Derivationes Polynomialium, Trigonometricarum et Logarithmicarum
Derivatio est instrumentum centrale calculi differentialis, cum applicationibus fundamentalibus in scientiis, ingenieria et oeconomia. Hic articulus praebet ducem progressivam ad derivationem functionum domandam, a polynomialibus usque ad functiones trigonometricae et logarithmicae. Per demonstrationes et exempla concreta quaeritur intellectus tam applicationis regularum quam fundamentorum earum.
Metas Discendi
- Intellegere notionem generalem derivationis eiusque proprietates fundamentales.
- Adhibere definitionem formalem derivationis ad calculandas derivationes basicas.
- Demonstrāre per limites derivationem functionum constantium et functionis identitatis.
- Obtinēre regulas ad derivandas functiones trigonometricae ex derivationibus fundamentalibus sinus et cosinus.
- Computāre derivationes functionum trigonometricae compositarum utens regulis algebraicis.
- Demonstrāre formaliter derivationem logarithmi naturalis per limites.
INDEX RERUM:
Derivatio Functionum Algebraicarum
Derivationes Functionum Transcendentalium
Hactenus solum recensuimus quid sit derivatio et quasdam eius proprietates algebraicas; nihil tamen diximus de modis eius computandi. Hic problema illud solvemus ostendendo singulas technicas derivationis atque modum earum obtinendi pro singulis casibus.
Derivatio Functionum Algebraicarum
Functio Constans
Si f(x) = c, cum c quavis sit constans realis, tunc habebitur
\displaystyle \frac{df(x)}{dx} =\frac{d}{dx}c = 0
DEMONSTRATIO: Revera haec demonstratio uno tantum passu perficitur:
\begin{array}{rll} (1) &\displaystyle \dfrac{d}{dx}c &=\displaystyle \lim_{\Delta x \to 0} \dfrac{c - c}{\Delta x} \quad \text{; Definitio derivationis pro $f(x)=c$} \\ \\ & &=\displaystyle \lim_{\Delta x \to 0} \dfrac{0}{\Delta x} = 0 \end{array}
Functio Identitatis
\displaystyle \frac{df(x)}{dx} =\frac{dx}{dx}=1
DEMONSTRATIO: Fere idem ac prius, etiam uno passu obtinetur:
\begin{array}{rll} (1) & \dfrac{d}{dx}x &= \displaystyle \lim_{\Delta x\to 0} \dfrac{(x+\Delta x) - x}{\Delta x} \quad \text{; Definitio derivationis pro $f(x) = x$}\\ \\ & &=\displaystyle \lim_{\Delta x\to 0} \dfrac{\Delta x}{\Delta x} = 1 \end{array}
Potestates Naturales
Si f(x) = x^n, ubi n est naturale quodlibet, tunc habebitur:
\displaystyle \frac{df(x)}{dx} =\frac{dx^n}{dx} =nx^{n-1}
DEMONSTRATIO: Ad hoc theorema demonstrandum, oportet uti theoremate binomii Newtoniani
