Chapter 15 - Section 15.2 - Partial Derivatives - Exercises - Page 1106: 13

$\frac{∂f}{∂x}=e^{x+y}$ $\frac{∂f}{∂y}=e^{x+y}$ $\frac{∂f}{∂x}|_{(1,-1)}=1$ $\frac{∂f}{∂y}|_{(1,-1)}=1$

We find the partial derivatives as follows: $f(x,y)=e^{x+y}$ $\frac{∂f}{∂x}=e^{x+y}(1)=e^{x+y}$ $\frac{∂f}{∂y}=e^{x+y}(1)=e^{x+y}$ $\frac{∂f}{∂x}|_{(1,-1)}=e^{1+(-1)}=e^0=1$ $\frac{∂f}{∂y}|_{(1,-1)}=e^{1+(-1)}=e^0=1$