Question

Let

\[ f(x, y) = \begin{cases} \frac{x^3 + y^3}{x^2 - y^2}, & x^2 - y^2 \neq 0 \\ 0, & x^2 - y^2 = 0 \end{cases} \]

Then the directional derivative of \( f \) at \( (0, 0) \) in the direction of \( \frac{4}{5} \hat{i} + \frac{3}{5} \hat{j} \) is ............

Hint:

The directional derivative exists if the function is differentiable at the point, and it is computed as the dot product of the gradient and the direction vector.

Answer 1

Correct Answer: 2.6

Step 1: Compute the gradient of \( f \).
The gradient of a function \( f \) is given by: \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right). \] We need to compute the partial derivatives of \( f(x, y) \) at the point \( (0, 0) \). - At \( (0, 0) \), since \( x^2 - y^2 = 0 \), \( f(0, 0) = 0 \). - Now, we compute the partial derivatives. First, for \( x \neq 0 \) and \( y \neq 0 \), apply the quotient rule for both partial derivatives. But since \( f(x, y) \) is undefined at \( (0, 0) \), we need to use the definition of the directional derivative.
Step 2: Compute the directional derivative.
The directional derivative \( D_{\mathbf{u}} f \) in the direction of a unit vector \( \mathbf{u} = \left( \frac{4}{5}, \frac{3}{5} \right) \) is: \[ D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}. \] At the point \( (0, 0) \), since \( f(0, 0) = 0 \), we compute the limit of the difference quotient as: \[ D_{\mathbf{u}} f = \lim_{h \to 0} \frac{f(0+h\mathbf{u}) - f(0)}{h}. \] Since \( f(0) = 0 \), we evaluate the above limit as: \[ \boxed{0}. \]