# Higher order derivative From $f:\mathbb{R}^2 \to \mathbb{R}^2$

If function $f:\mathbb{R}^2 \to \mathbb{R}^2,\;(x,y) \mapsto (x^2+y,y^2+x)$ Then What about continuity and Derivative at $(0,0)$? And how to find $Df(x,y)$ ?

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Looks like homework... –  Alexei Averchenko Oct 2 '12 at 12:54

You can just compute the partial derivatives and apply theorems known to you ;-) ... or you can have fun and apply directly the definition:

Given $x \in \mathbb R^2$, in order to find $Df(x)$ we have to find a linear $A\colon \mathbb R^2 \to \mathbb R^2$ such that $f(x+h) = f(x) + Ah + o(h), \quad h \to 0$ Now, \begin{align*} f(x+h) &= \bigl((x_1 + h_1)^2 + x_2 + h_2, (x_2 + h_2)^2 + x_1 + h_1\bigr)\\ &= (x_1^2 + x_2, x_2^2 + x_1) + (2x_1h_1 + h_2, 2x_2h_2 + h_1) + (h_1^2, h_2^2)\\ &=: f(x) + Ah + (h_1^2, h_2^2) \end{align*} We have $\frac{\|(h_1^2, h_2^2)\|}{\|h\|} \le 2\frac{\|h\|^2}{\|h\|} \to 0, \quad h \to 0$ and therefore $(h_1^2, h_2^2) = o(h), h \to 0$. Hence $f$ is differentiable in $x$ with $f'(x)h = (2x_1h_1 + h_2, 2x_2 + h_1)$ As the title of your question asks for higher order derivatives, we will compute $f''(x)$. So let $h, k \in \mathbb R^2$, then \begin{align*} f'(x+k)h &= \bigl(2(x_1 + k_1)h_1 + h_2, 2(x_2 + k_2)h_2 + h_2\bigr)\\ &= f'(x)h + (2k_1h_1, 2k_2h_2)\\ &=: f'(x)h + \bigl(f''(x)k\bigr)h \end{align*} There is no nonlinear term, so $f$ is twice differentiable. As $f''\colon \mathbb R^2 \to L\bigl(\mathbb R^2, L(\mathbb R^2, \mathbb R^2)\bigr)$ is constant, $f$ is smooth with $f^{(n)} = 0$ for $n \ge 3$. So $f$ is also continuous.

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