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\title{Derivatives}
\date{April 1, 2022}

\begin{document}
\ \vspace{-.25in}

\maketitle

\vspace{-.15in}



\dfn{
Let $I$ be an interval and let $f: I \to \RR$ and $c \in I$.
The \textbf{derivative} of $f$ at $c$ is $\displaystyle \boxed{\lim_{x \to c} \frac{f(x) - f(c)}{x-c}}$.
If the limit converges, then we say $f$ is differentiable at $c$. 
If $f$ is differentiable at every $c \in I$, then we say that $f$ is differentiable.
The quantity $\displaystyle \frac{f(x) - f(c)}{x-c}$ is called the difference quotient.
}

\prob{
Let $f: \RR \to \RR$ be defined by $f(x) = x^2$.
Prove that $f$ is differentiable (let $c \in \RR$ and prove that $\lim_{x \to c} \frac{f(x) - f(c)}{x-c}$ converges).
}

\prob{
Let $f: \RR \to \RR$ be defined by $f(x) = \lvert x \rvert$.
Prove that $f$ is not differentiable at $0$.
}

\newpage

\prob{
Prove the product rule: if $f$ and $g$ are both differentiable at $c$, then $(fg)'(c) = g(c)f'(c) + f(c)g'(c)$ (and thus $fg$ is differentiable at $c$).
}

\dfn{
Let $f: S \to \RR$.
The function $f$ has a \textbf{relative maximum} at $c\in S$ if there is $\delta > 0$ such that for all $x \in S \cap (c-\delta, c + \delta)$, $f(x) \leq f(c)$.
Reverse the last inequality to get the definition of a \textbf{relative minimum}.
}

\prob{Suppose $f: [a,b] \to \RR$ is differentiable at $c \in (a,b)$ and $f$ has a relative minimum at $c$.
Prove that $f'(c) = 0$.
Hint: break the interval $(c-\delta, c+\delta)$ into $(c-\delta, c]$ and $[c, c+\delta)$.
}

\end{document}