Derivata vt limes fvnctionis

Svmmarivm

Haec lectio derivatam fvnctionis in pvncto vt limitem definit: limitem declivitatvm rectarvm secantivm, qvi est declivitas rectae tangentis, et limitem velocitatvm mediarvm, qvi est velocitas momentanea. Accvrate statvit in qvibvs pvnctis definitio sensvm habeat, derivatas laterales, rectas tangentem et normalem, fvnctionem derivatam notationemqve Leibnitianam introdvcit, et demonstrat derivatam tres formvlationes aeqvivalentes admittere: limitem qvotientis incrementalis, approximationem per rectam cvm errore neglegibili et factorizationem Carathéodorii. Ex hac vltima colligitvr omnem fvnctionem derivabilem continvam esse; valor absolvtvs et radix cvbica ostendvnt conversam falsam esse.

Proposita discendi

Derivatam fvnctionis in pvncto vt limitem qvotientis incrementalis definire, eamqve vt declivitatem rectae tangentis et vt velocitatem momentaneam interpretari.
Derivatas laterales compvtare et per eas decernere vtrvm fvnctio in pvncto derivabilis sit.
Aeqvivalentiam inter qvotientem incrementalem, approximationem linearem et characterismvm Carathéodorii demonstrare.
Demonstrare omnem fvnctionem in pvncto derivabilem ibi continvam esse, et fvnctiones continvas agnoscere qvae derivabiles non svnt.

Recta tangens et velocitas

In lectione de parabolis recta cvrvae in pvncto tangens est cvm eam in eo solo attingit neqve traicit. Ea notio conicis svfficit, non avtem cvilibet cvrvae. In origine cvrva $y = x^3$ a sinistra infra axem horizontalem iacet, a dextra svpra: omnis recta per originem dvcta eam traicit, et rectae $y = mx$ cvm $m < 0$ eam in sola origine attingvnt, qvia $x^3 = mx$ postvlat $x = 0$ vel $x^2 = m$ . Nvlla recta condicioni satisfacit, et tamen axis horizontalis, vt ocvlis apparet, est recta qvae prope originem cvrvam optime comitatvr. Alia igitvr definitio reqviritvr, qvae non a forma globali cvrvae pendeat, sed solvm ab eo qvod prope pvnctvm accidit.

Notio antiqva est: recta tangens est positio ad qvam rectae secantes tendvnt. Consideremvs fvnctionem $f$ et pvnctvm $(a, f(a))$ delineationis eivs. Pro $h \neq 0$ , recta secans qvae per $(a, f(a))$ et per $(a + h, f(a + h))$ transit declivitatem habet

\frac{f(a + h) - f(a)}{h}

Cvm $h$ ad $0$ accedit, altervm pvnctvm ad primvm accedit, et si declivitates limitem habent, recta qvae per $(a, f(a))$ cvm ea declivitate transit candidata est ad rectam tangentem.

ExemplvmSecantes parabolae

Sint $f(x) = x^2$ et $a = 1$ . Declivitates qvarvndam secantivm per pvnctvm $(1, 1)$ compvtamvs, deinde earvm limitem. Figvra singvlas secantes ostendit et, in fine, rectam tangentem.

Demonstratio

$\frac{f(2) - f(1)}{1} = \frac{4 - 1}{1} = \resaltar{3}$
Cvm $h = 1$ , secans per $(1, 1)$ et $(2, 4)$ transit.
$\frac{f\left(\frac{3}{2}\right) - f(1)}{\frac{1}{2}} = \frac{\frac{9}{4} - 1}{\frac{1}{2}} = \resaltar{\frac{5}{2}}$
Cvm $h = \frac{1}{2}$ , altervm pvnctvm est $\left(\frac{3}{2}, \frac{9}{4}\right)$ et declivitas decrescit.
$\frac{(1 + h)^2 - 1}{h} = \frac{2h + h^2}{h} = \resaltar{2 + h}$
Generaliter, pro omni $h \neq 0$ , positivo vel negativo, declivitas simplificatvr; cvm $h = \frac{1}{4}$ valet $\frac{9}{4}$ .
$\lim_{h \to 0} (2 + h) = 2, \qquad y = 1 + 2(x - 1) = \resaltar{2x - 1}$
Cvm $h$ ad $0$ tendit, declivitas ad $2$ tendit. Recta qvae per $(1, 1)$ cvm declivitate $2$ transit est recta tangens; eadem est qvam lemma de tangente in lectione de parabolis dedit, cvm $x^2 = 4 \cdot \frac{1}{4} \, y$ .

Idem limes in motv investigando apparet. Si $s(t)$ est positio, tempore $t$ , pvncti qvod in recta movetvr, qvotiens $\frac{s(t_0 + h) - s(t_0)}{h}$ est eivs velocitas media inter tempora $t_0$ et $t_0 + h$ : motvs divisvs per tempvs consvmptvm. Velocitas tempore $t_0$ , qvam index velocitatis monstraret, est limes velocitatvm mediarvm cvm $h$ ad $0$ tendit. Exempli gratia, si $s(t) = t^2 - 4t$ , velocitas media inter $1$ et $1 + h$ est $\frac{(1 + h)^2 - 4(1 + h) + 3}{h} = h - 2$ , qvae ad $-2$ tendit: tempore $1$ pvnctvm sinistrorsvm movetvr, dvabvs vnitatibvs longitvdinis in vnitate temporis. Dvae qvaestiones, de recta tangente et de velocitate, ad evndem limitem dvcvnt, et is ipse limes nvnc definitvr.

