160 lines
5.1 KiB
TeX
160 lines
5.1 KiB
TeX
\langchapter{Différentiabilité}{Differentiability}
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%TODO Complete chapter
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Une fonction $\function{f}{I}{\K}$ est dérivable en $a \in I$ si le taux d'accroissement $T = \lim\limits_{x \to a}\frac{f(x) - f(a)}{x - a}$, si la limite existe $f'=T$
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\label{definition:derivative}
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$\lim\limits_{x \to a}\frac{f(x) - f(a)}{x - a} \equivalence \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}$
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\begin{proof}
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Let $f = \lim\limits_{x \to a}\frac{f(x) - f(a)}{x - a}$ and $h = x - a$
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$\equivalence \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}$
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\end{proof}
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\langsection{Propriétés}{Proprieties}
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%TODO Complete section
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$f$ est dérivable sur un intervale $I \implies f \in C^1$.
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Remarque : La réciproque est fausse, voir les fonctions de Weierstrass.
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\langsection{Dérivé de fonctions usuelles}{Dérivative of usuals functions}
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\langsubsection{Somme/Soustraction}{Sum/Difference}
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Let $\function{f,g}{I}{\K}$ est dérivable en $a \in I$, $(f + g)' = f' + g'$ ainsi que $(f - g)' = f' - g'$
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\begin{proof}
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$(f + g)' = \lim\limits_{h \to 0} \frac{[f(x + h) + g(x + h)] - [f(x) + g(x)]}{h} = \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} + \lim\limits_{h \to 0} \frac{g(x + h) - g(x)}{h} = f' + g'$
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\end{proof}
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With a similar argument, $(f - g)' = f' - g'$
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\langsubsection{Produit}{Product}
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Let $\function{f,g}{I}{\K}$ est dérivable en $a \in I$, $(fg)' = f'g + fg'$
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\begin{proof}
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$\implies (fg)' = \lim\limits_{h \to 0} \frac{f(x + h)g(x + h) - f(x)g(x)}{h}$
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$\implies \lim\limits_{h \to 0} \frac{f(x + h)g(x + h) - f(x)g(x) + f(x + h)g(x) - f(x + h)g(x)}{h}$
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$\implies \lim\limits_{h \to 0} \frac{f(x + h)[g(x + h) - g(x)] + g(x)[f(x + h) - f(x)]}{h}$
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$\implies \lim\limits_{h \to 0} f(x + h) \frac{g(x + h) - g(x)}{h} + g(x) \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}$
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$\implies f'g + fg'$
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\end{proof}
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\langsubsection{Division}{Quotient}
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Soit $\function{f,g}{I}{\K}$ est dérivable en $a \in I$, $(\frac{f}{g})' = \frac{f'g + fg'}{g^2}$
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\begin{proof}
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$\implies (\frac{f}{g})' = \lim\limits_{h \to 0} \frac{\frac{f(x + h)}{g(x + h)} - \frac{f(x)}{g(x)}}{h}$
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$\implies \lim\limits_{h \to 0} \frac{g(x)f(x + h) - f(x)g(x + h)}{g(x)g(x + h)h}$
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$\implies \lim\limits_{h \to 0} \frac{1}{g(x)g(x + h)} [ \frac{g(x)f(x + h) - f(x)g(x + h)}{h} ]$
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$\implies \lim\limits_{h \to 0} \frac{1}{g(x)g(x + h)} [ \frac{g(x)f(x + h) - f(x)g(x + h) + f(x)g(x) - f(x)g(x)}{h} ]$
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$\implies \lim\limits_{h \to 0} \frac{1}{g(x)g(x + h)} [ \frac{g(x)[f(x + h) - f(x)] + f(x)[g(x + h) - g(x)]}{h} ]$
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$\implies \lim\limits_{h \to 0} \frac{1}{g(x)g(x + h)} [ g(x)\frac{[f(x + h) - f(x)]}{h} + f(x)\frac{[g(x + h) - g(x)]}{h} ]$
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$\implies \frac{f'g + fg'}{g^2}$
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\end{proof}
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\subsection{Composition}
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Soit $\function{f,g}{I}{\K}$ est dérivable en $a \in I$, $(f \composes g(x))' = f' \composes g(x) + g'(x)$
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\begin{proof}
