How do I go about finding the directional derivative of?

Question

g(x, y) = 2*x*y + y^4 @ (1, 1) in the direction u = (4, -3)

and also the directional derivative of g(x, y) = x^2 - 2*x*y^2 @ (2,1) along a line which makes an angle of (pi/3) to the + x axis.

Thanks in advance.
Dave

Answer 1

The directional derivative is just
∇g . u
= (∂g/∂x, ∂g/∂y) . (4, -3)
= (2y, 2x + 4y^3) . (4, -3)
= (2, 6) . (4, -3)
= -10.

For the second one we have ∇g = (2x - 2y^2, -4xy) = (2, -8) and the directional derivative is
(2, -8) . (cos π/3, sin π/3)
= (2, -8) . (1/2, √3/2)
= 1 - 4√3.