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\lecture{2: Derivatives and Integrals}{Week 2}
This is the second week of the Mathematics Subject Test GRE prep course; here, we review the concepts of \textbf{derivatives} and \textbf{integrals}!
\section{Bestiary of Functions}
For convenience's sake, we list the definitions, integrals, derivatives, and key values of several functions here.
\begin{align*}
\begin{array}{|c|c|c|c|c|}
\hline
\textrm{Name} & \textrm{Domain} & \textrm{Derivative} & \textrm{Integral} & \textrm{Key Values} \\
\hline
\ln(x) & (0,\infty) & 1/x & x\cdot\ln(x) -x + C & \ln(1) = 0,\\
& & & & \ln(e) = 1 \\
\hline
e^x & \mathbb{R} & e^x & e^x +C & e^0 = 1, \\
& & & & e^1 = e \\
\hline
\sin(x) & \mathbb{R}& \cos(x) & -\cos(x) + C & \sin(0) = 0, \\
& & & & \sin\left(\frac{\pi}4\right) = \frac{\sqrt{2}}{2},\\
& & & & \sin\left(\frac{\pi}2\right) = 1\\
\hline
\cos(x) & \mathbb{R}& -\sin(x) & \sin(x) + C & \cos(0) = 1, \\
& & & &\cos\left(\frac{\pi}4\right) = \frac{\sqrt{2}}{2},\\
& & & & \cos\left(\frac{\pi}2\right) = 0\\
\hline
\tan(x) & x \neq \frac{2k+1}{2}\pi & \sec^2(x) & \ln|\sec(x)| + C & \tan(0) = 0, \\
& & & &\tan\left(\frac{\pi}4\right) = 1\\
\hline
\sec(x) & x \neq \frac{2k+1}{2}\pi & \sec(x)\tan(x) & \ln|\sec(x) + \tan(x)| + C & \sec(0) = 1, \\
& & & &\sec\left(\frac{\pi}4\right) = \sqrt{2}\\
\hline
\csc(x) & x \neq k\pi & -\csc(x)\cot(x) & \ln|\csc(x) - \cot(x)| + C & \csc\left(\frac{\pi}4\right) = \sqrt{2}, \\
& & & &\sec\left(\frac{\pi}2\right) = 1\\
\hline
\cot(x) & x \neq k\pi & -\csc^2(x) & \ln|\sin(x)| + C & \cot\left(\frac{\pi}4\right) = 1, \\
& & & &\cot\left(\frac{\pi}2\right) = 0\\
\hline
\arcsin(x) &(-1, 1) & \frac{1}{\sqrt{1-x^2}} & x\arcsin(x) +\sqrt{1-x^2} + C & \arcsin\left(0\right) = 0, \\
& & & &\arcsin\left(1\right) = \frac{\pi}{2} \\
\hline
\arccos(x) & (-1, 1) & -\frac{1}{\sqrt{1-x^2}} & x\arccos(x) -\sqrt{1-x^2} + C & \arccos\left(0\right) = \frac\pi2, \\
& & & &\arcsin\left(1\right) = 0 \\
\hline
\arctan(x) & \mathbb{R} & \frac{1}{1+x^2} & x\arctan(x) -\frac{\ln(1+x^2)}{2} + C & \arctan\left(0\right) = 0, \\
& & & &\arctan\left(1\right) = \frac{\pi}{2} \\
\hline
\end{array}
\end{align*}
Memorizing all of these is not necessary to do well on the GRE:\ as we'll discuss in class, you can derive almost all of these identities on the fly by using the product/chain rules or integration by parts/substitution! However, doing those calculations can take time, and memorizing these formulas will save you time on the test; consider studying them in the two weeks before you test with flashcards and the like!
\newpage
\section{The Derivative}
As always, we start with the formal definition:
\begin{defn}
For a function $f$ defined on some neighborhood $(a-\delta, a+\delta)$, we say that $f$ is \textbf{differentiable} at $a$ iff the limit
\begin{align*}
\lim_{h \to 0} \frac{f(a+h) - f(a)}{(a+h) - a}
\end{align*}
exists. If it does, denote this limit as $f'(a)$; we will often call this value the \textbf{derivative} of $f$ at $a$.
\end{defn}
Again, as before, if you find yourself directly using this definition to solve a GRE problem, something has likely gone wrong! Instead, you likely want to use one of several rules that we know for evaluating derivatives:
\subsection{Tools}
\begin{enumerate}
\item \textbf{Differentiation is linear}: For $f$, $g$ a pair of functions differentiable at $a$ and $\alpha, \beta$ a pair of constants,
\begin{align*}
(\alpha f(x) + \beta g(x))'\Big|_{a} =\alpha f'(a) + \beta g'(a).
\end{align*}
\item \textbf{Product rule}: For $f$, $g$ a pair of functions differentiable at $a$,
\begin{align*}
(f(x) \cdot g(x))'\Big|_{a} = f'(a) \cdot g(a) + g'(a) \cdot f(a).
