#### Answer

$\frac{1}{|2s+1|\sqrt( {( {s^2})+s})}$

#### Work Step by Step

Given that $y$ = $sec^{-1}(2s+1)$
on substituting ${2s+1}$ = $u$
on applying chain rule derivative
$\frac{dy}{ds}$ = $\frac{{dsec^{-1}{u}}}{du}$ $\times$ $\frac{du}{ds}$
or $\frac{dy}{ds}$ = $\frac{1}{{|u|}\sqrt ({u^2}-1)}$ $\times$ $\frac{du}{ds}$
so $\frac{dy}{ds}$ = $\frac{1}{|(2s+1)|\sqrt ({({2s+1})^2-1})}$ $\times$ $\frac{d {(2s+1)}}{ds}$
or $\frac{dy}{ds}$ = $\frac{1}{|2s+1|\sqrt( {(4 {s^2})+4s})}$ $\times$ $2$
The final answer is: $\frac{1}{|2s+1|\sqrt( {( {s^2})+s})}$