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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?

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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.

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  • $\begingroup$ I'm just curious...are my calculations correct so far? $\endgroup$ – emka Apr 8 '13 at 8:00
  • $\begingroup$ The calculations in the OP's question look correct, but in the final expression, one would need a find a way to recognise the numerator as (bounded by) something divisible by the denominator, so that the resulting quotient clearly goes to zero as $(x,y) \to (1,-1)$, and that doesn't look easy to me at all. $\endgroup$ – Branimir Ćaćić Apr 8 '13 at 8:05

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