The equation \( x y z-2 x \ln y-z^{2}=2 \) implicitly defines \( z=z(x, y) \) near \( (x, y, z)=(3,1,1) \). Find the tangent plane to \( z(x, y) \) at \( (3,1) \). a. \( z+x-3 y=1 \) b. \( 2 z-x+4 y=2 \) (c. \( z+x-3 y=-2 \) o d. \( z-x+3 y=1 \) e. \( z-x-3 y=0 \)

To find the tangent plane to the implicit surface defined by \( xy z - 2x \ln y - z^2 = 2 \) at the point \( (3, 1, 1) \), we can use the gradients. 1. Calculate the gradient of the left-hand side function \( F(x, y, z) = xyz - 2x \ln y - z^2 \). Compute partial derivatives: \[ F_x = yz - 2 \ln y, \quad F_y = xz - \frac{2x}{y}, \quad F_z = xy - 2z \] 2. Evaluate these derivatives at \( (3, 1, 1) \): - \( F_x(3, 1, 1) = 1 \cdot 1 - 2 \cdot 0 = 1 \) - \( F_y(3, 1, 1) = 3 \cdot 1 - 2 \cdot 3 = 3 - 6 = -3 \) - \( F_z(3, 1, 1) = 3 \cdot 1 - 2 \cdot 1 = 3 - 2 = 1 \) 3. The normal vector to the tangent plane is given by \( (F_x, F_y, F_z) = (1, -3, 1) \). 4. The equation of the tangent plane at this point can be formulated as: \[ F_x (x - 3) + F_y (y - 1) + F_z (z - 1) = 0 \] Substituting values yields: \[ 1(x - 3) - 3(y - 1) + 1(z - 1) = 0 \] Simplifying this expression gives: \[ z - x + 3y = 1 \] Thus, the answer is \( \text{d. } z - x + 3y = 1 \).