\newsectionNoPB \subsection{The differential} \setLabelNumber{all}{2} \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 $\displaystyle \lim_{\elementstack{x \rightarrow x_0}{x \neq x_0}} \frac{f(x) - f(x_0) - u(x - x_0)}{||x - x_0||} = 0$ 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 $X$. \newpage % ──────────────────────────────────────────────────────────────────── \stepLabelNumber{all} \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 % ──────────────────────────────────────────────────────────────────── \stepLabelNumber{all} \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 % ──────────────────────────────────────────────────────────────────── \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 \bi{differential is the Jacobi Matrix of $f$ at $x_0$}. This implies that most elementary functions are differentiable.\\ % ──────────────────────────────────────────────────────────────────── \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, i.e. multiply rows of first with cols of second matrix)\\ % ──────────────────────────────────────────────────────────────────── \setLabelNumber{all}{11} \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*} \dnrmvspace % ──────────────────────────────────────────────────────────────────── \stepLabelNumber{all} \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 $w$ % ──────────────────────────────────────────────────────────────────── \shortremark Because $X$ is open, the set $I$ contains an open interval $]-\delta, \delta[$ for some $\delta > 0$. % ──────────────────────────────────────────────────────────────────── \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$\dx f(x_0)(v)$ % ──────────────────────────────────────────────────────────────────── \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 $w_1 + w_2$ \shade{gray}{Computing a directional derivative} Always normalize the vector! We can compute a directional derivative using the differential $\limit{h}{0} \frac{f(x_0 + hv) - f(x_0)}{h}$ or using a $1$-dimensional helper function $g: h \mapsto f(x_0 + hv)$, calculating the derivative of it and evaluating $g'(0)$. That corresponds to the directional derivative. E.g. for function $f: x, y \mapsto x^2 + y^2$, we have $g: h \mapsto (x_0 + h)^2 + (y_0 + h)^2$. A final option is to compute it using a matrix-vector product: $D_v f(x_0) = J_f(x_0) v$ \rmvspace \begin{center} \begin{tikzpicture}[node distance = 0.5cm and 0.5cm, >={Classical TikZ Rightarrow[width=7pt]}] \node (contdiff) {$f$ cont. diff.}; \node (diff) [below=of contdiff] {$f$ differentiable}; \node (cont) [below=of diff] {$f$ continuous}; \node (diffcont) [right=of contdiff] {All $\partial_j f_i$ continuous}; \node (diffex) [below=of diffcont] {All $\partial_j f_i$ exist}; \node (notes) at (-5, -1) {Red arrows indicate no implication}; \draw[arrows = ->, double distance = 1.5pt] (contdiff) -- (diff); \draw[arrows = ->, double distance = 1.5pt] (diff) -- (cont); \draw[arrows = <->, double distance = 1.5pt] (contdiff) -- (diffcont); \draw[arrows = ->, double distance = 1.5pt] (diffcont) -- (diffex); \draw[arrows = ->, double distance = 1.5pt, transform canvas={yshift=0.2cm}] (diff) -- (diffex); \draw[arrows = ->, double distance = 1.5pt, transform canvas={yshift=-0.2cm}, color=red] (diffex) -- (diff); \draw[arrows = ->, double distance = 1.5pt, color=red] (diffex) -- (cont); \end{tikzpicture} \end{center} \drmvspace