Answer
3
Work Step by Step
(the derivative evaluated at $t=1$)
Use the product rule,
$\displaystyle \frac{d}{dt}[f(t)g(t)]=f^{\prime}(t)g(t)+f(t)g^{\prime}(t)$.
$f(t)=t^{2}-t^{0.5},\quad f^{\prime}(t)=2t-0.5t^{-0.5}$
$g(t)=t^{0.5}+t^{-0.5},\quad g^{\prime}(t)=0.5t^{-0.5}-0.5t^{-1.5}$
$\displaystyle \frac{dy}{dt}=(2t-0.5t^{-0.5})(t^{0.5}+t^{-0.5})+(t^{2}-t^{0.5})(0.5t^{-0.5}-0.5t^{-1.5})$
$\left.\frac{dy}{dt}\right|_{t=1}=(2-0.5)(1+1)+(1-1)(0.5-0.5)$
$=1.5\cdot 2+0=3$