Find the directional derivative?

Question

Find the directional derivative of f(x, y)=xye^(-xy^2) at the point (1, 1) in the direction <2/sqrt(5), 1/sqrt(5)>.

f=<ye^(-y^2), xe^(-2xy)>

f(1, 1)=<e, e^-2>

u=<2/sqrt(5), 1/sqrt(5)>/(sqrt(1^2+1^2))

=<2/sqrt(5), 1/sqrt(5)>/sqrt(2)

<e, e^-2>*<2/sqrt(5), 1/sqrt(5)>/sqrt(2)

=(2e/sqrt(5)-e^-2/sqrt(5))/sqrt(2)

=(2e/sqrt(10)-e^-2/sqrt(10))

I don't know how to simplify from here. Correct me if I'm wrong through steps.

Answer 1

By the definition, the derivative of f(x,y) along the direction defined by components u_x =2/√5 and u_y =1/√5, is equal to the dot product  ∇f • u, which is in componental form looks like this:

                                         ∇f • u = (∂f/∂x) u_x + (∂f/∂y)u_y

Take the derivatives

                   ∂f/∂x = yexp(-xy²)(1-xy²)               ∂f/∂y = xexp(-xy²)(1-2xy²)

For x=y=1 it makes   ∂f/∂x = 0, ∂f/∂y = -1/e. Thus your answer is

                                  ∇f • u = (∂f/∂y)u_y (1,1) = -1/(√5)e  

                            

                     

Answer 2

The directional derivative of f(x, y) = xye^(-xy^2) is

∇f • u, where u = <2/sqrt(5), 1/sqrt(5)>

= <e^(-xy^2) (y - xy^3), e^(-xy^2) (x - 2x^2 y^2)> • u

= <0, -1/e> • <2/sqrt(5), 1/sqrt(5)>, at (1, 1)

= -sqrt(5)/(5e) <==Answer