Derivata in pvncto

Vt in lectione de limite fvnctionis, primvm accvrate statvendvm est vbi limes sensvm habeat. Qvotiens valorem $f(a)$ reqvirit, ita vt $a$ ad dominivm pertinere debeat; et limes cvm $x$ ad $a$ tendit postvlat vt $a$ pvnctvm accvmvlationis dominii sit.

DefinitioDerivata fvnctionis in pvncto

Sint $f$ fvnctio realis in copia $A$ nvmerorvm realivm definita, et $a$ elementvm copiae $A$ qvod pvnctvm accvmvlationis copiae $A$ est. Qvotiens incrementalis fvnctionis $f$ in $a$ est fvnctio $\frac{f(x) - f(a)}{x - a}$ , definita pro iis $x$ copiae $A$ qvae a $a$ differvnt. $f$ in $a$ derivabilis dicitvr si qvotiens incrementalis limitem realem $L$ habet cvm $x$ ad $a$ tendit; is nvmervs est derivata fvnctionis $f$ in $a$ , et scribitvr $f'(a)$ :

f'(a) = L \leftrightarrow (\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in A)\left(0 < \abs{x - a} < \delta \rightarrow \abs{\frac{f(x) - f(a)}{x - a} - L} < \varepsilon\right)

Qvotiens in $A \setminus \{a\}$ definitvs est, et $a$ pvnctvm accvmvlationis eivs copiae est, qvia definitio pvncti accvmvlationis sola elementa a $a$ diversa respicit. Propter vnicitatem limitis derivata, si exsistit, vnica est. In pvncto solitario dominii derivata non definitvr, qvamqvam omnis fvnctio ibi continva est. Et si qvotiens ad $\infty$ vel ad $-\infty$ tendit, fvnctio in $a$ derivabilis non est: derivata nvmervs realis est. Definitio saepe scribitvr cvm incremento $h = x - a$ loco pvncti $x$ , vt in exemplo.

PropositioForma incrementalis derivatae

Svb hypothesibvs definitionis, et $f(a + h)$ solvm vbi definitvm est considerando:

f'(a) = L \Leftrightarrow \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} = L

Est mvtatio variabilis $x = a + h$ , qvam propositio de translatione et reflexione in lectione de limitibvs lateralibvs pro vno latere fecit, nvnc pro vtroqve simvl.

Demonstratio

$Q(a + h) = \frac{f(a + h) - f(a)}{h}, \qquad h \neq 0 \rightarrow a + h \neq a$
Sit $Q(x)$ qvotiens incrementalis. Fvnctio $g(h) = a + h$ , pro iis $h \neq 0$ qvibvs $a + h$ in $A$ est, ad $a$ tendit, valorem $a$ non assvmit, et eivs valores in dominio fvnctionis $Q$ svnt. Per theorema de compositione lectionis de algebra limitvm, si $Q$ ad $L$ in $a$ tendit, etiam $Q(a + h)$ ad $L$ in $0$ tendit.
$Q(x) = Q\left(a + (x - a)\right), \qquad x \neq a \rightarrow \resaltar{x - a \neq 0}$
Vicissim, fvnctio $x - a$ , pro iis $x$ copiae $A$ qvae a $a$ differvnt, ad $0$ tendit nec valorem $0$ assvmit; composita cvm $h \mapsto Q(a + h)$ , qvotientem $Q(x)$ dat, qvi ad evndem limitem tendit.

Primae derivatae directe compvtantvr. Si $f(x) = mx + b$ , qvotiens incrementalis valet $\frac{m(a + h) + b - (ma + b)}{h} = m$ pro omni $h \neq 0$ , et eivs limes est $m$ : derivata fvnctionis affinis est eivs declivitas in omni pvncto, et eivs recta tangens est ipsa recta. Praesertim derivata constantis est $0$ , et derivata fvnctionis identicae $f(x) = x$ est $1$ . Pro $f(x) = x^2$ , calcvlvs exempli, cvm $a$ loco $1$ , dat $\frac{(a + h)^2 - a^2}{h} = 2a + h$ , et derivata est $2a$ .

Derivatae laterales

In extremis intervalli, avt in pvncto vbi fvnctio formvlam mvtat, convenit qvotientem incrementalem in vtroqve latere separatim inspicere.

DefinitioDerivatae laterales

Sit $a$ elementvm copiae $A$ qvod pvnctvm accvmvlationis copiae $A \cap \ioo{a}{\infty}$ est. $f$ in $a$ derivabilis a dextra dicitvr si qvotiens incrementalis limitem realem a dextra in $a$ habet; is limes est derivata a dextra, $f'_+(a)$ :

f'_+(a) = L \leftrightarrow (\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in A)\left(a < x < a + \delta \rightarrow \abs{\frac{f(x) - f(a)}{x - a} - L} < \varepsilon\right)

Eodem modo, si $a$ pvnctvm accvmvlationis copiae $A \cap \ioo{-\infty}{a}$ est, derivata a sinistra, $f'_-(a)$ , est limes a sinistra qvotientis incrementalis, cvm condicione $a - \delta < x < a$ .

PropositioDerivata et derivatae laterales

Sit $a$ elementvm copiae $A$ .

Si $a$ pvnctvm accvmvlationis copiae $A \cap \ioo{a}{\infty}$ et copiae $A \cap \ioo{-\infty}{a}$ est, tvm $f$ in $a$ derivabilis est si et solvm si ambae derivatae laterales exsistvnt et aeqvales svnt; qvo casv $f'(a) = f'_+(a) = f'_-(a)$ .
Si $a$ pvnctvm accvmvlationis est pvnctorvm copiae $A$ ex vno tantvm latere, tvm $f$ in $a$ derivabilis est si et solvm si ab eo latere derivabilis est, et derivata est derivata lateralis.

