The rate of change of \( f(x, y)=e^{x y}+y^{2}-x^{2}+3 \) at \( (2,0) \) in the direction from \( (2,0) \) to \( (8,8) \) is a. 8 b. -4 c. \( -\frac{4}{5} \) d. -8 e. \( -\frac{2}{5} \)

To find the rate of change of the function \( f(x, y) = e^{xy} + y^2 - x^2 + 3 \) at the point \( (2, 0) \) in the given direction, we first need to compute the gradient of the function \( \nabla f \) and then determine the unit vector in the specified direction. 1. The gradient \( \nabla f(x, y) \) is computed as follows: \[ f_x = \frac{\partial f}{\partial x} = y e^{xy} - 2x, \] \[ f_y = \frac{\partial f}{\partial y} = x e^{xy} + 2y. \] 2. At \( (2, 0) \), we evaluate the gradient: \[ f_x(2, 0) = 0 \cdot e^{2 \cdot 0} - 2 \cdot 2 = -4, \] \[ f_y(2, 0) = 2 \cdot e^{2 \cdot 0} + 0 = 2. \] Thus, the gradient at \( (2, 0) \) is \( \nabla f(2, 0) = (-4, 2) \). 3. Next, we find the direction from \( (2, 0) \) to \( (8, 8) \): \[ \mathbf{d} = (8-2, 8-0) = (6, 8). \] 4. The magnitude of \( \mathbf{d} \) is: \[ ||\mathbf{d}|| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10. \] 5. The unit vector in the direction of \( \mathbf{d} \) is: \[ \mathbf{u} = \left( \frac{6}{10}, \frac{8}{10} \right) = \left( \frac{3}{5}, \frac{4}{5} \right). \] 6. Finally, the rate of change of \( f \) in the direction of \( \mathbf{u} \) is given by the dot product: \[ D_{\mathbf{u}} f(2, 0) = \nabla f(2, 0) \cdot \mathbf{u} = (-4, 2) \cdot \left( \frac{3}{5}, \frac{4}{5} \right) = -4 \cdot \frac{3}{5} + 2 \cdot \frac{4}{5} = -\frac{12}{5} + \frac{8}{5} = -\frac{4}{5}. \] Therefore, the rate of change of \( f(x, y) \) at \( (2, 0) \) in the specified direction is \( -\frac{4}{5} \). The correct answer is: c. \( -\frac{4}{5} \)