]> Haese & Harris - Chapter 21 Notes

21: Differential Calculus

Courtesy of Isaac Newton's fluxions and Gottfried Leibniz's infinitesimals, we are able to consider the slope between two points on a curve, and we have reason to believe that as the points approach one another, the slope between them settles down to a single number.

…but that's as rigorous as they got; despite their brilliance, the mathematicians of the late 17^th century had reached their limits. It took another century or so for someone to make some progress towards a rigorous definition.

A: The Idea of a Limit

Take a curve (call it $f (x)$ )…take a point on the curve (call it $(x, f (x))$ ). Now, consider another point on the curve at $(x + h, f (x + h))$ , where h is a (very) small number. The slope between the two points is $\frac{f (x + h) - f (x)}{(x + h) - x} = \frac{f (x + h) - f (x)}{h}$ . As long as h is non-zero, this gives the slope of the chord connecting the points (the average speed, to continue that analogy/application). As the points get closer together (and h gets closer to zero), the slope between them seems to approach a single value…but to find the slope at a single point (to find instantaneous speed), h must be zero, and the formula breaks down…how can the formula work so well, then just vanish?

It happens for a lot of functions, actually. Consider $g (x) = \frac{x^{2} + x - 2}{x - 1}$ . The domain of this function includes all real numbers except for 1. Let's look at the graph!

An Image

Looks like a straight line…and it looks like there is a point at x = 1! It's possible that your calculator shows a small gap at x = 1…which makes sense, since the domain does not include 1.

Back to our conundrum: how can this function march along like a good little line, with the minor exception of a single value?

A young Frenchman named Augustin-Louis Cauchy finally realized that sometimes, it isn't the destination that's important, but the journey.

(stick with me on this one)

Sometimes, a function approaches a limiting value without actually achieving the value. Sometimes, a function has a limit.

We've already seen this in the previous chapter. Remember how the slope approached 4? Let h be zero and the slope disappears entirely. But as h gets smaller, the slope function approaches a limiting value (4).

…but what does that really mean? Cauchy figured it out!

A function $f (x)$ has a limiting value of L as x approaches c if for any (small) real number ε, there exists another (small) real number δ so that any value of x within δ of c produces a function value ( $f (x)$ ) that is within ε of L.

Whew!

Try this: if you think that the function $f (x)$ is approaching a limiting value of L at x = c, then when you tell me how close to L you want to get (ε), I must be able to find how close to c you have to be (δ) in order to make it happen.

No? Graphically: Give me some region around L (call it the destination)…

An Image

…and I can find a region around c (call it the source)…

An Image

…where every function value coming from the source lands within the destination.

The notation: $lim_{x \to c} f (x) = L$ means that $\forall ε \in ℝ, \exists δ \in ℝ$ so that $x \in (c - δ, c + δ) ⇒ f (x) \in (L - ε, L + ε)$ .

This is not saying that the function ever achieves the value L…what it is saying is that you can get as close to that value as you could possibly desire…you can get arbitrarily close.

(so we whisper: for all intents and purposes, $f (c) = L$ ).

Let's return to our function $g (x) = \frac{x^{2} + x - 2}{x - 1}$ , and algebraically find the limit as x approaches 1.

$lim_{x \to 1} \frac{x^{2} + x - 2}{x - 1} = lim_{x \to 1} \frac{(x + 2) (x - 1)}{x - 1}$ . Pause for a moment…the only way we can legally remove/cancel the x − 1 terms is if they are non-zero. Fortunately, this is precisely what we mean when we work with limits: we're considering values close to c (1), but we're not going to allow a value of precisely c (1). So we can continue, and obtain $lim_{x \to 1} (x + 2)$ . At this point, allowing x to have a value of 1 doesn't cause any problems…so we do it, and get 3.

$lim_{x \to 1} \frac{x^{2} + x - 2}{x - 1} = 3$

Before moving on, here are some properties of limits:

For these definitions, assume that $lim_{x \to a} f (x) = L$ , $lim_{x \to a} g (x) = K$ , and b is a real number.

Constants can be factored out: $lim_{x \to a} [b \cdot f (x)] = b \cdot lim_{x \to a} [f (x)] = b \cdot L$

Operations can be distributed: $lim_{x \to a} [f (x) \pm g (x)] = lim_{x \to a} f (x) \pm lim_{x \to a} g (x) = L \pm K$ , $lim_{x \to a} [f (x) \cdot g (x)] = lim_{x \to a} f (x) \cdot lim_{x \to a} g (x) = L \cdot K$ , $lim_{x \to a} \frac{f (x)}{g (x)} = \frac{lim_{x \to a} f (x)}{lim_{x \to a} g (x)} = \frac{L}{K}$ (provided that K ≠ 0), and $lim_{x \to a} {[f (x)]}^{n} = L^{n}$ .

