If g(x, y) = x² y² − 4x, find the gradient vector ∇g(1, 4) and use it to find the tangent line to the level curve g(x, y) = 13 at the point (1, 4).
a) ∇g(1, 4) = (0, 0); Tangent line: y = 4x + 9
b) ∇g(1, 4) = (4, 8); Tangent line: y = 8x + 3
c) ∇g(1, 4) = (0, 8); Tangent line: y = 8x - 3
d) ∇g(1, 4) = (4, 0); Tangent line: y = 4x - 9