Answer
$-\displaystyle \frac{47}{4}$
Work Step by Step
$x=x(t)$ and we are given $x(1)=2$
By the chain rule,
$\left.\dfrac{dy}{dt}\right|_{t=1}=\left. \dfrac{dy}{dx}\dfrac{dx}{dt}\right|_{t=1}$... (evaluated at t=1)
$\left.\dfrac{dy}{dx}\right|_{t=1} = \left.\dfrac{dy}{dx}\right|_{x=2}\qquad$... because x(1)=2
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[x^{3}+x^{-1}]=3x^{2}-x^{-2}$
$\displaystyle \left.\frac{dy}{dx}\right|_{x=2}=3(4)-\frac{1}{4}=\frac{48-1}{4}=\frac{47}{4}$
$\displaystyle \left.\frac{dy}{dt}\right|_{t=1}=\left. \frac{dy}{dx}\frac{dx}{dt}\right|_{t=1}=\frac{47}{4}\cdot(-1)=-\frac{47}{4}$