B: Derivatives at a Given x-value

Let's apply the limiting process to our slope problem—that should give us a fairly rigorous definition of the slope at a point (or instantaneous speed).

Equation 1 - Slope of a Curve at a Point

$slope at c = lim_{x \to c} \frac{f (x) - f (c)}{x - c}$

(I changed a few variables there, but nothing of substance has changed)

This is called the derivative of $f (x)$ at $x = c$ . The notation is $f^{'} (c)$ . This is the slope of the line tangent to the curve at that point, and it is the instantaneous rate of change of y with respect to x at x = c.

Here's another limit definition of the derivative:

Equation 2 - The Derivative at a Point

$f^{'} (c) = lim_{h \to 0} \frac{f (c + h) - f (c)}{h}$

C: The Derivative Function

The derivative can be considered in a more general form: just don't fix the point!

Equation 3 - The Derivative of a Function

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

When considered this way, the result is not a single number, but an equation—specifically, the slope equation.

There are several other ways to denote a derivative: $\frac{d y}{d x}$ , $y^{'}$ , $\frac{d}{d x} [f (x)]$ …you should be familiar with all of them.

D: Simple Rules of Differentiation

If $f (x)$ is a constant function, then $f (x) = c$ (where c is some real number). Let's look at the derivative in this case…

$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = lim_{h \to 0} \frac{c - c}{h} = lim_{h \to 0} (0) = 0$

So for any constant function, $f^{'} (x) = 0$ .

Now let $f (x)$ be a simple power function— $f (x) = x^{n}$ , where n is a rational number. Through careful use of the Binomial Expansion of Polynomials, you can prove that $f^{'} (x) = n \cdot x^{n - 1}$ . We'll do this in class.

Factoring allows us to show that $\frac{d}{d x} [c \cdot f (x)] = c \cdot \frac{d}{d x} [f (x)]$ . You should be able to prove this on your own.

Grouping (the associative property) allows us to show that $\frac{d}{d x} [f (x) + g (x)] = \frac{d}{d x} [f (x)] + \frac{d}{d x} [g (x)]$ and $\frac{d}{d x} [f (x) - g (x)] = \frac{d}{d x} [f (x)] - \frac{d}{d x} [g (x)]$ . Again, you should be able to prove this.

Put together, these simple rules allow us to find the derivative of any polynomial!

E: The Chain Rule

Now that we can find the derivative of any polynomial, let's expand our horizons a bit—let's consider functions that are defined in terms of other functions. We'll begin with the case where one function is inside of the other function—a composition of functions.

In this case, we use the Chain Rule to find the derivative of the composition.

Equation 4 - The Chain Rule

$\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)$

Since we are currently restricting our attention to polynomials, we can focus the chain rule to the case of a polynomial raised to a power (a polynomial within a polynomial)—this results in the General Power Rule:

Equation 5 - The General Power Rule

$\frac{d}{d x} {[f (x)]}^{n} = n \cdot {[f (x)]}^{n - 1} \cdot f^{'} (x)$

F: Product and Quotient Rules

Functions can be combined by other methods! What about the product of two functions?

Equation 6 - The Product Rule

$\frac{d}{d x} [f (x) \cdot g (x)] = f^{'} (x) \cdot g (x) + f (x) \cdot g^{'} (x)$

How about the quotient of two functions?

Equation 7 - The Quotient Rule

$\frac{d}{d x} [\frac{f (x)}{g (x)}] = \frac{f^{'} (x) \cdot g (x) - f (x) \cdot g^{'} (x)}{{[g (x)]}^{2}}$

G: Tangents and Normals

The derivative can be used to find the slope of the line tangent to a curve at a point—once we have the slope of a line, all that is needed to write the equation of the line is a point on the line…but that was given! That was where we found the derivative!

The line tangent to $f (x)$ at the point $(c, f (c))$ has slope $f^{'} (c)$ and passes through the same point—so the equation of the tangent line is $y - f (c) = f^{'} (c) (x - c)$ .

If you can find the equation of the line tangent to the curve, then you can also find the equation of the line normal (perpendicular) to the curve—since you know that perpendicular lines have negative reciprocal slopes…

H: The Second Derivative

What we've been investigating is really the first derivative. You could take the slope equation and finds its derivative, and that would give you the second derivative.

(and you could do it again and find the third derivative…ad infinitum)

The second derivative is denoted $f^{″} (x)$ , $\frac{d}{d x} [\frac{d y}{d x}] = \frac{d^{2} y}{{d x}^{2}}$ , or $y^{″}$ . This is the rate of change of the slope…if we go back to the old speed example, it is the rate of change of speed. Of course, you know another word for that!

Acceleration!

Page last validated 2010-08-15