\end{align*}
\item \textbf{Quotient rule}: For $f$, $g$ a pair of functions differentiable at $a$, $g(a) \neq 0$, we have
\begin{align*}
\left(\frac{f(x)}{ g(x)}\right)'\Bigg|_{a} = \frac{f'(a) \cdot g(a) - g'(a) \cdot f(a)}{(g(a))^2}
\end{align*}
\item \textbf{Chain rule}: For $f$ a function differentiable at $g(a)$ and $g$ a function differentiable at $a$,
\begin{align*}
(f(g(x)))'\Big|_{a} = f'(g(a)) \cdot g'(a).
\end{align*}
\item \textbf{Inverse function rule}: Suppose that $f(x)$ is a bijective function with inverse $f^{-1}(x)$, and that $f^{-1}(x)$ is differentiable at some point $a$. Then we have that
\begin{align*}
\left(f^{-1}(x)\right)\Big|_a = \frac{1}{f'(f^{-1}(a))}
\end{align*}
\end{enumerate}
\subsection{Theorems and Interpretations}
The derivative has a number of useful interpretations and associated theorems. We state a few here:
\begin{enumerate}
\item \textbf{Physical phenomena}: if $f(x)$ is a function that calculates distance with respect to some time $t$, you can think of the derivative $f'(t)$ as denoting the velocity of $f$ at time $t$, and $f''(t)$ as denoting the acceleration of $f$ at time $t$.
\item \textbf{Tangents}: if $f(x) = y$ is a curve, the slope of the tangent to this curve at any point $x_0$ is given by $f'(x_0)$.
\begin{center}
\includegraphics[width=3in]{tangent.pdf}
\end{center}
\item \textbf{Mean Value Theorem}: The Mean Value Theorem (abbreviated MVT) is the following result. Suppose that $f$ is a continuous function on the interval $[a,b]$ that's differentiable on $(a,b)$. Then there is some value $c$ such that
\begin{align*}
f'(c) = \frac{f(b) - f(a)}{b-a}.
\end{align*}
In other words, there is some point $c$ between $f(a)$ and $f(b)$ such that the derivative at that point is equal to the slope of the line segment connecting $(a,f(a))$ and $(b,f(b))$. The following picture illustrates this claim:
\begin{center}
\includegraphics[width=4.5in]{math8_wk6_mvt.pdf}
\end{center}
\item \textbf{Classification of extrema}: You can use the derivative to find minima and maxima of functions! Specifically, recall the following two definitions:
\begin{defn}
A function $f$ has a \textbf{critical point} at some point $x$ if either of the two properties hold:
\begin{itemize}
\item $f$ is not differentiable, or
\item $f'(x) = 0$.
\end{itemize}
\end{defn}
\begin{defn}
A function $f$ has a \textbf{local maxima}(resp. \textbf{local minima}) at some point $a$ iff there is a neighborhood $(a-\delta, a+\delta)$ around $a$ such that $f(a) \geq f(x)$ (resp. $f(a) \leq f(x)$,) for any $x \in (a-\delta, a+\delta)$.
\end{defn}
Derivatives relate to these properties as follows:
\begin{prop}
If $f$ is a function that has a local minima or maxima at some point $t$, $t$ is a critical point of $f$. As a corollary, if $f$ is a continuous function defined on some interval $[a,b]$, $f$ adopts its global minima and maxima at either $f$'s critical points in $[a,b]$, or at the endpoints $\{a,b\}$ themselves.
Furthermore, if $f''(x)$ is defined and positive at one of these critical points, $f$ adopts a local minima at $x$; conversely, if $f''(x)$ exists and is negative, $f$ adopts a local maxima at $x$.
\end{prop}
\end{enumerate}
To illustrate these ideas, we work some sample problems:
\subsection{Worked examples.}
\begin{quest}
Choose $n \in \mathbb{N}$. Where does the function
\begin{align*}
f(x) = x^n - xn^n
\end{align*}
take its local and global minima and maxima in the interval $[-2n,2n]$?
\end{quest}
\begin{proof}
First, note that if $n = 1$ our function is identically 0, and thus its local and global minima and maxima are uninteresting. We will focus on $n > 1$ for the rest of the proof.
By the above proposition, we know that $f$ will take on these minima and maxima at its critical points and endpoints. Because $f$ is differentiable everywhere, $f$'s only critical points come at places where $f'(x) = 0$. We examine these points here:
\begin{align*}
f'(x) &= nx^{n-1} - n^n = 0\\
\Leftrightarrow nx^{n-1} &= n^n \\
\Leftrightarrow ~ x^{n-1} &= n^{n-1}. \\
\end{align*}
There are two cases, here: if $n$ is odd, its critical points occur at $\pm n$; if $n$ is even, however, its only critical point is at $n$. In either situation, we have that $f''(x) = n(n-1)x^{n-2}$; thus, we have that $x=n$ is a local minima regardless of whether $n$ is odd or even, while $x=-n$ is a local maxima for $n$ odd.
