#### Answer

$-\frac{|t|}{{t}\sqrt( { {1}-{t^2}})}$

#### Work Step by Step

Given that $y$ = $sec^{-1}{\frac{1}{t}}$ where t lies between $0$ and $1$
on substituting ${\frac{1}{t}}$ = $u$
on applying chain rule derivative
$\frac{dy}{dt}$ = $\frac{{dsec^{-1}{u}}}{du}$ $\times$ $\frac{du}{dt}$
or $\frac{dy}{dt}$ = $\frac{1}{{|u|}\sqrt ({u^2}-1)}$ $\times$ $\frac{du}{dt}$
so $\frac{dy}{dt}$ = $\frac{1}{|(\frac{1}{t})|\sqrt ({({\frac{1}{t}})^2-1})}$ $\times$ $\frac{d {(\frac{1}{t})}}{dt}$
or $\frac{dy}{dt}$ = $\frac{1}{|\frac{1}{t}|\sqrt( {(\frac{1}{t^2})-1})}$ $\times$ $-\frac{1}{t^2}$
The final answer is: $-\frac{|t|}{{t}\sqrt( { {1}-{t^2}})}$