\documentclass[a4paper,12pt]{article}
\setlength\oddsidemargin{0pt} \setlength\evensidemargin{0pt}
\setlength\topmargin{0pt}
\begin{document}
\parindent=0pt
{\bf Exam Question

Topic: Chain Rule}

Find the partial derivatives $\displaystyle\frac{\partial
z}{\partial x}$ and $\displaystyle\frac{\partial z}{\partial y}$
where $$z=\sin(xy)+x\cos y.$$

A change of variables is specified by means of the equations
$$x=u^2+v^2,\ \ y=uv.$$ Use the chain rule to find
$\displaystyle{\frac{\partial z}{\partial u}}$ in terms of $u$ and
$v$.

Calculate the value of this partial derivative when $u=1,\ v=0.$

 \vspace{0.5in}

{\bf Solution}

\begin{eqnarray*}
z&=&\sin(xy)+x\cos y\\ \displaystyle\frac{\partial z}{\partial
x}&=&y\cos(xy)+\cos y\\ \displaystyle\frac{\partial z}{\partial y}
&=& x\cos(xy)-x\sin y
\end{eqnarray*}

The chain rule gives
\begin{eqnarray*}
\displaystyle\frac{\partial z}{\partial
u}&=&\displaystyle\frac{\partial z}{\partial
x}\displaystyle\frac{\partial x}{\partial
u}+\displaystyle\frac{\partial z}{\partial
y}\displaystyle\frac{\partial y}{\partial u}\\ &=&
\left(uv\cos\left((u^2+v^2)uv\right)+\cos(uv)\right).2u\\ &+&
\left((u^2+v^2)\cos\left((u^2+v^2)uv\right)-(u^2+v^2)\sin(uv)\right).v
\end{eqnarray*}

When $u=1,\ v=0$ the value of this expression is 2.

\end{document}
