x} }}$$
Work Step by Step
$$\eqalign{
& y = {\sec ^{ - 1}}\left( {2x + 1} \right) \cr
& {\text{differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\sec }^{ - 1}}\left( {2x + 1} \right)} \right] \cr
& {\text{Using the formula on the page 467}}\,\,\frac{d}{{dx}}\left[ {{{\sec }^{ - 1}}u} \right] = \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{dx}} \cr
& {\text{Considering }}u = 2x + 1{\text{ we have}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left| {2x + 1} \right|\sqrt {{{\left( {2x + 1} \right)}^2} - 1} }}\frac{d}{{dx}}\left[ {2x + 1} \right] \cr
& {\text{Then}}{\text{,}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left| {2x + 1} \right|\sqrt {{{\left( {2x + 1} \right)}^2} - 1} }}\left( 2 \right) \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{2}{{\left| {2x + 1} \right|\sqrt {4{x^2} + 4x + 1 - 1} }} \cr
& \frac{{dy}}{{dx}} = \frac{2}{{\left| {2x + 1} \right|\sqrt {4{x^2} + 4x} }} \cr
& \frac{{dy}}{{dx}} = \frac{2}{{2\left| {2x + 1} \right|\sqrt {{x^2} + x} }} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left| {2x + 1} \right|\sqrt {{x^2} + x} }} \cr} $$
