Answer
$\dfrac{21}{13}$
Work Step by Step
In order to find the partial derivative, we will differentiate
with respect to $x$, by keeping $y$ as a constant to find the x-coordinate of the gradient vector, and vice versa:
$g_x=\dfrac{(xy+2)(1)-(x-y)(y)}{(xy+2)^2}=\dfrac{y^2+2}{(xy+2)^2} \\ g_y=\dfrac{(xy+2)(-1)-(x-y)(x)}{(xy+2)^2}=-\dfrac{x^2+2}{(xy+2)^2} $
Write the gradient equation.
$\nabla f (1,-1)= \lt f_x,f_y \gt = \lt \dfrac{(-1)^2+2}{(-1+2)^2}, -\dfrac{1+2}{(-1+2)^2} \gt =\lt 3,-3 \gt$
Thus, $v=\dfrac{u}{|u|} = \dfrac{\lt 12, 5 \gt }{\sqrt {(12)^2+(5)^2}} =\lt \dfrac{12}{13}, \dfrac{5}{13} \gt$
The directional derivative at that direction is given as:
$D_v f=\nabla f \cdot v=\lt 3,-3 \gt \times\lt \dfrac{12}{13}, \dfrac{5}{13} \gt =\dfrac{21}{13}$
