The Derivative

Derivative of a Constant
Constant Multiples
The Rate of Change

At its core, calculus is the study of the relationships between functions. Before we rigorously examine these relationships, we begin by building some intuition, using this concept of "function relationships" as a guide.

Suppose we have two functions: ${d}$ and ${s.}$ These two functions could be anything. If we're talking physics, we might say the function ${f}$ is a function of distance, and the function ${s}$ is a function of speed. If we're talking graphs in pure mathematics, however, we can think of ${f}$ as how high or low the graph climbs, and ${s}$ as how steep — the slope — the graph moves up or down. That "steepness" can also be interpreted as how quickly we climb or descend.

Our first function, ${f,}$ must take inputs. We can denote those inputs with ${t,}$ or with ${x.}$ Let's use ${t.}$ This gives us the first function's name: ${f(t).}$ Let's say the second function is called ${s(t)}$ That function ${s(t)}$ has a variety of names:

\begin{aligned} &s(t) = \dfrac{df}{dt} \\[1em] &s(t) = \dfrac{d}{dt} f \\[1em] &s(t) = f' \end{aligned}

All this said, let's consider the simplest case of the relationship between these two functions. Suppose we're travelling at constant speed. If we're travelling at constant speed, then the graph of ${s(t)}$ would be flat. Say it's ${4 \text{ m/s}.}$ Then ${s(t) = 4.}$ This means that after ${1}$ second, we've travelled ${4}$ meters. After ${2}$ seconds, ${8}$ meters. And after ${3}$ seconds, ${12}$ meters. Where ${s(t) = 4,}$ we have:

f(t) = 4t

This reveals that ${f(t) = [s(t)]t,}$ or, more concisely, ${f = st.}$ Notice that when the slope is constant, algebra suffices. The slope is obtained by simply dividing the distance by the time:

s = s(t) = \dfrac{df}{dt} = \dfrac{d}{dt}f = f' = \dfrac{f(t)}{t} = \dfrac{f}{t}

And viewed in a different direction:

f = st = \dfrac{df}{dt}(t) = \left( \dfrac{d}{dt} f \right)t = f' \cdot t = \left( \dfrac{f(t)}{t} \right) \cdot t = \left( \dfrac{f}{t} \right) = st

Now suppose we wanted to find our speed along two particular times. Say our speed at ${t_0,}$ a particular starting point in time, and at ${t_1,}$ the end point in time. In that case, what we're asking for is really:

s = \dfrac{df}{dt} = f' = \dfrac{d}{dt} f = \dfrac{f(t_1) - f(t_0)}{t_1 - t_0} = \dfrac{\Delta f}{\Delta t}

Say our speed varies. We might imagine the graphs to appear as:

This is in line with our intuition. The graph of ${f}$ shows that we cover a lot of distance at first, then begin to plateau, eventually stopping. The graph of ${s}$ accurately describes this situation in terms of speed. We're moving pretty fast at first, but then slow down, eventually stopping. This discussion illustrates the purpose of differential calculus. It is the mathematics that allows to understand ${f}$ through ${s,}$ and ${s}$ through ${f.}$

There are several ways to define a derivative. One way is to define the derivative geometrically. Another way is to define the derivative analytically. Yet another way is to define the derivative in terms of physics. And yet another way is to define it in terms of measurements.

For the next few sections, we will focus on how to differentiate — computing the derivative of a function. Question: How do we find the tangent line to a point ${P(x_0, y_0)}$ on some line ${y = f(x)?}$ We could just draw it¹:

Now, we can draw the above easily by hand, but how do we get a computer to do it? This is an overarching theme for the materials that follow: How do we answer these questions computationally? This is an imperative inquiry — it asks a "how to" question. And to even begin answering this question, we must understand the underlying propositions. What do we know is true? What is false? These are declarative inquiries — they ask "what is" questions.

