Answer
$\dfrac{dx}{dt} = \boxed{\dfrac{1}{3}-\dfrac{1}{3t^2}}$
Work Step by Step
$x = \dfrac{t^2+1}{3t}$
Dividing through by $3t$, we have,
$x = \dfrac{1}{3}t+\dfrac{1}{3}t^{-1}$
Taking the derivative of each side,
$\dfrac{dx}{dt} = \dfrac{1}{3}t^{1-1}+\dfrac{1}{3}(-1)t^{-1-1}$
$\dfrac{dx}{dt} = \boxed{\dfrac{1}{3}-\dfrac{1}{3t^2}}$
