Second directional derivative of f(x,y) is $D^2_uf(x,y) = D_u[D_uf(x,y)]$
If f(x,y)= $ \frac{5}{3}x^3 + 5x^2y - 5y^3 $ and $u=\langle\frac35, \frac45\rangle $ Find $D^2_uf(2,1) $.
My work: $ gradient\nabla f(x,y)=\langle\frac{15}{3}x^2 + 10xy, 5x^2-15y^2\rangle$
$D_uf(x,y)=\langle\frac{15}{3}x^2 + 10xy, 5x^2-15y^2\rangle \cdot \langle\frac{3}{5},\frac{4}{5}\rangle $
$=\frac{1}{5}(15x^2+30xy+20x^2-60y^2)$ $=\frac{1}{5}(35x^2+30xy-60y^2)$ $=7x^2 +6xy-12y^2$ I'm stuck. I never did this question before.
$D^2_uf(x,y)=\frac{3}{5}(14x+6y)+\frac{4}{5}(6x-24y)$ $=\frac{66}{5}x -\frac{78}{5}y$
$D^2_uf(2,1) = \frac{132}{5} -\frac{78}{5} = \frac{54}{5} $