To begin, we know that the equation of a line is ${y - y_0 = m(x - x_0),}$ where ${m}$ is the slope of the line. There are two propositions we need to determine what the tangent line is. First, what are the coordinates of the point ${P(x_0, y_0)?}$ Well, we know that ${y_0 = f(x_0).}$ Second, what is the slope ${m?}$

Here we have our first bit of terminology. We will now part ways with ${m,}$ and use the more formal term for the slope: ${f'(x_0)}$ (read: "The derivative of ${f.}$ ") The answer to this question is some number ${f'(x_0),}$ and it is the focus of the materials that follow.

definition. ${f'(x_0),}$ the > derivative of ${f}$ at ${x_0,}$ is the slope ${m}$ of the tangent line > to ${y = f(x)}$ at the point ${P(x_0, y_0).}$

We still have a problem. Why did we draw the line the way we did in the diagram above? Why couldn't it have been this green line:

How do we know that the green line is not a tangent line, but the red line is a tangent line? The green line crosses at some other point, call it ${Q.}$ But, this is not what prevents the green line from being a tangent line, because the tangent line itself could very well cross multiple points.

The green line is what we call the secant line. If we think of the point ${Q}$ as getting closer and closer to ${P,}$ then the slope of the secant line will get closer and closer to the slope of the tangent line. If we draw the secant line close enough, then we get the tangent line.

definition. The tangent line is the limit of the secant lines ${\overleftrightarrow{\rm PQ}}$ as ${Q}$ tends to ${P,}$ where ${P}$ is fixed.

This is the geometric interpretation of the derivative. Now we turn to an analytic interpretation of the derivative (interpreting the derivative in terms of symbols and formulas). Let's first draw the points ${P}$ and ${Q}$ again:

Above, we see just a secant line. The length ${\Delta x}$ is the change in ${x,}$ and the length ${\Delta f}$ is the change in ${f.}$ Given these two lengths, we define the slope of the secant line as:

m_{secant} = \dfrac{\Delta f}{\Delta x}

From the definition of the tangent line above, we can also define the slope of the tangent line:

m_{tangent} = \lim\limits_{\Delta x \to 0} \dfrac{\Delta f}{\Delta x}

This is still too general. Let's make things much more explicit. First, let's write the numerator of ${m_{secant},}$ ${\Delta f,}$ more explicitly. Recall that the point ${P}$ has the coordinates ${(x_0, f(x_0)).}$ We also need a formula for the point ${P.}$ If ${P}$ has the ${x}$ -coordinate ${x_0,}$ then ${x}$ -coordinate of ${Q}$ must be ${(x_0 + \Delta x).}$ Thus, the point ${Q}$ has the coordinates ${(x_0 + \Delta x, f(x_0 + \Delta x)).}$ Now we can write another formula for the derivative:

definition. The derivative is defined by the following formula:

m = f'(x_0) = \lim\limits_{\Delta x \to 0} \dfrac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}

This formula is called the difference quotient. Above, we have some new notation, ${\lim\limits_{\Delta x \to 0}.}$ This is called a limit. Let's define this notation informally (a more formal definition is provided in a later section):

definition. The notation:
$\lim\limits_{x \to a} f(x) = L$

reads as, “The limit of ${f(x)}$ as ${x}$ approaches ${a}$ equals ${L.}$ ” If the values of ${f(x)}$ are arbitrarily close to ${L}$ (i.e., as close to ${L}$ as we'd prefer) by making ${x}$ sufficiently close to ${a}$ on either the left- or right-hand side of ${a,}$ but ${x \neq a}$ then we write ${\lim\limits_{x \to a} f(x) = L.}$

This is an extremely important formula, and it forms the foundation for the materials that follow. It is the formula that allows us to compute the value of ${m,}$ or more appropriately, ${f'(x).}$ All that said, let's see it in action.

Suppose we have the function ${f(x) = \dfrac{1}{x}.}$ What is the derivative of this function? All we're going to do is plug in the function to the previous formula. Visually, what we're doing is taking a point on the hyperbola and determine the tangent line through that point. This is a geometric objective, but we are accomplishing it algebraically.