\begin{array}{rll} (1) &\displaystyle \dfrac{d}{dx}x^n = \lim_{\Delta x \to 0} \frac{(x+\Delta x)^n -x^n}{\Delta x} &\text{ ; Definitio liminis pro $f(x)= x^n$} \\ \\ & \displaystyle \phantom{\displaystyle \dfrac{d}{dx}x^n} =\lim_{\Delta x \to 0} \dfrac{\displaystyle \left[\sum_{k=0}^n {{n}\choose{k}} x^{n-k}(\Delta x)^{k} \right] - x^n}{\Delta x} & \text{; Theorema binomii Newtoniani, ad $(1)$} \\ \\ & \displaystyle\phantom{\displaystyle \dfrac{d}{dx}x^n} =\lim_{\Delta x \to 0} \dfrac{\displaystyle x^n + \left[\sum_{k=1}^n {{n}\choose{k}} x^{n-k}(\Delta x)^{k} \right] - x^n}{\Delta x} & \text{; Separato primo termino summae} \\ \\ & \displaystyle\phantom{\displaystyle \dfrac{d}{dx}x^n} =\lim_{\Delta x \to 0} \dfrac{\displaystyle \left[\sum_{k=1}^n {{n}\choose{k}} x^{n-k}(\Delta x)^{k} \right]}{\Delta x} & \text{; Terminis similibus sublatis} \\ \\ & \displaystyle\phantom{\displaystyle \dfrac{d}{dx}x^n} =\lim_{\Delta x \to 0} \displaystyle \left[\sum_{k=1}^n {{n}\choose{k}} x^{n-k}(\Delta x)^{k-1} \right] & \\ \\ & \displaystyle\phantom{\displaystyle \dfrac{d}{dx}x^n} =\lim_{\Delta x \to 0} \displaystyle \left[ {{n}\choose{1}} x^{n-1}(\Delta x)^{0} + \sum_{k=2}^n {{n}\choose{k}} x^{n-k}(\Delta x)^{k-1} \right] & \text{; Excerpto primo termino summae} \\ \\ & \displaystyle \color{blue} {\displaystyle \dfrac{d}{dx}x^n} = n x^{n-1} & \color{black} \end{array}
Potestates Integrales
Demonstratio modo recensita solum valet pro casu quo exponentes sunt numeri naturales, sed potest extendi ad quoscumque numeros integrales. Si a\in \mathbb{Z}, tunc habebitur
\displaystyle \frac{dx^a}{dx} = ax^{a-1}
Iam scimus hoc valere pro numeris integris positivis; sufficit ergo videre quid fiat cum sumantur exponentes negativi. Sufficit ostendere, igitur, verum esse quod sequitur:
\displaystyle \frac{dx^{-n}}{dx} = {-n}x^{-n-1}
DEMONSTRATIO: Ad hoc demonstrandum, sufficit considerare derivationem quotiensis:
\begin{array}{rll} (1) & \dfrac{d}{dx}x^{-n} &= \dfrac{d}{dx} \left( \dfrac{1}{x^n}\right) \\ \\ & &= \dfrac{0 \cdot nx^{n-1} - nx^{n-1} \cdot 1}{x^{2n}}\\ \\ & &= -nx^{n-1-2n} \\ \\ & &= -nx^{-n-1} \end{array}
Derivationes Functionum Transcendentalium
Functiones Trigonometricae
Hae continent sequentia praecepta derivationis:
| \displaystyle \frac{d}{dx}\sin(x) = \cos(x) | \displaystyle \frac{d}{dx}\sec(x) = \sec(x)\tan(x) |
| \displaystyle \frac{d}{dx}\cos(x) = -\sin(x) | \displaystyle \frac{d}{dx}\csc(x) = -\csc(x)\cot(x) |
| \displaystyle \frac{d}{dx}\tan(x) = \sec^2(x) | \displaystyle \frac{d}{dx}\cot(x) = -\csc^2(x) |
Ad obtinendas has singulas regulas, optimum est initium facere a derivationibus sinus et cosinus; exinde, his utens et algebra derivationum, derivationes ceterarum functionum trigonometricarum obtineri possunt.
