## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{ds}} = \frac{1}{{\left| {2s + 1} \right|\sqrt {{s^2} + s} }}$$
\eqalign{ & y = {\sec ^{ - 1}}\left( {2s + 1} \right) \cr & {\text{find the derivative of }}y{\text{ with respect to }}s \cr & \frac{{dy}}{{ds}} = {\sec ^{ - 1}}\left( {2s + 1} \right) \cr & {\text{we can use the formula }}\cr & \frac{{d\left( {{{\sec }^{ - 1}}u} \right)}}{{ds}} = \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{ds}}\,\,\,\,\left| u \right| > 1.\,\,\,\left( {{\text{see table 7}}{\text{.3}}} \right). \cr & {\text{here }}u = 2s + 1,\,\,{\text{then}} \cr & \frac{{dy}}{{ds}} = \frac{1}{{\left| {2s + 1} \right|\sqrt {{{\left( {2s + 1} \right)}^2} - 1} }}\frac{{d\left( {2s + 1} \right)}}{{ds}} \cr & \frac{{dy}}{{ds}} = \frac{1}{{\left| {2s + 1} \right|\sqrt {4{s^2} + 4s + 1 - 1} }}\left( 2 \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{ds}} = \frac{2}{{\left| {2s + 1} \right|\sqrt {4{s^2} + 4s} }} \cr & \frac{{dy}}{{ds}} = \frac{2}{{\left| {2s + 1} \right|\left( 2 \right)\sqrt {{s^2} + s} }} \cr & \frac{{dy}}{{ds}} = \frac{1}{{\left| {2s + 1} \right|\sqrt {{s^2} + s} }} \cr}