Question

Let \( f: \mathbb{R}^2 \to \mathbb{R} \) be differentiable. Let \( D_u f(0,0) \) and \( D_v f(0,0) \) be the directional derivatives of \( f \) at \( (0,0) \) in the directions of the unit vectors \( u = \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) \) and \( v = \left( \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right) \), respectively. If \( D_u f(0,0) = \sqrt{5} \) and \( D_v f(0,0) = \sqrt{5} \), then \[ \frac{\partial f}{\partial x} (0,0) + \frac{\partial f}{\partial y} (0,0) = \(\underline{\hspace{1cm}}\). \]

Hint:

To solve for the sum of partial derivatives from directional derivatives, use the formula for directional derivatives and solve the system of equations.

Answer 1

Correct Answer: 4

The directional derivative of \( f \) in the direction of a unit vector \( u = (u_1, u_2) \) is given by: \[ D_u f(x,y) = \frac{\partial f}{\partial x} (x,y) u_1 + \frac{\partial f}{\partial y} (x,y) u_2. \] Using the values given for the directional derivatives, we have: \[ D_u f(0,0) = \sqrt{5} \text{and} D_v f(0,0) = \sqrt{5}. \] We can now solve for the sum of the partial derivatives: \[ \frac{\partial f}{\partial x} (0,0) + \frac{\partial f}{\partial y} (0,0) = 2. \] Thus, the value is \( \boxed{2} \).