Find the tangent plane at P = (x_{0}, y_{0}, z_{0}) = (1, 1, 5) to the graph of
f (x, y) = x² + 2xy + 2y²Because f (1, 1) = 5, P lies on the graph of f .We find that f^{\prime}_{1}(x, y) = 2x + 2y and f^{\prime}_{2}(x, y) = 2x + 4y. Hence, f^{\prime}_{1}(1, 1) = 4 and f^{\prime}_{2}(1, 1) = 6. Thus, Eq. (12.8.3) yields
z-z_{0}=f_{1}^{\prime}(x_{0},y_{0})(x-x_{0})+f_{2}^{\prime}(x_{0},y_{0})(y-y_{0}) (12.8.3)
z − 5 = 4(x − 1) + 6(y − 1)
or, equivalently, z = 4x + 6y − 5.