Published Jul 03, 2021
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Definition 5.2.1 (Differentiability) Let \(g : A \to \mathbb{R}\) be a function defined on an interval \(A\). Given \(c \in A\), the derivative of \(g\) at \(c\) is defined by $$ g^\prime(c) = \lim_{x \to c}\frac{g(x) - g(c)}{x - c} $$ provided this limit exists. In this case, we say \(g\) is differentiable at \(c\) if \(g^\prime\) exists for all points \(c \in A\), we say that \(g\) is differentiable on \(A\)
Theorem 5.2.3. If \(g : A \to \mathbb{R}\) is differentiable at a point \(c \in A\), then \(g\) is continuous at \(c\) as well.
Theorem 5.2.4 (Algebraic Differentiability Theorem).
Let \(f\) and \(g\) be functions defined on an interval \(A\), and assume
both are differentiable at some point \(c \in A\). Then,
Theorem 5.2.5 (Chain Rule). Let \(f : A \to \mathbb{R}\) and \(g : B \to \mathbb{R}\) satisfy \(f(A) \subseteq B\) so that the composition of \(g \circ f\) is defined. If \(f\) is differentiable at \(c \in A\) and \(g\) is differentiable at \(f(c) \in B\), then \(g \circ f\) is differentiable at \(c\) with \((g \circ f)^\prime(c) = g^\prime(f(c)) \cdot f^\prime(c)\)
Theorem 5.2.6 (Interior Extremum Theorem). Let \(f\) be differentiable on an open interval \((a, b).\) if \(f\) attains a maximum value at some point \(c \in (a, b)\), then \(f^\prime(c) = 0\). The same is true if \(f(c)\) is a minimum value.
Theorem 5.2.7 (Darboux's Theorem). If \(f\) is differentiable on an interval \([a, b]\), and if \(\alpha\) satisfies \(f^\prime(a) < \alpha < f^\prime(b)\), then there exists a point \(c \in (a, b)\) where \(f^\prime(c) = \alpha\)