<s>[INST] Showing a function is differentiable using definition of derivative Let $f: \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $f(x,y)=(x^2-xy, x+y^2)$.  Use the definition of the derivative of a function to show that $f$ is differentiable at the point $p=(1,-1)$. 
My attempt is as follows:
Definition of derivative: $\lim\limits_{x \to p}\frac{|f(x)-f(p)-df_p(x-p)|}{|x-p|}$.  So,
$$\lim\limits_{(x,y)\to(1,-1)}\frac{|f((x,y))-f((1,-1))-df|_{(1,-1)}((x,y)-(1,-1)}{|(x,y)-(1,-1)|}.$$
Where $df|_{(1,-1)}=\begin{bmatrix}
                     3 & -1 \\
                     1 & -2 
                    \end{bmatrix}.$
Then, we have:
$$\lim\limits_{(x,y) \to (1,-1)} \frac{|<x^2-xy, x+y^2>-<2,2>-<3x-y-4, x-2y-3>|}{|<x-1, y+1>|}$$
Simplified a bit:
$$\lim\limits_{(x,y) \to (1,-1)} \frac{|<x^2-xy-3x+y+2, x+y^2-x+2y+1>|}{|<x-1, y+1>|} $$
$$\lim\limits_{(x,y) \to (1,-1)} \frac{\sqrt{(x^2-xy-3x+y+2)^2+(x+y^2-x+2y+1)^2}}{\sqrt{(x-1)^2+(y+1)^2}}$$
I still noticed that the bottom goes to zero.  Is there something that I am messing up computationally?
 [/INST] It's easier to show that
$$
 \lim_{h \to 0}\frac{\|f((1,-1)+h) - f(1,1) - df_{(1,-1)}(h)\|}{\|h\|} = 0.
$$
Indeed, for $h = (h_1,h_2) \in \mathbb{R}^2$,
$$
 f((1,-1)+h) - f(1,1) - df_{(1,-1)}(h)\\
= ((1+h_1)^2-(1+h_1)(-1+h_2),(1+h_1)+(-1+h_2)^2) - (1^2-1(-1),1+(-1)^2) - (3h_1 -h_2,h_1-2h_2)\\
= (h_1^2 -h_1 h_2+3h_1-h_2+2,h_2^2+h_1-2h_2+2) -(2,2)-(3h_1-h_2,h_1-2h_2)\\
= (h_1^2-h_1h_2,h_2^2) = (h_1(h_1-h_2),h_2^2),
$$
so that
$$
 \|f((1,-1)+h) - f(1,1) - df_{(1,-1)}(h)\|^2 = \|(h_1(h_1-h_2)h_2,h_2^2)\|^2\\
= h_1^2(h_1-h_2)^2 + h_2^4 = h_1^2(h_1^2-2h_1h_2+h_2^2) + h_2^4\\
\leq h_1^2(h_1^2 + (h_1^2+h_2^2)+h_2^2) + h_2^4\\
= 2h_1^4 + 2h_1^2h_2^2 + h_2^4 \leq 2(h_1^2+h_2^2)^2 = 2\|h\|^4,
$$
and hence
$$
 \lim_{h \to 0}\frac{\|f((1,-1)+h) - f(1,1) - df_{(1,-1)}(h)\|}{\|h\|}\\
\leq \lim_{h \to 0} \frac{\sqrt{2}\|h\|^2}{\|h\|} = \sqrt{2}\lim_{h\to 0}\|h\| = 0,
$$
as required.
</s>