Directional Derivatives?

The easy way to find a directional derivative is to take the dot product of the gradient with a unit vector in the desired direction.

Your unit direction vector is û = <cos 5π/4, sin 5π/4> = <-1/√2, -1/√2>.

The gradient of z(x,y) is:

     ∇z = <∂z/∂x , ∂z/∂y>

...with:

    ∂z/∂x = e^(2x - 3y) + (x) (2) e^(2x - 3y) = (1 + 2x) e^(2x - 3y)
    ∂z/∂y = (-3x) e^(2x - 3y)
    ∇z = <1 + 2x, -3x> e^(2x - 3y)      [after factoring out that common exponential]

So the direction derivative in the direction of û is

    D_û z(x,y) = ∇z • û = [-(1 + 2x)/√2 - (-3x)/√2] e^(2x - 3y)
                      = (-1 - 5x) e^(2x - 3y) / √2

At the point (x,y) = (1,-2), that's 9 e^8 / √2....Show more