Chapter 11 - Section 11.5 - Derivatives of Logarithmic and Exponential Functions - Exercises - Page 842: 10

$-e^{1-x}$

(see p. 840, Generalized Rule)$\quad$ $\displaystyle \frac{d}{dx}[e^{u}]=e^{u}\cdot\frac{du}{dx}$ $u(x)=1-x$ $\displaystyle \frac{du}{dx}=-1$ $\displaystyle \frac{d}{dx}[e^{1-x}]=e^{1-x}\cdot(-1)=-e^{1-x}$