Nvllo novo argvmento opvs est: derivatae laterales svnt limites laterales qvotientis incrementalis, et lectio de limitibvs lateralibvs eos iam cvm limite conivnxit.

Demonstratio

$\lim_{x \to a} \frac{f(x) - f(a)}{x - a} = L \Leftrightarrow f'_+(a) = L = \resaltar{f'_-(a)}$
In primo casv theorema de limite et limitibvs lateralibvs, ad qvotientem incrementalem adhibitvm, aeqvivalentiam dat.
$\left(\forall x \in A\right)\left(0 < \abs{x - a} < r \rightarrow x > a\right) \rightarrow f'(a) = \resaltar{f'_+(a)}$
In altero eadem lectio animadvertit, cvm $a$ pvnctvm accvmvlationis sit pvnctorvm dominii ex vno tantvm latere, limitem et limitem lateralem eivs lateris congrvere: exsistit $r > 0$ tale vt omnia $x$ copiae $A$ cvm $0 < \abs{x - a} < r$ in eo latere sint.

Exempli gratia, valor absolvtvs in $0$ qvotientem incrementalem $\frac{\abs{x}}{x}$ habet, qvi valet $1$ si $x > 0$ et $-1$ si $x < 0$ : eivs derivatae laterales svnt $f'_+(0) = 1$ et $f'_-(0) = -1$ , diversae, et $\abs{x}$ in $0$ derivabilis non est. Contra, si fvnctio in $\icc{0}{1}$ definita est, eivs derivata in $0$ est eivs derivata a dextra, et in $1$ eivs derivata a sinistra: definitio iam extrema dominii vt latera vnica tractat. Exempli gratia, $f(x) = x$ ad $\icc{0}{1}$ restricta in $0$ derivabilis est, cvm $f'(0) = 1$ .

Rectae tangens et normalis

Per derivatam recta tangens desinit esse figvra qvae ocvlis agnoscitvr et fit recta qvae compvtatvr.

DefinitioRecta tangens et recta normalis

Sit $f$ in $a$ derivabilis. Recta tangens delineationi fvnctionis $f$ in pvncto $(a, f(a))$ est recta qvae per id cvm declivitate $f'(a)$ transit:

y = f(a) + f'(a)(x - a)

Recta normalis in eodem pvncto est recta rectae tangenti perpendicvlaris qvae per id transit: si $f'(a) \neq 0$ , est recta $y = f(a) - \frac{1}{f'(a)}(x - a)$ , et si $f'(a) = 0$ , est recta verticalis $x = a$ .

Declivitas rectae normalis ex theoremate de rectis perpendicvlaribvs lectionis de aeqvatione rectae oritvr: dvae rectae non verticales perpendicvlares svnt si et solvm si prodvctvm declivitatvm earvm est $-1$ . Ad cvrvam $y = x^3$ redevntes, eivs qvotiens incrementalis in $0$ est $x^2$ , qvi ad $0$ tendit: recta tangens in origine est axis horizontalis, qvamqvam cvrvam traicit. Cvm $f$ in $a$ continva est et qvotiens incrementalis ad $\infty$ vel ad $-\infty$ tendit, secantes ad verticalem accedvnt; tvm dicitvr delineationem in $(a, f(a))$ habere rectam tangentem verticalem, rectam $x = a$ , qvamqvam $f$ in $a$ derivabilis non est.

Eodem modo, si $s(t)$ est positio pvncti qvod in recta movetvr, eivs velocitas tempore $t_0$ est $s'(t_0)$ , si exsistit. Eivs signvm sensvm motvs indicat, et eivs valor absolvtvs celeritatem. Lectio de rationibvs mvtationis hanc interpretationem explicabit.

Fvnctio derivata et notatio Leibnitiana

Qvodqve pvnctvm in qvo $f$ derivabilis est nvmervm affert, derivatam in eo. Ex omnibvs his nvmeris nova fvnctio formatvr.

DefinitioFvnctio derivata

Sit $f$ fvnctio in $A$ definita. Fvnctio derivata fvnctionis $f$ est fvnctio $f'$ qvae vnicviqve pvncto $x$ copiae $A$ in qvo $f$ derivabilis est nvmervm $f'(x)$ attribvit; eivs dominivm est copia horvm pvnctorvm. Si $f$ in omnibvs pvnctis copiae $S$ in $A$ contentae derivabilis est, dicitvr derivabilis in $S$ ; et si in toto dominio svo derivabilis est, simpliciter derivabilis.

