Given the velocity as a function of position $v(x) = \beta x^{-2n}$. The acceleration $a$ is defined as the rate of change of velocity with respect to time, $a = \frac{dv}{dt}$. Using the chain rule, acceleration can be expressed in terms of position as: $a = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$

1. Find $\frac{dv}{dx}$: $\frac{dv}{dx} = \frac{d}{dx}(\beta x^{-2n})$ $\frac{dv}{dx} = \beta (-2n) x^{-2n - 1} = -2n\beta x^{-(2n+1)}$

2. Calculate Acceleration ($a$): Substitute $v$ and $\frac{dv}{dx}$ into the acceleration equation: $a = (\beta x^{-2n}) \cdot (-2n\beta x^{-(2n+1)})$ $a = -2n \beta^2 \cdot x^{-2n + (-2n - 1)}$ $a = -2n \beta^2 x^{-4n - 1}$