\newsection
\subsection{The differential}
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\compactdef{Differentiable function} We have function $f: X \rightarrow \R^m$, linear map $u : \R^n \rightarrow \R^m$ and $x_0 \in X$. $f$ is differentiable at $x_0$ with differential $u$ if
$\displaystyle \lim_{\elementstack{x \rightarrow x_0}{x \neq x_0}} \frac{1}{||x - x_0||} (f(x) - f(x_0) - u(x - x_0) = 0$ where the limit is in $\R^m$.
We denote $\dx f(x_0) = u$.
If $f$ is differentiable at every $x_0 \in X$, then $f$ is differentiable on $X$.
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\shortproposition
\rmvspace Let $f: X \rightarrow \R^m$ be differentiable on $X$
\begin{itemize}[noitemsep]
    \item $f$ is continuous on $X$
    \item $f$ admits partial derivatives on $X$ with respect to each variable
    \item Assume $m = 1$, let $x_0 \in X$ and let $u(x_1, \ldots, x_n) = a_1 x_1 + \ldots + a_n x_n$ be diff. of $f$ at $x_0$.
          Then $\partial_{x_i} f(x_0) = a_i$ for $1 \leq i \leq n$
\end{itemize}
\rmvspace
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\shortproposition Let $f, g : X \rightarrow \R^m$ with $X \subseteq \R^n$ open
\rmvspace
\begin{itemize}[noitemsep]
    \item The function $f + g$ is differentiable with differential $\dx (f + g) = \dx f + \dx g$. If $m = 1$, then $fg$ is differentiable
    \item If $m = 1$ and if $g(x) \neq 0 \forall x \in X$, then $f \div g$ is differentiable
\end{itemize}
\rmvspace
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\shortproposition If $f$ as above has all partial derivatives on $X$ and if they are all continuous on $X$, then $f$ is differentiable on $X$.
The differential is the Jacobi Matrix of $f$ at $x_0$.
This implies that most elementary functions are differentiable.\\
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\compactproposition{Chain Rule} For $X \subseteq \R^n$ and $Y \subseteq \R^m$ both open and $f: X \rightarrow Y$ and $g : Y \rightarrow \R^p$ are both differentiable.
Then $g \circ f$ is differentiable on $X$ and for any $x \in X$, its differential is given by
$\dx (g \circ f)(x_0) = \dx g(f(x_0)) \circ \dx f(x_0)$.
The Jacobi matrix is $J_{g \circ f}(x_0) = J_g(f(x_0)) J_f(x_0)$ (RHS is a matrix product)\\
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\compactdef{Tangent space} The graph of the affine linear approximation $g(x) = f(x_0) + u(x - x_0)$, or the set
\vspace{-0.75pc}
\begin{align*}
    \{ (x, y) \in \R^n \times \R^m : y = f(x_0) + u(x - x_0) \}
\end{align*}

\drmvspace\rmvspace
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\compactdef{Directional derivative} $f$ has a directional derivative $w \in \R^m$ in the direction of $v \in \R^n$,
if the function $g$ defined on the set $I = \{ t \in \R : x_0 + tv \in X \}$ by $g(t) = f(x_0 + tv)$ has a derivative at $t = 0$ and is equal to $w$
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\shortremark Because $X$ is open, the set $I$ contains an open interval $]-\delta, \delta[$ for some $\delta > 0$.
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\shortproposition Let $f$ as previously be differentiable. Then for any $x \in X$ and non-zero $v \in \R^n$,
$f$ has a directional derivative at $x_0$ in the direction of $v$, given by$\dx f(x_0)(v)$
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\shortremark The values of the above directional derivative are linear with respect to the vector $v$.
Suppose we know the dir. der. $w_1$ and $w_2$ in directions $v_1$ and $v_2$, then the directional derivative in direction $v_1 + v_2$ is $w_1 + w_2$