Exempli gratia, derivata fvnctionis $x^2$ est fvnctio $2x$ , in toto $\mathbb{R}$ definita; derivata valoris absolvti valet $1$ in $\ioo{0}{\infty}$ et $-1$ in $\ioo{-\infty}{0}$ , et eivs dominivm $0$ non continet. Notatio $f'$ , cvm apice, Lagrangio debetvr. Altera notatio vsitata Leibnitio debetvr: si $y = f(x)$ , derivata scribitvr $\frac{\mathrm{d}y}{\mathrm{d}x}$ , vel $\frac{\mathrm{d}f}{\mathrm{d}x}$ , et operatio derivandi secvndvm $x$ scribitvr $\frac{\mathrm{d}}{\mathrm{d}x}$ ante expressionem qvae derivatvr:

\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right) = 2x, \qquad \frac{\mathrm{d}}{\mathrm{d}x}\left(mx + b\right) = m

Notatio in memoriam revocat vnde derivata veniat. Si $\Delta x$ est incrementvm variabilis et $\Delta y = f(x + \Delta x) - f(x)$ incrementvm respondens fvnctionis, qvotiens incrementalis est $\frac{\Delta y}{\Delta x}$ , et derivata eivs limes est:

\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}

Instandvm est $\frac{\mathrm{d}y}{\mathrm{d}x}$ interim vnvm signvm esse, non qvotientem: $\mathrm{d}y$ et $\mathrm{d}x$ nondvm separatim sensvm habent; habebvnt in lectione de derivatis svccessivis et differentiali. Notatio variabilem secvndvm qvam derivatvr manifestam facit: si positio a tempore pendet, velocitas scribitvr $s'(t)$ vel $\frac{\mathrm{d}s}{\mathrm{d}t}$ .

Tres characterismi derivatae

Definitio dicit qvotientem incrementalem ad $f'(a)$ tendere. Dvo alii modi svnt idem dicendi, qvisqve ad alivm finem vtilis: secvndvs fvnctionem cvm recta tangente sva comparat; tertivs dividere vitat.

TheoremaCharacterismi derivatae

Sint $f$ fvnctio in $A$ definita, $a$ elementvm copiae $A$ qvod pvnctvm accvmvlationis copiae $A$ est, et $L$ nvmervs realis. Tres affirmationes qvae svbicivntvr aeqvivalentes svnt:

Qvotiens incrementalis. $f$ in $a$ derivabilis est et $f'(a) = L$ .
Approximatio linearis. $f(a + h) = f(a) + Lh + \operatorname{o}(h)$ cvm $h$ ad $0$ tendit; id est, residvvm $R(h) = f(a + h) - f(a) - Lh$ condicioni $\lim_{h \to 0} \frac{R(h)}{h} = 0$ satisfacit.
Carathéodory. Exsistit fvnctio $\varphi \colon A \to \mathbb{R}$ , in $a$ continva, cvm $\varphi(a) = L$ , talis vt $f(x) - f(a) = \varphi(x)(x - a)$ pro omni $x$ copiae $A$ .

Symbolvm $\operatorname{o}$ parvvm idem est atqve in lectione de infinitesimis et infinitis aeqvivalentibvs, cvm $f(a + h)$ solvm vbi definitvm est considerato. Fvnctio $\varphi$ est declivitas secantis, in $a$ per derivatam completa. Figvra ostendit, cvm $f(x) = \frac{x^2 + 1}{2}$ et $a = 1$ , residvvm inter cvrvam et rectam tangentem, secantem et, in fine, delineationem fvnctionis $\varphi$ , qvae hoc casv est recta $\varphi(x) = \frac{x + 1}{2}$ .

Demonstratio

$\frac{R(h)}{h} = \frac{f(a + h) - f(a)}{h} - L \to 0 \Leftrightarrow \resaltar{\frac{f(a + h) - f(a)}{h} \to L}$
Prima et secvnda: qvotiens residvi per $h$ est qvotiens incrementalis minvs $L$ . Per lemma de limite vt distantia ad nihilvm tendente, lectionis de algebra limitvm, ad $0$ tendit si et solvm si qvotiens incrementalis ad $L$ tendit, et per formam incrementalem id aeqvivalet condicioni $f'(a) = L$ .
$\varphi(x) = \begin{cases} \frac{f(x) - f(a)}{x - a} & x \neq a \\ L & x = a \end{cases}, \qquad \lim_{x \to a} \varphi(x) = L = \resaltar{\varphi(a)}$
A prima ad tertiam: $\varphi$ definitvr vt qvotiens incrementalis extra $a$ et vt $L$ in $a$ . Aeqvalitas valet in omni $x \neq a$ per definitionem, et in $x = a$ qvia ambo eivs membra valent $0$ . Cvm $a$ pvnctvm accvmvlationis copiae $A$ sit et limes fvnctionis $\varphi$ in $a$ sit $L = \varphi(a)$ , propositio de continvitate et limite lectionis de limite et continvitate continvitatem fvnctionis $\varphi$ in $a$ dat.
$x \neq a \rightarrow \frac{f(x) - f(a)}{x - a} = \varphi(x) \to \varphi(a) = \resaltar{L}$
A tertia ad primam: si $x \neq a$ , per $x - a$ dividitvr, et qvotiens incrementalis est restrictio fvnctionis $\varphi$ ad $A \setminus \{a\}$ . Cvm $\varphi$ in $a$ continva sit, per eandem propositionem ad $\varphi(a) = L$ tendit, et per lemma de restrictione, lectionis de algebra limitvm, etiam qvotiens.

Fvnctio $\varphi$ tertii characterismi, si exsistit, vnica est: extra $a$ est qvotiens incrementalis, et in $a$ eivs valor est limes eivs qvotientis. Secvndvs characterismvs explicat qvo sensv recta tangens sit recta qvae cvrvam optime comitatvr. Si pro $f(a + h)$ ponitvr $f(a) + mh$ , cvm declivitate $m$ a $f'(a)$ diversa, error est $(f'(a) - m)h + \operatorname{o}(h)$ , cvivs qvotiens per $h$ ad $f'(a) - m \neq 0$ tendit: eivsdem ordinis est ac $h$ . Solvm cvm declivitate $f'(a)$ error prae $h$ neglegibilis est. Tertivs characterismvs commodvm technicvm habet: est aeqvalitas inter prodvcta, sine divisionibvs. Lectiones de regvlis derivandi et de regvla catenae eo vtentvr ne per qvantitates dividant qvae evanescere possvnt. Prima conseqventia statim patet.

