1. The problem is to find $\frac{dy}{dx}$ for the parametric equations $y = t^6 - 5$ and $x = 4^3 - 1$. 2. First, note that $x = 4^3 - 1 = 64 - 1 = 63$ is a constant, independent of $t$. 3. Since $x$ is constant, $\frac{dx}{dt} = 0$. 4. Calculate $\frac{dy}{dt}$: $$\frac{dy}{dt} = \frac{d}{dt}(t^6 - 5) = 6t^5$$ 5. The derivative $\frac{dy}{dx}$ for parametric equations is given by: $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ 6. Substitute the values: $$\frac{dy}{dx} = \frac{6t^5}{0}$$ 7. Division by zero is undefined, so $\frac{dy}{dx}$ does not exist for this parametric curve. 8. Geometrically, since $x$ is constant, the curve is a vertical line at $x=63$, and the slope of a vertical line is undefined. Final answer: $\frac{dy}{dx}$ is undefined because $x$ is constant.