Local extrema is not defined on the boundaries of an interval.

Differentiation

Suppose $f: [a, b] \rightarrow \R$ is a function and fix $x \in [a, b]$. For $t \neq x$, define the difference quotient $\varphi(t)$ as $\frac{f(t) - f(x)}{ t- x}$. The derivative of $f$ at $x$ is defined below as:

$$ f'(x) :=\lim _{t \rightarrow x} \varphi(t) = \lim _{t \rightarrow x}\frac{f(t) - f(x)}{t-x} $$

If this limit exists, $f$ is said to be differentiable at $x$ and if it exists for all $x \in E \sube [a, b]$, $f$ is said to be differentiable on $E$. Note that if $x = a$, then we take $f'(x) = \varphi(x^+)$ and similarly for $x=b$.

Differentiability Implies Continuity. Suppose that $f: [a,b] \rightarrow \R$ is differentiable at a point $x\in [a,b]$. Then $f$ is continuous at $x$.

Proof </aside>

Linearity, Product, Quotient Rules. Suppose that $f, g : [a, b] \rightarrow \R$ are differentiable at a point $x \in [a, b]$. Then $f+g, fg, \frac{f}{g}$ are differentiable at $x$ and are given by:

$$ \textbf{Linearity. } (f+g)'(x) = f'(x) + g'(x) \hspace{131pt} \\ \textbf{Product Rule. } (fg)'(x) = f'(x)g(x) + f(x)g'(x) \hspace{84pt} \\ \textbf{Quotient Rule. } (\frac{f}{g})'(x) = \frac{1}{g(x)^2} (f'(x)g(x)- f(x)g'(x)) \hspace{15pt} g(x) \neq 0 $$

The proofs for these follow from the definition of the derivative and offer no insight.

</aside>

Chain Rule. Suppose that $f : [a,b] \rightarrow \R$ is continuous on $[a,b]$ and differentiable at a point $x \in [a,b]$ and suppose that $g : I \rightarrow \R$ is defined such that $f([a,b]) \sube I$ and $g$ is differentiable at $f(x)$. Then the function $h := g \circ f$ is differentiable at $x$ and is given by:

$$ h'(x) = (g\circ f)'(x) = g'(f(x)) \cdot f'(x) $$

Proof </aside>

L’Hopital’s Rule. Suppose $f,g : [a, b] \rightarrow \R$ are differentiable on $(a, b)$ and $g'(x) \neq 0$ for all $x \in (a,b)$. Also, suppose that $\lim _{x \rightarrow a} \frac{f'(x)}{g'(x)} \rightarrow A$. Then,

$$ [\lim _{x \rightarrow a} f(x) = \lim _{x \rightarrow a} g(x) = 0 \text{ or }
\lim _{x \rightarrow a } g(x) = +\infty ] \implies \lim _{x \rightarrow a} \frac{f(x)}{g(x)} = A $$

Proof </aside>

Growth Theorems

Local Extrema are Stationary. Suppose that $f : [a, b] \rightarrow \R$ . If $x \in (a,b)$ is a local maximum or local minimum, then if $f$ is differentiable at $x$, $f'(x) = 0$.

Proof </aside>

Generalized Mean Value Theorem. Suppose $f, g : [a, b] \rightarrow \R$ are continuous on $[a, b]$ and differentiable on $(a,b)$. Then there exists $x \in (a,b)$ such that:

$$ (f(b) - f(a))g'(x) = (g(b) - g(a)) f'(x) $$

Proof </aside>

Sign of Derivative Implies Monotonicity. Suppose $f : [a,b] \rightarrow \R$ is differentiable on $(a, b)$. Then, we have:

$$ \forall x \in (a, b), f'(x) \geq 0 \implies f \text{ is monotonic and increasing} \\

\forall x \in (a, b), f'(x) = 0 \implies f \text{ is constant } \hspace{71pt} \\

\hspace{1pt} \forall x \in (a, b), f'(x) \geq 0 \implies f \text{ is monotonic and decreasing} $$

Proof </aside>

Continuity Theorems

Derivatives have no Simple Discontinuities. Suppose that $f: [a, b] \rightarrow \R$ is differentiable on $[a, b]$ and that $\lambda \in (f'(a), f'(b))$. Then, there exists $x \in (a, b)$ such that $f'(x) = \lambda$.

An equivalent statement of this theorem is that derivatives satisfy the conclusion of the intermediate value theorem, even if discontinuous.

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Differentiable Functions have no Simple Discontinuities. Suppose that $f :[a,b ] \rightarrow \R$ is differentiable on $[a,b]$. Then $f$ cannot have any simple discontinuities.

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Higher-Order Derivatives

For $f^{(n)}$ to exist at a point $x \in \R$, $f^{(n-1)}$ must exist in a neighborhood of $x$ and $f^{(n-1)}$ must be differentiable at $x$. However, for $f^{(n-1)}$ to exist in a neighborhood of $x$, $f^{(n-2)}$ must be differentiable in that neighborhood.