\lim\limits_{\Delta x \to 0}\dfrac{\Delta f}{\Delta x} = \lim\limits_{\Delta x \to 0}\dfrac{\dfrac{1}{x_0 + \Delta x} - \dfrac{1}{x_0}}{\Delta x}

Now we simplify:

\begin{aligned} f'\left(\dfrac{1}{x}\right) &= \lim\limits_{\Delta x \to 0} \dfrac{\Delta f}{\Delta x} \\ &= \lim\limits_{\Delta x \to 0}\dfrac{\dfrac{1}{x_0 + \Delta x} - \dfrac{1}{x_0}}{\Delta x} \\ &= \lim\limits_{\Delta x \to 0}\dfrac{1}{\Delta x} \left( \dfrac{x_0 - (x_0 + \Delta x)}{x_0(x_0 + \Delta x)} \right) \\ &= \lim\limits_{\Delta x \to 0}\dfrac{1}{\Delta x} \left( \dfrac{\cancel{x_0} - (\cancel{x_0} + \Delta x)}{x_0(x_0 + \Delta x)} \right) \\ &= \lim\limits_{\Delta x \to 0}\dfrac{1}{\Delta x} \left( \dfrac{- \Delta x}{x_0(x_0 + \Delta x)} \right) \\ &= \lim\limits_{\Delta x \to 0}\dfrac{1}{\cancel{\Delta x}} \left( \dfrac{- \cancel{\Delta x}}{x_0(x_0 + \Delta x)} \right) \\ &= \lim\limits_{\Delta x \to 0}-\dfrac{1}{x_0(x_0 + \Delta x)} \\ &= -\dfrac{1}{x_0^2} \end{aligned}

Notice what the final output of ${f'}$ is. First, it's a negative value. This conforms to the tangent line sloping down. Second, as ${x}$ tends towards infinity, ${f'}$ gets less and less steep. ${f'}$ corresponds to this characteristic as well. As ${x_0}$ becomes a very large number, ${-\dfrac{1}{x_0^2}}$ becomes a smaller and smaller in magnitude.

Let's go back to some notation. So far, we've seen several:

${y = f(x)}$
${\Delta y = \Delta f}$

We also saw this notation for the derivative:

f'

The notation above is called Newton's notation. Another set of notations for the derivative (all of the notations below mean the same thing):

\dfrac{df}{dx} \equiv \dfrac{dy}{dx} \\[2em] \dfrac{d}{dx}f \equiv \dfrac{d}{dx}y

This notation is called Leibniz's notation. Newton's notation is by far the most concise, but of course, that comes at the cost of omitting a great deal of information. Both notations, however, are used extensively, and often together.

Arguably, the simplest functions to consider the derivatives for are the following: (1) constant functions; (2) power functions; (3) polynomial functions; and (4) exponential functions. Exploring the derivatives for these functions provides clear applications of the preceding section's concepts, as well as several key insights to differentiation.

Derivative of a Constant

Consider the function ${f(x) = C,}$ where ${C}$ is a constant. Applying the definition of a derivative:

\begin{aligned} f'(x) &= \lim\limits_{h \to 0} \dfrac{f(x + h) - f(x)}{h} \\[1em] &= \lim\limits_{h \to 0} \dfrac{c - c}{h} \\[1em] &= \lim\limits_{h \to 0} 0 \\[1em] &= 0 \end{aligned}

We can support this algebraic analysis by examining the graphs of constant functions:

Above, we see the graphs of ${f(x) = 2,}$ ${f(x) = 3,}$ and ${f(x) = -4.}$ These functions do not "slope." They stay constant. Based on our analyses , we have the following theorem:

theorem. Derivative of a Constant. Given the constant function ${f(x) = C,}$
$\dfrac{d}{dx} (c) = 0$

