Tangent Planes & Linear Approximation - Calculus 3

The surface $z = x^2 + 2xy - y^2 + 3$ has a tangent plane at the point $(1, -1, 2)$ . A particle moves along this tangent plane from the point $(1, -1, 2)$ to the point $(1.1, -0.9, h)$ . What is the value of $h$ ?
Answer: $h = 2.4$ First, find the partial derivatives: $f_x = 2x + 2y$ and $f_y = 2x - 2y$ . At $(1, -1)$ : $f_x(1, -1) = 2(1) + 2(-1) = 0$ and $f_y(1, -1) = 2(1) - 2(-1) = 4$ . The tangent plane equation is $z - 2 = 0(x - 1) + 4(y - (-1))$ , which simplifies to $z = 4y + 6$ . At the point $(1.1, -0.9)$ , the tangent plane gives $z = 4(-0.9) + 6 = 2.4$ .