TheoremaDerivabilis ergo continva

Si $f$ in $a$ derivabilis est, tvm $f$ in $a$ continva est. Factorizatio Carathéodorii $f$ per operationes inter fvnctiones in $a$ continvas scribit.

Demonstratio

$f(x) = f(a) + \varphi(x)(x - a)$
Per theorema de characterismis exsistit $\varphi$ , in $a$ continva, talis vt $f(x) = f(a) + \varphi(x)(x - a)$ pro omni $x$ copiae $A$ .
$\lim_{x \to a} f(x) = f(a) + \varphi(a) \cdot 0 = \resaltar{f(a)}$
Constans $f(a)$ et polynomivm $x - a$ in $a$ continva svnt, et $\varphi$ qvoqve. Per theorema de operationibvs cvm fvnctionibvs continvis, lectionis de limite et continvitate, svmma constantis et prodvcti in $a$ continva est.

Conversa falsa est. Continvitas postvlat vt $f(x) - f(a)$ ad $0$ tendat; derivabilitas plvs postvlat: vt eivs qvotiens per $x - a$ limitem habeat, id est vt ad $0$ tendat saltem tam celeriter qvam $x - a$ et sine oscillatione inter declivitates diversas.

PropositioContinvitas sine derivata

Valor absolvtvs et radix cvbica in $0$ continva svnt et in $0$ derivabilia non svnt: primvs derivatas laterales diversas habet, et qvotiens incrementalis secvndae ad $\infty$ tendit, ita vt eivs delineatio in origine rectam tangentem verticalem habeat. Ambo continva svnt per corollarivm de fvnctionibvs continvis notis, lectionis de limite et continvitate. Figvra ostendit, in valore absolvto, dvas declivitates laterales et, in radice cvbica, secantes magis magisqve ardvas.

Demonstratio

$f'_+(0) = \lim_{x \to 0^+} \frac{\abs{x}}{x} = 1 \neq -1 = \lim_{x \to 0^-} \frac{\abs{x}}{x} = \resaltar{f'_-(0)}$
Qvotiens incrementalis valoris absolvti in $0$ valet $1$ a dextra et $-1$ a sinistra. Derivatae laterales diversae svnt et, per propositionem de derivata et derivatis lateralibvs, $\abs{x}$ in $0$ derivabilis non est.
$\frac{\sqrt[3]{x} - 0}{x - 0} = \frac{1}{\left(\sqrt[3]{x}\right)^2} \to \resaltar{\infty}$
Qvotiens incrementalis radicis cvbicae in $0$ est inversvm ipsivs $\left(\sqrt[3]{x}\right)^2$ , qvod positivvm est si $x \neq 0$ et ad $0$ tendit propter continvitatem radicis. Per lemma inversi, lectionis de limitibvs infinitis, qvotiens ad $\infty$ tendit, nec limitem realem habet.

Ambae fvnctiones in vno tantvm pvncto deficivnt, sed exsistvnt fvnctiones in tota recta continvae qvae in nvllo pvncto derivabiles svnt; Weierstrassivs anno 1872 exemplvm celeberrimvm protvlit, cvivs constrvctio hvnc cvrsvm excedit.

Problemata solvta

Problemata difficvltate crescvnt: primvm derivatam compvtat qvae in ora dominii exsistere desinit; secvndvm recta tangente vtitvr ad proprietatem hyperbolae detegendam; tertivm ostendit derivatas laterales posse non exsistere; qvartvm, fvnctionem in vno tantvm pvncto derivabilem esse posse; et vltimvm factorizationem Carathéodorii cvm derivatis lateralibvs conivngit.

Problema solvtvm 1Radix qvadrata in ora dominii svi

Derivabilitatem fvnctionis $f(x) = \sqrt{x}$ , in $\ico{0}{\infty}$ definitae, investigare, et rectam tangentem in pvncto abscissae $4$ invenire.

Solvtio

$\frac{\sqrt{a + h} - \sqrt{a}}{h} = \frac{(a + h) - a}{\left(\sqrt{a + h} + \sqrt{a}\right) h} = \frac{1}{\sqrt{a + h} + \sqrt{a}}$
Si $a > 0$ , qvotiens incrementalis est forma $\frac{0}{0}$ ; per expressionem conivgatam mvltiplicatvr et dividitvr.
$f'(a) = \resaltar{\frac{1}{2\sqrt{a}}}, \qquad y = 2 + \frac{1}{4}(x - 4) = \frac{x}{4} + 1$
Propter continvitatem radicis denominator ad $2\sqrt{a} \neq 0$ tendit, et per theorema de limite inversi, lectionis de algebra limitvm, qvotiens ad $\frac{1}{2\sqrt{a}}$ tendit. In $a = 4$ derivata est $\frac{1}{4}$ et recta tangens per $(4, 2)$ transit.
$\lim_{h \to 0^+} \frac{\sqrt{h} - 0}{h} = \lim_{h \to 0^+} \frac{1}{\sqrt{h}} = \resaltar{\infty}$
In $a = 0$ pvncta dominii solvm a dextra svnt, et per propositionem de derivata et derivatis lateralibvs derivata est derivata a dextra. Qvotiens est $\frac{1}{\sqrt{h}}$ , et cvm $\sqrt{h}$ positivvm sit et ad $0$ tendat, lemma inversi dat $\infty$ .
$f \colon \ico{0}{\infty} \to \mathbb{R}, \qquad f' \colon \resaltar{\ioo{0}{\infty}} \to \mathbb{R}$
Ergo $\sqrt{x}$ in $\ioo{0}{\infty}$ derivabilis est, non avtem in $0$ , vbi continva est et eivs delineatio rectam tangentem verticalem $x = 0$ habet. Fvnctio derivata minvs dominivm habet qvam fvnctio.

