Day 10, Mat129: Calculus I

• For some of you, the Intermediate Value Theorem (IVT) is so obvious that you don't get it: what is there to say? Well, whenever you want to use a theorem, you need to satisfy the conditions of the theorem. So you'll need a continuous function $f$, whose value will be $N$ at some value $x=c$ in the domain; and you'll need to show that $f$ takes values both below and above $N$, so that $N$ is intermediate between them!

  Let's look at #54: it's kind of like the Tibetan monk problem, in that we had to create a single function from the two functions (up and down).

• If $f$ is differentiable at $a$, then $f$ is continuous at $a$.
• But the converse is not true: just because $f$ is continuous at $a$ does not mean that it is differentiable there.
  

• Differentiability on an interval implies continuity and smoothness.

Let's use the definition of the derivative function,

\[ f^\prime(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \]

to determine the derivative functions of these important functions:

1. Constant functions: $f(x)=c$
2. Linear functions: $f(x)=mx+b$
3. Quadratic functions: $f(x)=ax^2+bx+c$

4. An important application of the tangent line

The first two derivatives are easy: each function has a straight line graph, so the tangent line is the graph of the function itself. The slope of each of those graphs is easy to see (either 0 or m). The derivatives are hence on full display. Not so much for the quadratic, and it's the first function that has an explicit dependence on $x$.

How well can you recognize derivatives and the functions that give rise to them? Let's try #3, p. 122.

• What causes breaks in derivative functions?
• How is symmetry preserved in derivatives?
• How does the degree of a polynomial change under differentiation?

Let's think of a quadratic motion, e.g. the motion of an eraser thrown across the room. Let $s(t)$ be the height of the eraser at time $t$:

\[s(t)=at^2+bt+c\]

Each of the coefficients has an important, intuitive role to play:

• $c$ is the initial height, at time $t=0$;
• $b$ is the initial velocity, the rate of change of height at time $t=0$;
• $2a$ is the acceleration (due to gravity) of the height (the derivative of the velocity is acceleration).
  • It is constant across this motion, if we neglect other forces (e.g. air resistance), or the fact that gravity is actually a function of the distance from the center of the Earth.
  • Forces result in accelerations (and vice versa, per Newton's equation $f=ma$).

Think about the parabola which is this function's graph: it's an umbrella. The ball will rise, reach its apex, then fall back to Earth.

The parentheses are supposed to tip you off to the fact that these aren't powers, but rather derivatives.

\[ f^\prime(x) = \frac{dy}{dx} = \frac{df}{dx} = \frac{d}{dx}\left(f(x)\right)=\left(f(x)\right)'=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \]

There are several different ways of writing the derivative, and you need to get used to them. The form \(\frac{d}{dx}\left(f(x)\right)\) reminds us that the operation of differentiation is itself a function: it takes a function in its domain and returns another function -- the derivative function.

• #2, p. 122 (so, from slopes, one constructs a derivative function)
• #25 (using "the most important definition in calculus"; note the slightly different domains)
• #48, p. 125.

• Rules that help us avoid having to use the definition each time (with Proofs):
  • The constant multiple rule (makes sense)
    $(cf(x))'=cf'(x)$

  • The sum rule

    works as we'd hope: "The derivative of a sum is the sum of the derivatives."

    $(f(x) + g(x))'=f'(x)+g'(x)$

  • The product rule (doesn't work the way we'd hope):
    $(f(x) \cdot g(x))'=f'(x)g(x)+f(x)g'(x)$

    To prove it we use one of my favorite tricks: the addition of a special form of 0.

  • The power rule:
    $(x^n)'=n x^{n-1}$

    where $n$ is a positive integer.

    We can prove this via dominoes -- i.e., mathematical induction -- or via the binomial theorem)

  • NOTE: at this point we have all the rules necessary to differentiate all polynomials, without needing to resort to the definition!

    The derivative of the monomial $x^n$ is $nx^{n-1}$, and

    The derivative of the monomial $c x^n$ is $nc x^{n-1}$ (by constant multiple).

    A polynomial is just a sum of these. So we apply the sum rule, and the power rule, and the constant multiple rule to the flight of the eraser, to get

    $s'(t)=(at^2+bt+c)'=2at+b$

    and

    $s''(t)=(2at+b)'=2a$

• Other rules:
  • The quotient rule:
    low d high minus high d low,
    over the denominator squared they go.

    $\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$

• #13, p. 136
• #22, p. 136 (expand first)
• #48, p. 137
• #58, p. 137