Answer
$$\frac{{dy}}{{dt}} = - \frac{1}{{2v\sqrt {v - 1} }}$$
Work Step by Step
$$\eqalign{
& y = {\sin ^{ - 1}}\left( {\frac{1}{{\sqrt v }}} \right),\,\,\,\,\,v > 1 \cr
& y = {\sin ^{ - 1}}\left( {{v^{ - 1/2}}} \right) \cr
& {\text{find the derivative of }}y{\text{ with respect to }}v \cr
& \frac{{dy}}{{dv}} = \frac{{d\left( {{{\sin }^{ - 1}}\left( {{v^{ - 1/2}}} \right)} \right)}}{{dv}} \cr
& {\text{we can use the formula }}\cr
&\frac{{d\left( {{{\sin }^{ - 1}}u} \right)}}{{dt}} = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dt}}.\,\,\,\left( {{\text{see table 7}}{\text{.3}}} \right). \cr
& {\text{here }}u = {v^{ - 1/2}}t,\,\,{\text{then}} \cr
& \frac{{dy}}{{dt}} = \frac{1}{{\sqrt {1 - {{\left( {{v^{ - 1/2}}} \right)}^2}} }}\frac{{d\left( {{v^{ - 1/2}}} \right)}}{{dt}} \cr
& \frac{{dy}}{{dt}} = \frac{1}{{\sqrt {1 - {{\left( {{v^{ - 1/2}}} \right)}^2}} }}\left( { - \frac{1}{2}{v^{ - 3/2}}} \right) \cr
& {\text{simplifying, we get:}} \cr
& \frac{{dy}}{{dt}} = \frac{1}{{\sqrt {1 - {v^{ - 1}}} }}\left( { - \frac{1}{{2{v^{3/2}}}}} \right) \cr
& \frac{{dy}}{{dt}} = \frac{{{v^{1/2}}}}{{\sqrt {v - 1} }}\left( { - \frac{1}{{2{v^{3/2}}}}} \right) \cr
& \frac{{dy}}{{dt}} = - \frac{1}{{2v\sqrt {v - 1} }} \cr} $$