Problema solvtvm 2Recta tangens hyperbolae

Demonstrare rectam tangentem cvrvae $y = \frac{1}{x}$ in pvncto $\left(a, \frac{1}{a}\right)$ , cvm $a > 0$ , cvm axibvs triangvlvm areae $2$ formare, qvodcvmqve sit $a$ , et pvnctvm contactvs esse pvnctvm medivm segmenti qvod axes in ea abscindvnt.

Solvtio

$\frac{\frac{1}{a + h} - \frac{1}{a}}{h} = \frac{-h}{a(a + h) \, h} = -\frac{1}{a(a + h)} \to \resaltar{-\frac{1}{a^2}}$
Qvotiens incrementalis fractionibvs svbtrahendis simplificatvr; per theorema de limite inversi eivs limes est $-\frac{1}{a^2}$ .
$y = \frac{1}{a} - \frac{1}{a^2}(x - a) = \resaltar{\frac{2}{a} - \frac{x}{a^2}}$
Recta tangens per $\left(a, \frac{1}{a}\right)$ cvm ea declivitate transit.
$\dato{c}{(2a, 0), \quad \left(0, \frac{2}{a}\right)}, \qquad \left(\frac{2a + 0}{2}, \frac{0 + \frac{2}{a}}{2}\right) = \resaltar{\left(a, \frac{1}{a}\right)}$
Cvm $y = 0$ intersectio cvm axe horizontali obtinetvr, et cvm $x = 0$ intersectio cvm verticali. Pvnctvm medivm inter ambas est pvnctvm contactvs.
$\frac{1}{2} \cdot 2a \cdot \frac{2}{a} = \resaltar{2}$
Triangvlvm cathetos $2a$ et $\frac{2}{a}$ in axibvs habet. Eivs area non pendet a $a$ : figvra dvo talia triangvla ostendit, formis diversis et eadem area.

Problema solvtvm 3Continva et sine derivatis lateralibvs

Sint $f(x) = x \sin \frac{1}{x}$ et $g(x) = x^2 \sin \frac{1}{x}$ pro $x \neq 0$ , et $f(0) = g(0) = 0$ . Demonstrare $f$ in $0$ continvam esse nec in $0$ derivatam nec derivatas laterales habere, et $g$ in $0$ derivabilem esse, cvm $g'(0) = 0$ .

Solvtio

$\frac{f(x) - f(0)}{x - 0} = \frac{x \sin \frac{1}{x}}{x} = \resaltar{\sin \frac{1}{x}}$
Per problema solvtvm 3 lectionis de limite et continvitate, $f$ in $0$ continva est. Eivs qvotiens incrementalis in $0$ est $\sin \frac{1}{x}$ .
$\sin \frac{1}{x_n} = 1, \quad \sin \frac{1}{y_n} = -1, \qquad x_n = \frac{1}{\frac{\pi}{2} + 2n\pi}, \quad y_n = \frac{1}{\frac{3\pi}{2} + 2n\pi}$
Problema propositvm 3 lectionis de limite demonstravit $\sin \frac{1}{x}$ limitem in $0$ non habere per dvas familias pvnctorvm positivorvm, in qvibvs valet $1$ et $-1$ . Idem argvmentvm limitem a dextra negat; cvm pvnctis oppositis, in qvibvs valet $-1$ et $1$ , negatvr limes a sinistra.
$\frac{g(x) - g(0)}{x - 0} = x \sin \frac{1}{x} \to 0, \qquad g'(0) = \resaltar{0}$
Contra, qvotiens incrementalis fvnctionis $g$ est $x \sin \frac{1}{x}$ , prodvctvm fvnctionis circvmscriptae et alterivs qvae ad $0$ tendit; per theorema lectionis de theoremate compressionis, ad $0$ tendit.
$\abs{g(x)} \leq x^2 = \resaltar{\operatorname{o}(x)}$
Ambae fvnctiones prope $0$ infinities oscillant. Oscillatio fvnctionis $f$ per $\abs{x}$ circvmscripta est, qvod eivsdem ordinis est ac denominator $x - 0$ , et qvotiens oscillare pergit; oscillatio fvnctionis $g$ per $x^2$ circvmscripta est, qvod prae $x$ neglegibile est, et qvotiens exstingvitvr.

Problema solvtvm 4Derivabilis in vno tantvm pvncto

Sit $f$ fvnctio qvae valet $x^2$ si $x$ rationalis est et $0$ si $x$ irrationalis est. Demonstrare $f$ in $0$ derivabilem esse, cvm $f'(0) = 0$ , nec in vllo alio pvncto derivabilem esse.

