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I came across the following problem that says:

Consider the map $f \colon \Bbb R^2 \rightarrow \Bbb R^2$ defined by $$f(x,y)=(3x-2y+x^2,4x+5y+y^2)$$ Then I have to determine whether the following statements are true or not?
1. $f$ is continuous at $(0,0)$ and all directional derivatives exist at $(0,0).$
2. $f$ is differentiable at $(0,0)$ and the derivative $Df(0,0)$ is invertible.

The problem is that I can not compute $Df(0,0)$.Can someone provide me the formula by means of which I can compute it. With regards and thanks in advance for your time.

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  • $\begingroup$ $f$ is not from $\Bbb {R}$ to $\Bbb {R}$, instead, $f$ is from $\Bbb {R}^2$ to $\Bbb {R}^2$, i.e $f \colon \Bbb {R}^2 \rightarrow \Bbb {R}^2$ $\endgroup$
    – Paul
    Commented Mar 24, 2013 at 5:36
  • $\begingroup$ @Paul yes.You are right. Thanks for pointing out. $\endgroup$
    – learner
    Commented Mar 24, 2013 at 7:01

1 Answer 1

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Denote $$f_1(x,\; y)=3x-2y+x^2,\\ f_2(x,\; y)=4x+5y+y^2.$$ Then $$Df(x,\;y)=\begin{pmatrix} \dfrac{\partial{f_1(x,\; y)}}{\partial{x}} && \dfrac{\partial{f_1(x,\; y)}}{\partial{y}} \\ \dfrac{\partial{f_2(x,\; y)}}{\partial{x}} && \dfrac{\partial{f_2(x,\; y)}}{\partial{y}} \end{pmatrix}$$

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