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\lipsum[2]
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% TODO Complete proof
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\end{proof}
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\langsubsection{Exponentiel}{Exponential}
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\subsubsection{Base e}
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Soit $x \in R$ et $\function{f}{\R}{\R}, (e^{f(x)})' = f'(x)e^{f(x)}$
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\begin{proof}
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\lipsum[2]
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% TODO Complete proof
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\end{proof}
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\langsubsubsection{Base arbitraire}{Arbitrary base}
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Soit $x \in \R, b \in \R^*$ et $\function{f}{\R}{\R}, (b^{f(x)})' = f'(x)b^{f(x)}$
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\begin{proof}
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Soit $x \in \R, b \in \R^*$ et $\function{f}{\R}{\R}$
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Il y a plusieurs manières de prouver l'égalité en voici certaines :
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\textbf{Preuve par calcul de limite}
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$$(b^{f(x)})' = \lim\limits_{h \to 0} \frac{b^{x + h} - b^x}{h} = b^x \lim\limits_{h \to 0} \frac{b^h - 1}{h}$$ %TODO Complete proof (find ln trhough limits)
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\textbf{Preuve par la dérivé de $e^{f(x)}$}
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$$(b^{f(x)})' = (e^{f(x)\ln(b)})'$$
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By the chain rule
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$$e^{f(x)\ln(b)} f(x)'\ln(b) = f(x)'\ln(b)b^{f(x)}$$
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\end{proof}
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\langsubsection{Logarithme}{Logarithm}
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\subsubsection{Base e}
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Soit $x \in R^*_+, (\ln(x))' = \frac{1}{x}$
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\begin{proof}
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Il y a plusieurs manières de prouver l'égalité en voici certaines :
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\textbf{Preuve par instantiation}
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Soit $x \in R^*_+$, posons $\frac{d}{dx} x = 1$
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$$\implies \frac{d}{dx} e^{\ln(x)} = 1$$
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By the chain rule
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$$e^{\ln(x)} \frac{d}{dx} \ln(x) = 1 \implies x \frac{d}{dx} \ln(s) = 1 \implies \frac{d}{dx} \ln(s) = \frac{1}{x}$$
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\textbf{Preuve par dérivé implicite}
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Soit $x \in R^*_+$, posons $y = \ln{x}$
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$$\implies e^y = x \implies \frac{dy}{dx} e^y = 1 \implies \frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{e^{\ln(x)}} \implies \frac{d}{dx} \ln(x) = \frac{1}{x}$$
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\end{proof}
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\langsubsubsection{Base arbitraire}{Arbitrary base}
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Soit $x \in R^*_+, (\log_b(x))' = \frac{1}{x \ln(b)}$
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\begin{proof}
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Il y a plusieurs manières de prouver l'égalité en voici certaines :
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\textbf{Preuve par instantiation}
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Soit $x \in R^*_+$, posons $\frac{d}{dx} x = 1$
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$$\implies \frac{d}{dx} b^{\log_b(x)} = 1$$
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By the chain rule
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$$\ln(s) b^{\log_b(x)} \frac{d}{dx} \log_b(x) = 1 \implies x \ln(b) \frac{d}{dx} \log_b(s) = 1 \implies \frac{d}{dx} \log_b(s) = \frac{1}{x\ln(b)}$$
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\textbf{Preuve par dérivé implicite}
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Soit $x \in R^*_+$, posons $y = \log_b{x}$
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$$\implies b^y = x \implies \ln(b) \frac{dy}{dx} b^y = 1 \implies \frac{dy}{dx} = \frac{1}{\ln(b)b^y} = \frac{1}{\ln(b)b^{\log_b(x)}} \implies \frac{d}{dx} \log_b(x) = \frac{1}{x \ln(b)}$$
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\end{proof}
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\section{Extremums}
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%TODO Complete section
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