Solvtio

$x \in \mathbb{Q} \rightarrow x \, g(x) = x^2, \qquad x \notin \mathbb{Q} \rightarrow x \, g(x) = 0$
Sit $g$ fvnctio problematis solvti 5 lectionis de limite et continvitate, qvae valet $x$ in rationalibvs et $0$ in irrationalibvs. Tvm $f(x) = x \, g(x)$ pro omni $x$ , in vtroqve casv.
$\frac{f(x) - f(0)}{x - 0} = \frac{x \, g(x)}{x} = g(x) \to g(0) = 0, \qquad f'(0) = \resaltar{0}$
In $0$ qvotiens incrementalis est $g(x)$ , qvi ad $g(0) = 0$ tendit qvia illvd problema demonstravit $g$ in $0$ continvam esse.
$\lim_{x \to a} f(x) = f(a) \rightarrow \lim_{x \to a} \frac{f(x)}{x} = \frac{f(a)}{a} = \resaltar{g(a)}$
Sit $a \neq 0$ . Si $f$ in $a$ continva esset, etiam $g(x) = \frac{f(x)}{x}$ continva esset, per regvlam qvotientis operationvm cvm fvnctionibvs continvis, cvm denominatore in $a$ non nvllo; sed idem problema demonstravit $g$ in $a$ discontinvam esse. Ergo $f$ in $a$ discontinva est, et per theorema «derivabilis ergo continva» in $a$ derivabilis non est.

Problema solvtvm 5Valor absolvtvs fvnctionis derivabilis

Sit $f$ fvnctio in intervallo aperto qvod $a$ continet definita, in $a$ derivabilis et cvm $f(a) = 0$ . Demonstrare $F = \abs{f}$ in $a$ derivatas laterales $\abs{f'(a)}$ et $-\abs{f'(a)}$ habere, ita vt in $a$ derivabilis sit si et solvm si $f'(a) = 0$ . Id ad $\abs{\sin x}$ et ad $\abs{x^3}$ in $0$ adhibere.

Solvtio

$\frac{F(x) - F(a)}{x - a} = \frac{\abs{f(x)} - \abs{f(a)}}{x - a} = \frac{\abs{\varphi(x)}\abs{x - a}}{x - a} = \begin{cases} \abs{\varphi(x)} & x > a \\ -\abs{\varphi(x)} & x < a \end{cases}$
Per characterismvm Carathéodorii, $f(x) = \varphi(x)(x - a)$ , cvm $\varphi$ in $a$ continva et $\varphi(a) = f'(a)$ . Valoribvs absolvtis svmptis, factor $\abs{x - a}$ est $x - a$ a dextra et $a - x$ a sinistra.
$F'_+(a) = \abs{f'(a)}, \qquad F'_-(a) = -\abs{f'(a)}, \qquad \abs{f'(a)} = -\abs{f'(a)} \Leftrightarrow \resaltar{f'(a) = 0}$
Fvnctio $\abs{\varphi}$ in $a$ continva est, per compositionem fvnctionvm continvarvm, et ad $\abs{f'(a)}$ tendit. Per propositionem de derivata et derivatis lateralibvs, $F$ in $a$ derivabilis est si et solvm si ambae derivatae laterales congrvvnt, id est si $\abs{f'(a)} = -\abs{f'(a)}$ .
$\frac{\sin x}{x} \to 1 \neq 0, \qquad \frac{x^3}{x} = x^2 \to \resaltar{0}$
Sinvs condicioni $\sin 0 = 0$ satisfacit, et eivs qvotiens incrementalis in $0$ est $\frac{\sin x}{x}$ , qvi ad $1$ tendit per limitem fvndamentalem trigonometriae: $\abs{\sin x}$ in $0$ derivatas laterales $1$ et $-1$ habet, nec derivabilis est. Cvbvs in $0$ qvotientem $x^2$ habet, qvi ad $0$ tendit: $\abs{x^3}$ in $0$ derivabilis est, cvm derivata $0$ .

Problemata proposita

Problemata difficvltate crescvnt: primvm derivatam in geometria circvli et sphaerae legit; secvndvm rectas normales qvaerit qvae per pvnctvm datvm transevnt; tertivm ostendit derivatam lateralem exsistere posse vbi fvnctio salit; qvartvm rectam tangentem derivatae cvm recta tangente circvmferentiae comparat; et vltimvm qvotientem investigat qvi derivatae similis est neqve tamen derivata est. Vnvmqvodqve indicivm habet, qvod aperire convenit solvm postqvam problema temptatvm est.

Problema propositvm 1Area circvli et volvmen sphaerae

Sint $A(r) = \pi r^2$ area circvli radii $r$ et $V(r) = \frac{4}{3}\pi r^3$ volvmen sphaerae radii $r$ . Per definitionem $\frac{\mathrm{d}A}{\mathrm{d}r}$ et $\frac{\mathrm{d}V}{\mathrm{d}r}$ compvtare, et qvodqve eventvm agnoscere.

Indicivm

Evolve $(r + h)^2$ et $(r + h)^3$ et qvotientem incrementalem simplifica anteqvam limitem svmas. Incrementvm areae est area coronae circvlaris latitvdinis $h$ .

Responsvm

\frac{\mathrm{d}A}{\mathrm{d}r} = 2\pi r, \qquad \frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2

Svnt perimetrvm circvmferentiae et area svperficiei sphaericae.

Problema propositvm 2Rectae normales per pvnctvm

Omnia pvncta parabolae $y = x^2$ invenire qvorvm recta normalis per pvnctvm $(0, 1)$ transit, et eas rectas normales scribere.

Indicivm

Recta normalis in $(a, a^2)$ declivitatem $-\frac{1}{2a}$ habet si $a \neq 0$ ; impone vt per $(0, 1)$ transeat. Noli oblivisci casvm $a = 0$ , in qvo derivata evanescit et recta normalis verticalis est.

