Chapter 12 - Section 12.3 - Higher Order Derivatives: Acceleration and Concavity - Exercises - Page 901: 3

$\displaystyle \frac{4}{x^{3}}$

$f''(x)=\displaystyle \frac{d}{dx}\left[\frac{df}{dx}\right]=\frac{d^{2}f}{dx^{2}}$ $\displaystyle \frac{dy}{dx}=\frac{d}{dx}[2x^{-1}]= -2x^{-2}$ $\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}[ -2x^{-2}]=-2(-2x^{-3})=\displaystyle \frac{4}{x^{3}}$