Answer
$\frac{\sqrt2}{\sqrt( {1}-{2t^2})}$
Work Step by Step
Given that $y$ = $sin^{-1}(\sqrt{2}{t})$
on substituting $\sqrt{2}{t}$ = $u$
on applying chain rule derivative
$\frac{dy}{dt}$ = $\frac{{dsin^{-1}{u}}}{du}$ $\times$ $\frac{du}{dt}$
or $\frac{dy}{dt}$ = $\frac{1}{\sqrt (1-{u^2})}$ $\times$ $\frac{du}{dt}$
so $\frac{dy}{dt}$ = $\frac{1}{\sqrt (1-{(\sqrt2{t})^2})}$ $\times$ $\frac{d\sqrt {2}{t}}{dt}$
or $\frac{dy}{dt}$ = $\frac{1}{\sqrt( {1}-{2t^2})}$ $\times$ $\sqrt2$
The final answer is: $\frac{\sqrt2}{\sqrt( {1}-{2t^2})}$