Responsvm

Tria pvncta: $(0, 0)$ , cvm recta normali $x = 0$ , et $\left(\pm\frac{\sqrt{2}}{2}, \frac{1}{2}\right)$ , cvm rectis normalibvs $y = 1 \mp \frac{x}{\sqrt{2}}$ .

Problema propositvm 3Derivatae laterales partis integrae

Demonstrare partem integram, $E(x) = \lfloor x \rfloor$ , derivabilem esse in omni nvmero reali non integro, cvm derivata $0$ , et in qvoqve integro $n$ derivatam a dextra, $0$ , habere, sed non derivatam a sinistra. Nvm hoc theoremati «derivabilis ergo continva» repvgnat?

Indicivm

In integro $n$ , a dextra fvnctio valet $n$ , et a sinistra $n - 1$ : qvotiens incrementalis est $0$ a dextra et $\frac{-1}{x - n}$ a sinistra. Ad vltimam qvaestionem, hypothesim theorematis cvm eo qvod in $n$ accidit compara.

Responsvm

E'(x) = 0 \quad (x \notin \mathbb{Z}), \qquad E'_+(n) = 0, \qquad \lim_{x \to n^-} \frac{E(x) - E(n)}{x - n} = \infty

Nvlla est repvgnantia: fvnctio in $n$ derivabilis non est; derivata a dextra solvm continvitatem a dextra dat (theorema, ad restrictionem fvnctionis $E$ ad $\ico{n}{\infty}$ adhibitvm), qvae valet.

Problema propositvm 4Recta tangens circvmferentiae

Sit $f(x) = \sqrt{1 - x^2}$ , semicircvlvs svperior centri $(0, 0)$ et radii $1$ . $f'(a)$ pro $-1 < a < 1$ compvtare, demonstrare rectam tangentem perpendicvlarem esse radio qvi originem cvm $(a, f(a))$ ivngit, vt in lectione de circvmferentia, et investigare qvid in $a = \pm 1$ accidat.

Indicivm

Mvltiplica et divide qvotientem incrementalem per $\sqrt{1 - (a + h)^2} + \sqrt{1 - a^2}$ . Si $a \neq 0$ , radivs declivitatem $\frac{f(a)}{a}$ habet; si $a = 0$ , verticalis est. In $a = 1$ pvncta dominii solvm a sinistra svnt.

Responsvm

f'(a) = -\frac{a}{\sqrt{1 - a^2}}, \qquad -\frac{a}{\sqrt{1 - a^2}} \cdot \frac{\sqrt{1 - a^2}}{a} = -1 \quad (a \neq 0)

In $a = 0$ recta tangens horizontalis est et radivs verticalis. In $a = 1$ qvotiens ad $-\infty$ tendit, et in $a = -1$ ad $\infty$ : $f$ in $\pm 1$ derivabilis non est, et eivs delineatio ibi rectas tangentes verticales habet.

Problema propositvm 5Qvotiens symmetricvs

Sit $f$ in intervallo aperto qvod $a$ continet definita. Demonstrare: si $f$ in $a$ derivabilis est, tvm $\lim_{h \to 0} \frac{f(a + h) - f(a - h)}{2h} = f'(a)$ . Demonstrare per $f(x) = \abs{x}$ in $a = 0$ evm limitem exsistere posse qvin $f$ derivabilis sit.

Indicivm

Scribe approximationem linearem pro $h$ et pro $-h$ et eas svbtrahe: residva dant $\operatorname{o}(h)$ . Pro valore absolvto nvmerator est $\abs{h} - \abs{-h}$ .

Responsvm

\frac{f(a + h) - f(a - h)}{2h} = f'(a) + \frac{R(h) - R(-h)}{2h} \to f'(a), \qquad \frac{\abs{h} - \abs{-h}}{2h} = 0

Valor absolvtvs qvotientem symmetricvm $0$ in $0$ habet, nec ibi derivabilis est.

Solvtio explicata

Solvtio

$f(a + h) = f(a) + f'(a)h + R(h), \qquad f(a - h) = f(a) - f'(a)h + R(-h)$
Per approximationem linearem, cvm $L = f'(a)$ , residvvm $R$ condicioni $\frac{R(h)}{h} \to 0$ satisfacit. Mvtato $h$ in $-h$ , per theorema de compositione, etiam $\frac{R(-h)}{-h} \to 0$ .
$\frac{f(a + h) - f(a - h)}{2h} = f'(a) + \frac{1}{2} \cdot \frac{R(h)}{h} + \frac{1}{2} \cdot \frac{R(-h)}{-h} \to \resaltar{f'(a)}$
Dvae aeqvalitates svbtrahvntvr et per $2h$ dividitvr; per theorema de limite combinationis linearis, dvo qvotientes residvorvm dant $0$ .
$\frac{\abs{h} - \abs{-h}}{2h} = \resaltar{0}, \qquad f'_+(0) = 1 \neq -1 = f'_-(0)$
Pro valore absolvto in $0$ nvmerator nvllvs est pro omni $h$ , et qvotiens symmetricvs ad $0$ tendit; tamen derivatae laterales svnt $1$ et $-1$ . Qvotiens symmetricvs dvas declivitates laterales in medivm redigit et angvlvm occvltat.

Derivata vt limes fvnctionis

Svmmarivm

Proposita discendi

Recta tangens et velocitas

Derivata in pvncto

Derivatae laterales

Rectae tangens et normalis

Fvnctio derivata et notatio Leibnitiana

Tres characterismi derivatae

Problemata solvta

Problemata proposita

Dvbitatio tva de hac lectione

Qvaestiones de hac lectione