\be
\item Compute the gradient of the \CMC objective function as a function of
$\ds \vecx=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$.
\item Use Definition \ref{Th:diff_Gradient_diff_def} to show that
$\ds f(x_1,x_2)=x_1^3-2x_1-x_2$ is differentiable
at $\ds \vecx_0=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
\item Suppose $S \subseteq \R^n$, $f:S \to \R$ and $\vecx_0$ belongs
to the interior of $S$. If $\vecd$ is a nonzero vector in
$\R^n$, we define the
\emph{directional derivative of $f$\index{directional derivative} at
$\vecx_0$ in the direction of $\vecd$} as
\begin{equation}\label{E:Exercises_diff_gradient_direct_deriv_def}
f'_{\vecd}\left(\vecx_0\right)=\lim_{h \to 0^+} \frac{f\left(\vecx_0+h\vecd\right)-f(\vecx_0)}{h},
\end{equation}
provided this limit exists.
	\be
	\item Show that if $f$ is differentiable at $\vecx_0$, then
the limit in (\ref{E:Exercises_diff_gradient_direct_deriv_def})
exists and that
\begin{equation*}
f'_{\vecd}\left(\vecx_0\right)=\nabla f(\vecx_0)^t \vecd .
\end{equation*}
	\item Show that the preceding quantity $\ds f'_{\vecd}\left(\vecx_0\right)$ depends only upon
the direction of $\vecd$, not its magnitude. Hence, $\vecd$ may always be assigned a unit vector value.
\item For the \CMC objective function $f$, determine the directional derivative of $f$ in
the direction of the origin at $(x_1,x_2)=(10,20)$.
	\ee
\item Suppose $S \subseteq \R^n$, $f:S \to \R$, and $f$ is differentiable at $\vecx_1$ in $S$.
Fix $\vecx_2$ in $S$. (Note: The differentiability assumption implies $\ds \vecx_1+h\vecx_2$ also belongs to $S$ for $h$ sufficiently close to zero.)
Show that $\ds \phi(h)= f\left(\vecx_1+h\vecx_2\right)$ is differentiable at $h=0$ and satisfies
\begin{equation*}
\phi'(0)=\nabla f(\vecx_1)^t \vecx_2.
\end{equation*}
\ee
\endinput
