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

To compute the directional derivative of the function \( f(x, y)=e^{-x-y} \) at the point \( P(\ln 2, \ln 3) \) in the direction of the vector \( \langle 1, 2 \rangle \), follow these steps: 1. **Calculate the Gradient**: The gradient of \( f \) is given by \[ \nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right). \] Calculating the partial derivatives: \[ \frac{\partial f}{\partial x} = -e^{-x-y}, \quad \frac{\partial f}{\partial y} = -e^{-x-y}. \] Thus, \[ \nabla f(x, y) = \left(-e^{-x-y}, -e^{-x-y}\right). \] 2. **Evaluate the Gradient at Point \( P \)**: Substitute \( P(\ln 2, \ln 3) \): \[ \nabla f(\ln 2, \ln 3) = \left(-e^{-\ln 2 - \ln 3}, -e^{-\ln 2 - \ln 3}\right) = \left(-\frac{1}{6}, -\frac{1}{6}\right). \] 3. **Normalize the Direction Vector**: The direction vector \( \mathbf{v} = \langle 1, 2 \rangle \) has a magnitude of \[ \|\mathbf{v}\| = \sqrt{1^2 + 2^2} = \sqrt{5}. \] The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is \[ \mathbf{u} = \left\langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right\rangle. \] 4. **Compute the Directional Derivative**: The directional derivative \( D_u f \) is given by \[ D_u f = \nabla f \cdot \mathbf{u} = \left(-\frac{1}{6}, -\frac{1}{6}\right) \cdot \left\langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right\rangle. \] Now compute the dot product: \[ D_u f = -\frac{1}{6} \cdot \frac{1}{\sqrt{5}} - \frac{1}{6} \cdot \frac{2}{\sqrt{5}} = -\frac{1}{6\sqrt{5}} - \frac{2}{6\sqrt{5}} = -\frac{3}{6\sqrt{5}} = -\frac{1}{2\sqrt{5}}. \] Thus, the directional derivative is \[ \boxed{-\frac{1}{2\sqrt{5}}}. \]