This accomplished, we can then evaluate our function at these points along with the endpoints, and use this to find its global maxima and minima:
For $n$ odd:
\begin{align*}
f(-2n) &= (-2n)^n - (-2n)\cdot n^n = n^n\left(2n - 2^n\right),\\
f(-n) &= (-n)^n - (-n)\cdot n^n = n^n\left(n-1\right),\\
f(n) &= (n)^n - (n)\cdot n^n = n^n\left(1-n\right),\\
f(2n) &= (2n)^n - (2n)\cdot n^n = n^n\left(2^n-2n\right),\\
\end{align*}
So: if $n > 2$, we know that $2^n > 2n$; consequently, we have that $f(-2n)$ is the global minima and $f(2n)$ is the global maxima. Because every odd number other than 1 is $>2$, we've thus resolved our question of $n$ odd.
For $n$ even:
\begin{align*}
f(-2n) &= (-2n)^n - (-2n)\cdot n^n = n^n\left(2n + 2^n\right),\\
f(n) &= (n)^n - (n)\cdot n^n = n^n\left(1-n\right),\\
f(2n) &= (2n)^n - (2n)\cdot n^n = n^n\left(2^n-2n\right).\\
\end{align*}
For any even value of $n$, this function has its global maxima at $f(-2n)$ and its global minima at $f(n)$. Thus, we've classified $f$'s local and global minima and maxima for any value of $n$: so we're done!
\end{proof}
\begin{quest}
Let $p(t)$ denote the current location of a particle moving in a one-dimensional space. Call this particle ``nice'' if $p(0) = 0, p(1) = 1$, $p'(0) = p'(1) = 0$, and $p(t)$ is continuous.
What is
\begin{align*}
\inf_{\textrm{``nice'' particles}} \left( \sup_{t \in [0,1]} |p''(t)|\right) ?
\end{align*}
\end{quest}
\begin{proof}
To start studying the above claim, let's assume that there is \textbf{some} answer $M$: in other words, there is some $M$ such that
\begin{enumerate}
\item $M \geq |p''(t)|$, for any nice particle $p$ and any $t \in [0,1]$.
\item $M$ is the smallest such number that the above claim holds for.
\end{enumerate}
What can we do from here? Well: we have some boundary conditions (niceness tell us that $p(0) = 0, p(1) = 1, p'(0) = 0, p'(1) = 0$) and one global piece of information ($|p''(t)| < M$). How can we turn this knowledge of the second derivative into information about rest of the function?
Well: if we apply the mean value theorem to the function $p''(t)$, what does it say? It tells us that on any interval $[a,b]$, there is some $c \in (a,b)$ such that
\begin{align*}
\frac{p'(b) - p'(a)}{b-a} = (p')'(x) = p''(c).
\end{align*}
In other words, it relates the first and second derivatives to each other! So, if we apply our known bound $|p''(t)| < M, \forall t \in [0,1]$, we've just shown that
\begin{align*}
\left| \frac{p'(b) - p'(a)}{b-a} \right| = |p''(c)| < M,
\end{align*}
for any $a** -\frac{M}{2}t^2 + 1, \forall t \in(0,1).
\end{align*}
But what happens if we plug in $t = \frac{1}{2}$? In our first bound, we have $p\left( \frac{1}{2} \right) < \frac{M}{2}\left(\frac{1}{2}\right)^2 = \frac{M}{8}$. Conversely, in our second bound we have $p\left(1 - \frac{1}{2} \right) > -\frac{M}{2}\left(\frac{1}{2}\right)^2 + 1 = 1 - \frac{M}{8}$.
In other words, we have $\frac{M}{8} < p(1/2) < 1 - \frac{M}{8}$, which forces $M \leq 4$. So we know an upper bound on our answer!
Moreover, it is an attainable bound, whose answer is suggested by our work here: if we actually set $M = 4$, we get the piecewise function
\begin{align*}
p(t) = \left\{ \begin{array}{cc} 2t^2, & t \in (-\infty, 1/2] \\ 1-2(1-t)^2, & t \in [1/2, \infty)\\ \end{array} \right.
\end{align*}
This function is continuous, as it is piecewise made up of polynomials (which are continuous) and at their join we have $2(1/2)^2 = 1/2 = 1-2(1-1/2)^2 = 1/2$. As well, $p(0) = 0, p(1) = 1$, and the derivative of $p(t)$ is just
\begin{align*}
p(t) = \left\{ \begin{array}{cc} 4t, & t \in (-\infty, 1/2) \\ 4(1-t), & t \in (1/2, \infty)\\ \end{array} \right.,
\end{align*}
which has $p'(0) = 0 = p'(1) = 0$.
\end{proof}
\section{Integration}
As before, we start by defining our terms:
\begin{defn}
\textbf{The integral}: A function $f$ is \textbf{integrable}\footnote{To be specific, Riemann-integrable.} on the interval $[a,b]$ if and only if the following holds:
\begin{itemize}
\item For any $\epsilon > 0$,
\item there is a partition $a=t_0 < t_1 < \ldots < t_{n-1}**