Constant Multiples

Consider the function ${f(x) = \dfrac{3x^6}{2}.}$ What is the derivative of this function? To begin, we can rewrite this function as: ${f(x) = \dfrac{3}{2} x^6.}$ Rewriting it this way, we see that ${\dfrac{3}{2}}$ is really just a constant. Accordingly, we can construct a more general form for this function: ${f(x) = C \cdot g(x),}$ where ${C = \dfrac{3}{2}}$ and ${g(x) = x^6.}$

Thinking more generally, let's forget about the original problem and think about applying the derivative definition to a function of the form ${f(x) = C \cdot g(x).}$ We have:

\begin{align} f'(x) &= \lim\limits_{h \to 0} \dfrac{f(x + h) - f(x)}{h} \\[1em] &= \lim\limits_{h \to 0} \dfrac{C g(x + h) - C g(x)}{h} \\[1em] &= \lim\limits_{h \to 0} C \left[ \dfrac{g(x + h) - g(x)}{h} \right] \\[1em] &= C \lim\limits_{h \to 0} \dfrac{g(x + h) - g(x)}{h} \\[1em] &= C g'(x) \end{align}

The transition from line (3) to (4) follows from the limit law for multiplication. From the analysis above, we have the Constant Multiple Rule:

Theorem: Constant Multiple Rule. Given the function ${f(x) = C \cdot g(x),}$ where ${C}$ is a constant and ${C \in \reals,}$

f'(x) = C \cdot g'(x)

Accordingly, the derivative of ${f(x) = \dfrac{3x^6}{2}}$ is ${f'(x) = \dfrac{3}{2} \cdot \dfrac{d}{dx} (x^6) = \dfrac{3}{2} \cdot (6x^5).}$ In other words, ${f'(x) = 9x^5.}$

The Rate of Change

In the previous sections, we saw two interpretations of the derivative: (1) a geometric interpretation, and (2) an algebraic interpretation. Now we turn to another: a physical interpretation. By physical, we are referring to an interpretation of the derivative in terms of applied mathematics. Under the physical interpretation, we think of the derivative as a rate of change.

Referencing the diagram above, ${\dfrac{\Delta x}{\Delta y}}$ represents the average rate of change. The derivative ${\dfrac{dy}{dx},}$ however, represents the instantaneous rate of change. The concept of instantaneous rate of change underlies numerous fields, the most obvis of which are in physics:

Where ${q}$ is a charge and ${t}$ is time, ${\dfrac{dq}{dt}}$ represents current.
Where ${s}$ is a distance ${t}$ is time, ${\dfrac{ds}{dt}}$ represents speed.

For example, suppose that we drop a large block from a building 80 meters high. Using the acceleration due to gravity (9.807 ${m/s^2}$ ), we have the following formula:

h = 80 - 5t^2

where ${h}$ is the height and ${t}$ is time.

Thus, at ${t = 0,}$ we are at the very top of the building. ${h = 80 - 0 = 80.}$ At ${t = 4,}$ we are at the bottom ${h = 80 - 5(4)^2 = 80 - 80 = 0.}$ The average speed of the block falling is the change in ${h}$ divided by the change in ${t:}$

\begin{aligned} \dfrac{\Delta h}{\Delta t} &= \dfrac{h_f - h_0}{t_f - t_0} \\ &= \dfrac{0 - 80}{4 - 0} \\ &= -20 m/s \end{aligned}

Thus, the average speed of the block falling is ${20 m/s.}$ What is the speed of the block the moment it hits the pavement? To answer this question, we need to compute the instantaneous speed:

\begin{aligned} &h = 80 - 5t^2 \\ &\dfrac{d}{dt} h = 0 - 10t \\ &= -10t \end{aligned}

Thus, at the point of impact, we have ${t = 4,}$ so ${h' = -40 m/s.}$ Thus, the speed of the block at the moment of impact is ${40 m/s.}$ Physically interpreting the derivative is not just limited to physics. In meteorology, we use derivatives to compute temperature gradients, which are what cause airflows (which in turn are what cause winds and storm systems).

The word tangent is derived from the Latin tangens, meaning “touching.” ↩

The Derivative

Derivative of a Constant

Constant Multiples

The Rate of Change

Footnotes