Answer
$\frac{1}{|s|\sqrt( {( 25{s^2})-1})}$
Work Step by Step
Given that $y$ = $sec^{-1}(5s)$
on substituting ${5s}$ = $u$
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 $\frac{dy}{ds}$ = $\frac{1}{|(5s)|\sqrt ({({5s})^2-1})}$ $\times$ $\frac{d {(5s)}}{ds}$
or $\frac{dy}{ds}$ = $\frac{1}{|5s|\sqrt( {(25{s^2})-1})}$ $\times$ $5$
The final answer is: $\frac{1}{|s|\sqrt( {( 25{s^2})-1})}$
