Answer
$\frac{-1}{{2}{t}\sqrt( t-1)}$
Work Step by Step
Given that $y$ = $cot^{-1}(\sqrt{t}-1)$
on substituting $\sqrt{t-1}$ = $u$
on applying chain rule derivative
$\frac{dy}{dt}$ = $\frac{{dcot^{-1}{u}}}{du}$ $\times$ $\frac{du}{dt}$
or $\frac{dy}{dt}$ = $-\frac{1}{ (1+{u^2})}$ $\times$ $\frac{du}{dt}$
so $\frac{dy}{dt}$ = $-\frac{1}{ (1+{(\sqrt{t-1})^2})}$ $\times$ $\frac{d\sqrt {t-1}}{dt}$
or $\frac{dy}{dt}$ = $-\frac{1}{t}$ $\times$ $\frac{1}{2\sqrt{t-1}}$
The final answer is: $\frac{-1}{{2}{t}\sqrt( t-1)}$