Find the directional derivative?

Question

Find the directional derivative of f(x, y)=x^2*y+4y^2 at (2, 1) for u=<1/2, sqrt(3)/2>.

Answer: 2+6sqrt(3)

The gradient is <2xy, 8y> at (2, 1)=<4, 8>

And <4, 8>*<1/2, sqrt(3)/2>=2+4sqrt(3)

This doesn't match the answer in the book.

Please check my work and correct me through steps.

Answer 1

Your work is correct by all accounts except for the gradient of your function.

The gradient is supposed to be <2xy, x^2+8y> because of the y term in the function (x^2*y + 4y^2)

<2xy, x^2+8y> @ (2,1) = <4,12>

<4,12>*<1/2, sqrt(3)/2> = 2 + 12sqrt(3)/2 = 2 + 6sqrt(3)

hope that helped