Linear transformation from $R^2$ to $R^2$.

Question

Let $\vec{f}: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, where $\vec{f} (\vec{x}) = (x+y^2, x^3+5y)$ and $\vec{x} = (x,y) \in \mathbb{R}^2$. Let $\vec{h} = (h_1, h_2)$ and $\vec{a} = (1,1) \in \mathbb{R}^2$. How do I show that the derivative of $\vec{f}$ at $\vec{a}$ is the map

$A\vec{h} := \vec{f}' (\vec{a}) \vec{h} = (h_1 + 2h_2, 3h_1 + 5h_2)$?

Answer 1

\begin{eqnarray*} \mathbf{f(x)} &=&(x+y^{2},x^{3}+5y) \\ \partial _{x}\mathbf{f(x)} &=&(1,3x^{2}) \\ \partial _{y}\mathbf{f(x)} &=&(2y,5) \\ \partial _{\mathbf{x}}\mathbf{f(x)} &=&\left( \begin{array}{cc} 1 & 2y \\ 3x^{2} & 5% \end{array}% \right) \\ (\partial _{\mathbf{x}}\mathbf{f})\mathbf{(a)} &=&\left( \begin{array}{cc} 1 & 2 \\ 3 & 5% \end{array}% \right) \\ \{(\partial _{\mathbf{x}}\mathbf{f)(a)\}h} &=&\left( \begin{array}{cc} 1 & 2 \\ 3 & 5% \end{array}% \right) \left( \begin{array}{c} h_{1} \\ h_{2}% \end{array}% \right) =\left( \begin{array}{c} h_{1}+2h_{2} \\ 3h_{1}+5h_{2}% \end{array}% \right) \end{eqnarray*}