Find the directional derivative of the function at \( p \) in the direction of \( \mathbf{v} \). \[ f(x, y)=x^{3}-y^{3}, \quad P(6,5), \mathbf{v}=\frac{\sqrt{2}}{2}(\mathbf{i}+\mathbf{j}) \]

To find the directional derivative of the function \( f(x, y) = x^3 - y^3 \) at the point \( P(6, 5) \) in the direction of the vector \( \mathbf{v} = \frac{\sqrt{2}}{2}(\mathbf{i}+\mathbf{j}) \), we first need to compute the gradient of \( f \). The gradient \( \nabla f(x, y) \) is given by: \[ \nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] Calculating the partial derivatives: \[ \frac{\partial f}{\partial x} = 3x^2, \quad \frac{\partial f}{\partial y} = -3y^2 \] Now, evaluating the gradient at the point \( P(6, 5) \): \[ \nabla f(6, 5) = \left( 3(6)^2, -3(5)^2 \right) = \left( 3 \times 36, -3 \times 25 \right) = (108, -75) \] Next, we need to find the directional derivative \( D_{\mathbf{u}} f(P) \), where \( \mathbf{u} \) is the unit vector in the direction of \( \mathbf{v} \). Given that \( \mathbf{v} \) is already a unit vector: \[ D_{\mathbf{u}} f(P) = \nabla f(P) \cdot \mathbf{u} \] Calculating the dot product: \[ \nabla f(6, 5) \cdot \mathbf{u} = (108, -75) \cdot \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \] Calculating: \[ = 108 \cdot \frac{\sqrt{2}}{2} + (-75) \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}(108 - 75) = \frac{\sqrt{2}}{2}(33) \] Thus, the directional derivative of \( f \) at \( P \) in the direction of \( \mathbf{v} \) is: \[ D_{\mathbf{u}} f(P) = \frac{33\sqrt{2}}{2} \]