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{\bf Question}

Verify that $f_{xy} = f_{yx}$ for:
\begin{description}
\item[(a)] $f(x,y) = \exp\left( -(x^2 + y^2 - 0.25)^2\right)$
\item[(b)] $f(x,y) = x^2 \cos(x + 2y) + y^2 \cos(2x +y)$
\end{description}
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{\bf Answer}



\begin{description}


\item[(a)]


\begin{eqnarray*} h  & = &\exp\left( -(x^2 + y^2 - 0.25)^2\right) \\
\frac{\partial h}{\partial x} & = & -4x(x^2 + y^2 -0.25)\exp\left(
-(x^2 + y^2 - 0.25)^2\right) \\ \frac{\partial h}{\partial y} & =
& -4y(x^2 + y^2 -0.25) \exp\left( -(x^2 + y^2 -
0.25)^2\right)\end{eqnarray*}

$\displaystyle\frac{\partial^2 h}{\partial x
\partial y}  =  \frac{\partial}{\partial x}\left[ \frac{\partial
h}{\partial y}\right]$

$ = \exp\left( -(x^2 + y^2 - 0.25)^2\right) ( -8xy + 4x(x^2 +
y^2-0.25)^2 \times 2 \times 2y )$

$\displaystyle\frac{\partial^2 h}{\partial y\partial x}  =
\frac{\partial}{\partial y}\left[ \frac{\partial h}{\partial
x}\right]$

$ =  \exp\left( -(x^2 + y^2 - 0.25)^2\right) ( -8xy + 4x(x^2 +
y^2-0.25)^2 \times 2 \times 2y )$


So $\displaystyle\frac{\partial^2 h}{\partial x \partial y} =
\frac{\partial^2 h}{\partial y \partial x}$ as expected


\item[(b)]


\begin{eqnarray*}
f(x,y) & = & x^2 \cos(x + 2y) + y^2 \cos (2x + y) \\
\frac{\partial f}{\partial x} & = & 2x \cos(x+ 2y) - x^2 \sin (x +
2y) - 2y^2 \sin (2x + y) \\ \frac{\partial f}{\partial y} & = &  -
2 x^2 \sin (x + 2y) - 2y \cos(2x+ y) + y^2 \sin (2x +
y)\end{eqnarray*}

$\displaystyle\frac{\partial^2 f}{\partial y
\partial x} =\frac{\partial}{\partial y}\left[ \frac{\partial
f}{\partial x}\right]$

$ =  -4x \sin(x + 2y) - 2x^2 \cos (x + 2y) - 4y \sin (2x + y) -
2y^2 \cos (2x + y)$

$\displaystyle\frac{\partial^2 f}{\partial x \partial y} =
\frac{\partial}{\partial x}\left[ \frac{\partial f}{\partial
y}\right]$

$  =  -4x \sin(x + 2y) - 2x^2 \cos (x + 2y) - 4y \sin (2x + y) -
2y^2 \cos (2x + y)$


So $\displaystyle\frac{\partial^2 f}{\partial x \partial y} =
\frac{\partial^2 f}{\partial y \partial x}$  as expected
\end{description}



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