Chapter 15 - Section 15.2 - Partial Derivatives - Exercises - Page 1106: 14

$\frac{∂f}{∂x}=2e^{2x+y}$ $\frac{∂f}{∂y}=e^{2x+y}$ $\frac{∂f}{∂x}|_{(1,-1)}=2e$ $\frac{∂f}{∂y}|_{(1,-1)}=e$

We find the partial derivatives as follows: $f(x,y)=e^{2x+y}$ $\frac{∂f}{∂x}=e^{2x+y}(2)=2e^{2x+y}$ $\frac{∂f}{∂y}=e^{x+y}(1)=e^{2x+y}$ $\frac{∂f}{∂x}|_{(1,-1)}=2e^{2(1)+(-1)}=2e^1=2e$ $\frac{∂f}{∂y}|_{(1,-1)}=e^{2(1)+(-1)}=e^1=e$