64 lines
3.7 KiB
TeX
64 lines
3.7 KiB
TeX
\section{The Derivative}
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\begin{definition}
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Let $I \subseteq \R$ be an interval, let $f:I \rightarrow \R$, and let $ c \in I$. We say that a real number $L$ is the \textbf{derivative of $f$ at $c$} if given any $\varepsilon > 0$ there exists $\delta (\varepsilon) > 0$ such that if $x \in I$ satisfies $0 < |x-c|<\delta (\varepsilon)$, then
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\[\abs{\frac{f(x)-f(c)}{x-c}-L}<\varepsilon.\]
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In this case we say that $f$ is \textbf{differentiable} at $c$, and we write $f'(c)$ for $L$. In other words, the derivative of $f$ at $c$ is given by the limit
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\[f'(c) = \lim\limits_{x\to c} \frac{f(x)-f(c)}{x-c}\]
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provided this limit exists. (We allow the possibility that $c$ may be the endpoint of the interval.)
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\end{definition}
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\begin{theorem}
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If $f:I \rightarrow \R$ has a derivative at $c \in I$, then $f$ is continuous at $c$.
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\end{theorem}
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\begin{theorem}
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Let $I \subseteq \R$ be an interval, let $c \in I$ , and let $f:I \rightarrow \R$ and $g:I \rightarrow \R$ be functions that are differentiable at $c$. Then:
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\begin{enumerate}
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\item If $\alpha \in \R$, then the function $\alpha f$ is differentiable at $c$, and \[(\alpha f)'(c) = \alpha f'(c)\]
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\item The function $f+g$ is differentiable at $c$, and
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\[(f+g)'(c) = f'(c)+g'(c)\]
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\item (Product Rule) The function $fg$ is differentiable at $c$, and
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\[(fg)'(c) = f'(c)g(c) + f(c)g'(c).\]
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\item (Quotient Rule) If $g(c) \neq 0$, then the function $f/g$ is differentiable at $c$, and
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\[\left( \frac{f}{g}\right)'(c) = \frac{f'(c)g(c)-f(c)g'(c)}{(g(c))^2}\]
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\end{enumerate}
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\end{theorem}
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\begin{corollary}
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If $f_1, f_2, \dots, f_n$ are functions on an interval $I$ to $\R$ that are differentiable at $c \in I$, then:
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\begin{enumerate}
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\item The function $f_1 + f_2 + \dots + f_n$ is differentiable at $c$ and
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\[(f_1 + f_2 + \dots + f_n)'(c) = f_1'(c) + f_2'(c) + \dots + f_n'(c)\]
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\item The function $f_1f_2 \dots f_n$ is differentiable at $c$ and
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\[(f_1f_2 \dots f_n)'(c) = f_1'(c)f_2(c) \dots f_n(c)+f_1(c)f_2'(c) \dots f_n(c) + \dots + f_1(c)f_2(c) \dots f_n'(c).\]
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An important special case of the extended product rule occurs if the functions are equal, that is, $f_1 = f_2 = \dots = f_n = f$. Then the above becomes
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\[(f^n)'(c) = n(f(c))^{n-1}f'(c)\]
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\end{enumerate}
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\end{corollary}
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\begin{theorem}[\textbf{Carathéodory's Theorem}]
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Let $f$ be defined on an interval $I$ containing the point $c$. Then $f$ is differentiable at $c$ if and only if there exists a function $\varphi$ on $I$ that is continuous at $c$ and satisfies
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\[f(x)-f(c)=\varphi (x)(x-c)\ \ \ \ \text{for}\ \ \ \ x \in I\]
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In this case, we have $\varphi (c)=f'(c)$.
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\end{theorem}
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\begin{theorem}[\textbf{Chain Rule}]
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Let $I, J$ be intervals in $\R$, let $g:I \rightarrow \R$ and $f:J \rightarrow \R$ be functions such that $f(J) \subseteq I$, and let $c \in J$. If $f$ is differentiable at $c$ and if $g$ is differentiable at $f(c)$, then the composite function $g \circ f$ is differentiable at $c$ and
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\[(g \circ f)'(c) = g'(f(c)) \cdot f'(c).\]
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\end{theorem}
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\begin{theorem}
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Let $I$ be an interval in $\R$ and let $f:I \rightarrow \R$ be strictly monotone and continuous on $I$. Let $J:=f(I)$ and let $g:J \rightarrow \R$ be the strictly monotone and continuous function inverse to $f$. If $f$ is differentiable at $c \in I$ and $f'(c) \neq0$, then $g$ is differentiable at $d:=f(c)$ and
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\[g'(d)=\frac{1}{f'(c)}=\frac{1}{f'(g(d))}\]
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\end{theorem}
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\begin{theorem}
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Let $I$ be an interval and let $f:I \rightarrow \R$ be strictly monotone on $I$. Let $J:= f(I)$ and let $g:J \rightarrow \R$ be the function inverse to $f$. If $f$ is differentiable on $I$ and $f'(x) \neq 0$ for $x \in I$, then $g$ is differentiable on $J$ and
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\[g' = \frac{1}{f' \circ g}\]
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\end{theorem}
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