Demonstratio Derivationis Sini
\begin{array}{rll} (1) &\dfrac{d}{dx}\sin(x) = \displaystyle \lim_{\Delta x \to 0} \dfrac{\sin(x+\Delta x) - \sin(x)}{\Delta x} & \text{; Definitio derivationis sinus} \\ \\ &\phantom{\dfrac{d}{dx}\sin(x)} = \displaystyle \lim_{\Delta x \to 0} \dfrac{\sin(x)\cos(\Delta x) + \sin(\Delta x)\cos(x) - \sin(x)}{\Delta x} & \\ \\ &\phantom{\dfrac{d}{dx}\sin(x)} = \displaystyle \lim_{\Delta x \to 0} \dfrac{ \sin(x)\left[\cos(\Delta x) -1\right] + \sin(\Delta x)\cos(x) }{\Delta x} & \\ \\ &\phantom{\dfrac{d}{dx}\sin(x)} = \displaystyle \sin(x)\lim_{\Delta x \to 0} \left[\dfrac{\cos(\Delta x) - 1}{\Delta x} \right] + \cos(x) \lim_{\Delta x \to 0} \left[ \dfrac{\sin(\Delta x)}{\Delta x} \right] & \\ \\ (2)&\displaystyle \lim_{\Delta x\to 0} \dfrac{\sin(\Delta x)}{\Delta x} = 1 & \text{; Per theorema interpositum (Sandwich)}\\ \\ (3)&\displaystyle \lim_{\Delta x\to 0} \dfrac{\cos(\Delta x) - 1}{\Delta x} = \lim_{\Delta x\to 0} \dfrac{\cos(\Delta x) - 1}{\Delta x} \cdot \dfrac{\cos(\Delta x) + 1}{\cos(\Delta x) + 1} & \\ \\ &\displaystyle\phantom{\lim_{\Delta x\to 0} \dfrac{\cos(\Delta x) - 1}{\Delta x}} = \lim_{\Delta x\to 0} \dfrac{\cos^2(\Delta x) - 1}{\Delta x (\cos(\Delta x) + 1)} & \\ \\ &\displaystyle\phantom{\lim_{\Delta x\to 0} \dfrac{\cos(\Delta x) - 1}{\Delta x}} = \lim_{\Delta x\to 0} \dfrac{-\sin^2(\Delta x)}{\Delta x (\cos(\Delta x) + 1)} & \\ \\ &\displaystyle\phantom{\lim_{\Delta x\to 0} \dfrac{\cos(\Delta x) - 1}{\Delta x}} =- \lim_{\Delta x\to 0} \dfrac{\sin(\Delta x)}{\Delta x} \cdot \lim_{\Delta x\to 0} \dfrac{\sin(\Delta x)}{\cos(\Delta x) + 1} & \\ \\ &\displaystyle\phantom{\lim_{\Delta x\to 0} \dfrac{\cos(\Delta x) - 1}{\Delta x}} =- (1)\cdot(0) = 0 \\ \\ (4) &\color{blue}\dfrac{d}{dx}\sin(x) = \cos(x) \color{black} & \text{; Ex (1,2,3)} \end{array}
Demonstratio Derivationis Cosini
\begin{array}{rll} (1) & \dfrac{d}{dx}\cos(x) = \displaystyle \lim_{\Delta x \to 0} \dfrac{\cos(x + \Delta x) - \cos(x)}{\Delta x} & \text{; Definitio derivationis cosini} \\ \\ & \phantom{\dfrac{d}{dx}\cos(x)} = \displaystyle \lim_{\Delta x \to 0} \dfrac{\cos(x)\cos(\Delta x) - \sin(x)\sin(\Delta x) - \cos(x)}{\Delta x} \\ \\ & \phantom{\dfrac{d}{dx}\cos(x)} = \displaystyle \lim_{\Delta x \to 0} \dfrac{\cos(x) [ \cos(\Delta x) - 1] - \sin(x)\sin(\Delta x)}{\Delta x}\\ \\ & \phantom{\dfrac{d}{dx}\cos(x)} = \cos(x) \displaystyle \lim_{\Delta x \to 0} \dfrac{ [ \cos(\Delta x) - 1]}{\Delta x} - \sin(x) \lim_{\Delta x \to 0} \dfrac{\sin(\Delta x)}{\Delta x}\\ \\ & \phantom{\dfrac{d}{dx}\cos(x)} = \cos(x) \cdot(0) - \sin(x)\cdot (1)\\ \\ &\color{blue}\dfrac{d}{dx}\cos(x) = - \sin(x) \color{black} \end{array}
Derivationes Tangentis, Secantis, Cosescantis et Cotangentis
Habitis resultatis pro sinu et cosino, derivationes reliquarum functionum trigonometricarum facile nunc obtinentur.
