[math] \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow} \newcommand{\ddxof}[1]{\frac{d #1}{d x}} \newcommand{\ddx}{\frac{d}{dx}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\dydx}{\ddxof y} \newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}} \newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}} \newcommand{\dist}{\mathrm{distance}} \newcommand{\arccot}{\mathrm{arccot\:}} \newcommand{\arccsc}{\mathrm{arccsc\:}} \newcommand{\arcsec}{\mathrm{arcsec\:}} \newcommand{\arctanh}{\mathrm{arctanh\:}} \newcommand{\arcsinh}{\mathrm{arcsinh\:}} \newcommand{\arccosh}{\mathrm{arccosh\:}} \newcommand{\sech}{\mathrm{sech\:}} \newcommand{\csch}{\mathrm{csch\:}} \newcommand{\conj}[1]{\overline{#1}} \newcommand{\mathds}{\mathbb} [/math]

There are many examples of a function [math]f[/math] and a number [math]a[/math] such that [math]f(a)[/math] is defined ([math]a[/math] is in the domain of [math]f[/math]) but [math]f^\prime(a)[/math] does not exist. Another way of saying the same thing is that the domain of [math]f^\prime[/math] can be a ph proper subset of the domain of [math]f[/math]. It is equally possible for [math]f^\prime (a)[/math] to be defined and [math]f^{\prime\prime} (a)[/math] not to be. Let [math]f[/math] be the function defined by

[[math]] f(x) = \dilemma{\frac{x^2}2 & \mbox{if $x\geq 0$,}} {-\frac{x^2}2 & \mbox{if \ltmath\gtx\leq 0[[/math]]

.}} </math>

• Compute [math]f^\prime[/math].
• Is [math]f[/math] a differentiable function?
• Show that [math]f^{\prime\prime}(0)[/math] does not exist, and compute [math]f^{\prime\prime}(x)[/math] for [math]x\ne 0[/math].