\( 1<\begin{array}{l}\text { Compute the directional derivative of the following function at the given point } P \text { in the direction of the given vector. Be } \\ \text { sure to use a unit vector for the direction vector. } \\ f(x, y)=\ln \left(3+x^{2}+3 y^{2}\right) ; P(3,-1) ;\langle 2,1\rangle \\ \text { The directional derivative is } \\ \text { (Type an exact answer, using radicals as needed.) }\end{array} . \begin{array}{l}\text {. }\end{array} \)

To compute the directional derivative of the function \( f(x, y) = \ln(3 + x^2 + 3y^2) \) at the point \( P(3, -1) \) in the direction of the vector \( \langle 2, 1 \rangle \), we first need to find the gradient of the function. 1. **Calculate the gradient**: \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = \left( \frac{2x}{3 + x^2 + 3y^2}, \frac{6y}{3 + x^2 + 3y^2} \right) \] 2. **Evaluate the gradient at point \( P \)**: \[ \nabla f(3, -1) = \left( \frac{2(3)}{3 + 3^2 + 3(-1)^2}, \frac{6(-1)}{3 + 3^2 + 3(-1)^2} \right) = \left( \frac{6}{3 + 9 + 3}, \frac{-6}{3 + 9 + 3} \right) = \left( \frac{6}{15}, \frac{-6}{15} \right) = \left( \frac{2}{5}, -\frac{2}{5} \right) \] 3. **Unit vector in the direction of \( \langle 2, 1 \rangle \)**: \[ \mathbf{u} = \frac{\langle 2, 1 \rangle}{\sqrt{2^2 + 1^2}} = \frac{\langle 2, 1 \rangle}{\sqrt{5}} = \left( \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right) \] 4. **Find the directional derivative**: \[ D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u} = \left( \frac{2}{5}, -\frac{2}{5} \right) \cdot \left( \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right) = \frac{2}{5} \cdot \frac{2}{\sqrt{5}} + \left(-\frac{2}{5}\right) \cdot \frac{1}{\sqrt{5}} = \frac{4}{5\sqrt{5}} - \frac{2}{5\sqrt{5}} = \frac{2}{5\sqrt{5}} \] Therefore, the directional derivative is: \[ \frac{2}{5\sqrt{5}} \]