<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\setlength{\textwidth}{6.5in} \setlength{\textheight}{9in}
\setlength{\topmargin}{-.25in} \setlength{\oddsidemargin}{-.25in}
\newcommand\br{\textbf{R}}
\pagestyle{empty}
\begin{document}

\parindent=0pt

\bf{Question}

Write down the Jacobian matrix $DF(p)$ for
$F:\mathbf{R}^n\rightarrow\mathbf{R}^m$ at a typical point
$p\in\mathbf{R}^n$:

\begin{description}

\item[(i)]
$F:\mathbf{R}^2\rightarrow\mathbf{R}^3\ F(x_1,x_2)=(x_2^2+2x_2,
2\sin x_1x_2,(x_1-x_2)^2)$

\item[(ii)]
$F:\mathbf{R}^3\rightarrow\mathbf{R}^2\
F(x_1,x_2,x_3)=(x_1+2x_2+3x_3, 4x_1+5x_2+6x_3)$

\item[(iii)]
$F:\mathbf{R}^2\rightarrow\mathbf{R}\
F(x_1,x_2)=2x_1^2+x_1x_2-x_2^2.$

\end{description}



\bf{Answer}

\begin{description}
\item{(i)}

$$DF(p) = \left ( \begin{array}{cc} 0 &amp; 2x_2+2\\ 2x_2 \cos x_1x_2
&amp; 2x_1 \cos x_1 x_2\\ 2(x_1-x_2) &amp; -2 (x_1-x_2)
\end{array} \right ), \ \ \ p=(x_1,x_2).$$

\item{(ii)}

$$DF(p) = \left ( \begin{array}{ccc} 1 &amp; 2 &amp; 3 \\ 4 &amp; 5 &amp; 6
\end{array} \right ), \ \ \ p=(x_1, x_2, x_3).$$

[Here $F$ is linear and is therefore its own derivative: the same
at every point $p$.]

\item{(iii)}

$$DF(p) = ( 4x_1 + x_2, x_1 - 2x_2) \ \ (=dF(p)), \ \ \
p=(x_1,x_2).$$
\end{description}


\end{document}
</pre></body></html>