b) Find the directional derivalive \( D_{u} f(x, y, z) \) for \( f(x, y, z)=\sin (y z)+\ln x^{2} \) in the direction of the vector \( v=\{1,1,-1\rangle \). Hence find \( D_{u} f(1,1, z) \).

To find the directional derivative \( D_{u} f(x, y, z) \) for the function \( f(x, y, z) = \sin(yz) + \ln x^{2} \) in the direction of the vector \( v = \{ 1, 1, -1 \} \), we first need to compute the gradient \( \nabla f(x, y, z) \). The gradient is given by: \[ \nabla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \] Calculating the partial derivatives: 1. \( \frac{\partial f}{\partial x} = \frac{2}{x} \) 2. \( \frac{\partial f}{\partial y} = z \cos(yz) \) 3. \( \frac{\partial f}{\partial z} = y \cos(yz) \) Therefore, the gradient is: \[ \nabla f(x, y, z) = \left( \frac{2}{x}, z \cos(yz), y \cos(yz) \right) \] Next, we need to find the unit vector \( u \) in the direction of \( v \): \[ |v| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3} \] Thus, the unit vector \( u \) is: \[ u = \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right) \] Now, we compute the directional derivative: \[ D_{u} f(x, y, z) = \nabla f(x, y, z) \cdot u = \left( \frac{2}{x}, z \cos(yz), y \cos(yz) \right) \cdot \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right) \] Calculating this dot product gives: \[ D_{u} f(x, y, z) = \frac{1}{\sqrt{3}} \left( \frac{2}{x} + z \cos(yz) - y \cos(yz) \right) = \frac{1}{\sqrt{3}} \left( \frac{2}{x} + (z - y) \cos(yz) \right) \] Finally, evaluating \( D_{u} f(1, 1, z) \): \[ D_{u} f(1, 1, z) = \frac{1}{\sqrt{3}} \left( 2 + (z - 1) \cos(z) \right) \] So, the final result for \( D_{u} f(1, 1, z) \) is: \[ D_{u} f(1, 1, z) = \frac{1}{\sqrt{3}} \left( 2 + (z - 1) \cos(z) \right) \]