What is the derivative of this equation:
y= arcsin (radical(2t))
The chain rule is
$\displaystyle \frac{dy}{dt} = \frac{dy}{du}\times \frac{du}{dt}$
For your function $\displaystyle y = \arcsin(\sqrt{2t})$ make $\displaystyle u = \sqrt{2t}$ so $\displaystyle y = \arcsin(u)$
Now find $\displaystyle \frac{dy}{du}$ and $\displaystyle \frac{du}{dt}$