Second partial derivative with $f(x,y)=x^3+5x^2y+y^3$

Question

I have this problem:

I found $F_x=3x^2+10yx$ and $F_y=5x^2+3y^2$, then $D_uf=\frac{9}{5}x^2+6yx+4x^2$

$F_{x2}= \frac{58}{5}x + 6y, F_{y2}=6x+{24}{5}y$

$D_uf_2=\frac{174}{25}x+\frac{18}{5}y+\frac{32}{5}x+\frac{96}{25}y=>at(2,1)=>534/25$

The answer however is $\frac{774}{25}$

Answer 1

Your mistake was at the first derivative, you forgot to add $\frac{12}5y^2$. See this full calculation:

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$$f(x,y) = x^3+5x^2y+y^3$$

$$\nabla f = \left(3x^2+10xy \atop 5x^2+3y^2 \right) $$

$$\nabla f \cdot u = \left(3x^2+10xy \atop 5x^2+3y^2 \right)\cdot \left(\frac 3 5 \atop \frac 4 5\right) =\frac {29} 5x^2+6xy+ \frac{12}5y^2 $$

$$\nabla \left(\frac {29} 5x^2+6xy+ \frac{12}5y^2\right)\cdot u = \left( \frac{58}5x+6y\atop6x+\frac{24}5 y\right) \cdot \left(\frac 3 5 \atop \frac 4 5\right) $$ $$ = \frac{174}{25}x+\frac{18}{5}y+\frac{24}5x + \frac{96}{25}y$$ $$ = \frac{294}{25}x + \frac{186}{25}y $$ Evaluating at $(2,1)$ gives: $$\left(\frac{294}{25}x + \frac{186}{25}y\right)\left(2,1\right) =\frac{588}{25}+\frac{186}{25}=\frac{775}{25}$$

Answer 2

If $f(x,y)= x^3 + 5x^2y + y^3 $ and $u=\langle\frac35, \frac45\rangle $

$$\nabla f(x,y)=\langle 3 x^2 + 10xy, 5x^2 +3y^2\rangle$$

$$D_uf(x,y)=\langle 3 x^2 + 10xy, 5x^2 +3y^2\rangle \cdot \langle\frac{3}{5},\frac{4}{5}\rangle =\frac{1}{5}(29x^2 + 30xy +12y^2) $$

$$D^2_uf(x,y)=\frac{3}{5}\underbrace{\frac{1}{5}(58x+30y)}_{\frac{\partial D_uf(x,y)}{\partial x}}+\frac{4}{5}\underbrace{\frac{1}{5}(30x + 24y)}_{\frac{\partial D_uf(x,y)}{\partial y}} = \frac{1}{25}(174x + 90y + 120x + 96y) = \frac{294x+186y}{25}$$ which gives $$D^2_uf(2,1)=\frac{774}{25}$$