Answer
$sech^2 \sqrt t + \dfrac{\tanh \sqrt t}{\sqrt t}$
Work Step by Step
Given: $y=2 t^{1/2} \tanh (t^{1/2})$
Since, we know that $\dfrac{d}{dx} (\tanh x)=sech^2 x$
This implies:
$\dfrac{dy}{dt}= 2[t^{1/2} sech^2 (t^{1/2})( \dfrac{1}{2}t^{-1/2} )+ \tanh (t^{1/2}) (\dfrac{1}{2}t^{-1/2})] = sech^2 \sqrt t+ (t^{-1/2}) \tanh ( \sqrt t)$
or, $=sech^2 \sqrt t + \dfrac{\tanh \sqrt t}{\sqrt t}$
