1. **State the problem:** Find the first and second derivatives of the parametric functions $x=t$ and $y=t-\frac{1}{t}$ at $t=4$. 2. **Recall formulas:** For parametric equations $x(t)$ and $y(t)$, the first derivative $\frac{dy}{dx}$ is given by $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ The second derivative $\frac{d^2y}{dx^2}$ is $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$$ 3. **Calculate derivatives with respect to $t$:** $$\frac{dx}{dt} = 1$$ $$\frac{dy}{dt} = 1 - \left(-\frac{1}{t^2}\right) = 1 + \frac{1}{t^2}$$ 4. **Find the first derivative $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1} = 1 + \frac{1}{t^2}$$ 5. **Evaluate the first derivative at $t=4$:** $$\frac{dy}{dx}\bigg|_{t=4} = 1 + \frac{1}{4^2} = 1 + \frac{1}{16} = \frac{17}{16}$$ 6. **Differentiate $\frac{dy}{dx}$ with respect to $t$ to find $\frac{d}{dt}\left(\frac{dy}{dx}\right)$:** $$\frac{d}{dt}\left(1 + \frac{1}{t^2}\right) = 0 - \frac{2}{t^3} = -\frac{2}{t^3}$$ 7. **Calculate the second derivative $\frac{d^2y}{dx^2}$:** $$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{-\frac{2}{t^3}}{1} = -\frac{2}{t^3}$$ 8. **Evaluate the second derivative at $t=4$:** $$\frac{d^2y}{dx^2}\bigg|_{t=4} = -\frac{2}{4^3} = -\frac{2}{64} = -\frac{1}{32}$$ **Final answers:** - First derivative at $t=4$ is $\frac{17}{16}$. - Second derivative at $t=4$ is $-\frac{1}{32}$.