The value of $f_x((0,0))+f_y((0,0))+f_{xy}((0,0)) $ is?

Question

Let $f:\mathbb{R}^2 \Rightarrow \mathbb{R} $ defined by $$f((x,y)) = x^2 +2y^2 -3xy, \forall (x,y) \in \mathbb{R}^2$$ The value of $f_x((0,0))+f_y((0,0))+f_{xy}((0,0))$ is:

• $0$
• Undefined
• $-3$
• 3

I'm puzzled since in my opinion the value should be $0$, while the correct answer has been reported to be $-3$. To get $f_x$ you take the partial derivative for $x$ of $f_{xy}$ and the same for $y$. $$f(x) = \frac{\partial f(x,y)}{\partial x} \text{ and } f(y) = \frac{\partial f(x,y)}{\partial y}$$ So, if I haven't made my calculations wrong (they are fairly simple): $$ f(x) = 2x -3y \text{ and } f(y) = 4y -3x $$ $$ \Rightarrow f_x((0,0)) = 2\cdot0 -3\cdot0 = 0 \text{ and } f_y((0,0))= 4 \cdot 0 - 3 \cdot 0 = 0 \text{ and } f_{xy}((0,0)) = 0^2+2 \cdot0^2 -3 \cdot 0 \cdot 0 = 0$$ Each function is zero at $(0,0)$, or not? Am I missing something? The linear transformation shouldn't change anything since, by definition, the origin doesn't move.

Answer 1

The answer is $-3$.

Let $f:\begin{cases} \Omega\subseteq \mathbb{R}^{2}&\longrightarrow \mathbb{R},\\(x,y)&\longmapsto x^{2}+2y^{2}-3xy \end{cases}$ then clearly $f\in \mathcal{C}^{+\infty}(\Omega,\mathbb{R})$ and then we have

• $f_{x}(x,y)=2x-3y$ then $f_{x}(0,0)=2\cdot 0-3\cdot 0=0$ so $f_{x}(0,0)=0$.
• $f_{y}(x,y)=4y-3x$ then $f_{y}(0,0)=4\cdot 0-3\cdot 0=0$ so $f_{y}(0,0)=0$.
• $f_{xy}(x,y)=(f_{x})_{y}(x,y)=(2x-3y)_{y}=-3$ so $f_{xy}(0,0)=-3$.

Therefore, $$f_{x}(0,0)+f_{y}(0,0)+f_{xy}(0,0)=0+0-3=-3.$$

Answer 2

Note that $f_{xy}=-3$ since $f_x=2x-3y$ and then evaluating at $(0,0)$ gives $-3$.