\begin{array}{rl} \dfrac{d}{dx}\tan(x) &= \dfrac{d}{dx} \left( \dfrac{\sin(x)}{\cos(x)} \right) = \dfrac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} = \dfrac{1}{\cos^2(x)} = \color{blue}\sec^2(x) \color{black}\\ \\ \dfrac{d}{dx}\sec(x) &= \dfrac{d}{dx}\left(\dfrac{1}{\cos(x)} \right) = \dfrac{\sin(x)}{\cos^2(x)} =\color{blue}\sec(x)\tan(x) \color{black}\\ \\ \dfrac{d}{dx}\csc(x) &= \dfrac{d}{dx}\left(\dfrac{1}{\sin(x)}\right) = -\dfrac{\cos(x)}{\sin^2(x)} =\color{blue} - \csc(x)\cot(x)\color{black}\\ \\ \dfrac{d}{dx} \cot(x) &= \dfrac{d}{dx} \left(\dfrac{\cos(x)}{\sin(x)}\right) = \dfrac{-\sin^2(x)-\cos^2(x)}{\sin^2(x)} = -\dfrac{1}{\sin^2(x)} =\color{blue} -\csc^2(x)\color{black} \end{array}
Derivatio Functionum Logarithmicarum
Derivatio logarithmi naturalis data est per formulam
\displaystyle \frac{d}{dx}\ln(x) = \frac{1}{x}
DEMONSTRATIO: Ratiocinando ex definitione derivationis, habetur argumentatio sequens:
\begin{array}{rll} (1) & \dfrac{d}{dx} \ln(x) = \displaystyle \lim_{\Delta x \to 0} \left [\dfrac{\ln(x+\Delta x) - \ln(x)}{\Delta x} \right] &\text{; Definitio derivationis pro Logarithmo Naturali} \\ \\ & \phantom{\dfrac{d}{dx} \ln(x)} = \displaystyle \lim_{\Delta x \to 0} \left[ \dfrac{1}{\Delta x} \ln \left( \dfrac{x+\Delta x}{x} \right) \right] \\ \\ & \phantom{\dfrac{d}{dx} \ln(x)} = \displaystyle \lim_{\Delta x \to 0} \left[ \ln \left( \dfrac{x+\Delta x}{x} \right)^{\frac{1}{\Delta x} } \right] \\ \\ & \phantom{\dfrac{d}{dx} \ln(x)} = \displaystyle \lim_{\Delta x \to 0} \left[ \ln \left( \dfrac{x+\Delta x}{x} \right)^{\frac{1}{\color{red}x\color{black}} \frac{\color{red}x\color{black}}{\Delta x} } \right] \\ \\ & \phantom{\dfrac{d}{dx} \ln(x)} = \displaystyle \lim_{\Delta x \to 0} \left[ \dfrac{1}{x} \ln \left( 1 + \dfrac{\Delta x}{x} \right)^{ \frac{x}{\Delta x} } \right] \\ \\ & \phantom{\dfrac{d}{dx} \ln(x)} =\dfrac{1}{x} \ln \displaystyle \left[ \lim_{\Delta x \to 0} \left( 1 + \dfrac{\Delta x}{x} \right)^{ \frac{x}{\Delta x} } \right] \\ \\ (2) & n=\dfrac{x}{\Delta x} & \text{; substitutio}\\ \\ (3) & (\Delta x \to 0^+) \longrightarrow (n\to +\infty) \\ \\ (4) & \dfrac{d}{dx} \ln(x) = \dfrac{1}{x} \ln\left[ \displaystyle \lim_{n \to +\infty} \left(1 + \dfrac{1}{n} \right)^n \right] = \dfrac{1}{x} \ln(e) = \color{blue}\dfrac{1}{x} \color{black} & \text{; Ex (1,2,3)} \end{array}
His peractis, passu per passum percursae sunt derivationes fundamentales quas quisque discipulus cognoscere debet: a functionibus algebraicis fundamentalibus usque ad praecipuas functiones transcendentes, ut sunt trigonometricae et logarithmus naturalis. His demonstrationibus bene intellectis, poteris regulas derivationis adhibere, earum originem ac iustificationem formalem comprehendere. Haec cognitio fundamentum praebet ad tractanda confidenter problemata difficiliora, quae subtiliorem mutationis analysin exigunt.