Chapter 12 - Section 12.3 - Higher Order Derivatives: Acceleration and Concavity - Exercises - Page 901: 7

$ e^{-(x-1)}$

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