Section - 4.1 The Rate Of Change Of A Function

Before we embark on setting the groundwork for the derivative of a function, let's review some terminology and concepts. Remember that the slope of a line is defined as the quotient of the difference in y-values and the difference in x-values. Recall from Section 1.2 that a difference between two quantities is often denoted by the Greek symbol \(\Delta\) - read “delta” as shown next, where delta notation is being used when calculating and interpreting the slope of a line.

Calculating and Interpreting the Slope of a Line.

Suppose we are given two points \(\left(x_{1},y_{1}\right)\) and \(\left(x_{2},y_{2}\right)\) on the line of a linear function \(y = f(x)\text{.}\) Then the slope of the line is calculated by

\begin{equation*} m = \dfrac{\Delta y}{\Delta x} = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\text{.} \end{equation*}

We can interpret this equation by saying that the slope \(m\) measures the change in \(y\) per unit change in \(x\text{.}\) In other words, the slope \(m\) provides a measure of sensitivity .

For example, if \(y = 100x + 5\text{,}\) a small change in \(x\) corresponds to a change one hundred times as large in \(y\text{,}\) so \(y\) is quite sensitive to changes in \(x\text{.}\)

Next, we introduce the properties of two special lines, the tangent line and the secant line, which are pertinent for the understanding of a derivative.

Secant Line.

Secant is a Latin word meaning to cut, and in mathematics a secant line cuts an arbitrary curve described by \(y = f(x)\) through two points \(P\) and \(Q\text{.}\) The figure shows two such secant lines of the curve \(f\) to the right and to the left of the point \(P\text{,}\) respectively.

Since by necessity the secant line goes through two points on the curve of \(y = f(x)\text{,}\) we can readily calculate the slope of this secant line.

Definition 4.1. Slope of Secant Line — Average Rate of Change.

Suppose we are given two points \(\left(x_{1},y_{1}\right)\) and \(\left(x_{2},y_{2}\right)\) on the secant line of the curve described by the function \(y = f(x)\) as shown. Then the slope of the secant line is calculated by

\begin{equation*} m = \dfrac{\Delta y}{\Delta x} = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\text{.} \end{equation*}

Note that we may also be given the change in \(x\) directly as \(\Delta x\text{,}\) i.e the two points are given as \(\left(x,f(x)\right)\) and \(\left(x+\Delta x, f(x + \Delta x)\right)\text{,}\) and so

\begin{equation*} m = \dfrac{\Delta y}{\Delta x} = \dfrac{f(x + \Delta x)-f(x)}{\Delta x}\text{.} \end{equation*}

Note:

1. In the above figure, the value of \(\Delta x\) must be negative since it is on the left side of \(x\text{.}\)

2. The slope of the secant line is also referred to as the average rate of change of \(f\) over the interval \(\left[x,x+\Delta x\right]\text{.}\)

3. The expression \(\dfrac{f(x + \Delta x) - f(x)}{\Delta x}\) is referred to as the difference quotient.

Tangent Line.

Tangent is a Latin word meaning to touch, and in mathematics a tangent line touches an arbitrary curve described by \(y = f(x)\) at a point \(P\) but not any other points nearby as shown.

Since by definition the tangent line only touches one point on the curve of \(y = f(x)\text{,}\) we cannot calculate the slope of this tangent line with our slope formula for a line. In fact, up to now, we do not have any way of calculating this slope unless we are able to use some geometry.

Suppose that \(y\) is a function of \(x\text{,}\) say \(y = f(x)\text{.}\) Since it is often useful to know how sensitive the value of \(y\) is to small changes in \(x\text{,}\) let's explore this concept via an example, and see how this will inform us about the calculation of the slope of the tangent line.

Example 4.2. Small Changes in \(x\).

Consider \(\ds y=f(x)=\sqrt{625-x^2}\) (the upper semicircle of radius 25 centered at the origin), and let's compute the changes of \(y\) resulting from small changes of \(x\) around \(x=7\text{.}\)

Solution

When \(x=7\text{,}\) we find that \(\ds y=\sqrt{625-49}=24\text{.}\) Suppose we want to know how much \(y\) changes when \(x\) increases a little, say to 7.1 or 7.01.

Let us look at the ratio \(\Delta y/\Delta x\) for our function \(\ds y=f(x)=\sqrt{625-x^2}\) when \(x\) changes from 7 to \(7.1\text{.}\) Here \(\Delta x=7.1-7=0.1\) is the change in \(x\text{,}\) and

\begin{align*} \Delta y =f(x+\Delta x)-f(x)\amp =f(7.1)-f(7)\\ \amp =\sqrt{625-7.1^2}-\sqrt{625-7^2}\\ \amp \approx 23.9706-24=-0.0294\text{.} \end{align*}

Thus, \(\Delta y/\Delta x\approx -0.0294/0.1=-0.294\text{.}\) This means that \(y\) changes by less than one third the change in \(x\text{,}\) so apparently \(y\) is not very sensitive to changes in \(x\) at \(x=7\text{.}\) We say “apparently” here because we don't really know what happens between 7 and value at \(7\text{.}\) This is not in fact the case for this particular function, but we don't yet know why.

The quantity \(\Delta y/\Delta x\approx -0.294\) may be interpreted as the slope of the secant line through \((7,24)\) and \((7.1,23.9706)\text{.}\) In general, if we draw the secant line from the point \((7,24)\) to a nearby point on the semicircle \((7+\Delta x,\,f(7+\Delta x))\text{,}\) the slope of this secant line is the so-called difference quotient

\begin{equation*} \frac{f(7+\Delta x)-f(7)}{\Delta x}= \frac{\sqrt{625-(7+\Delta x)^2}-24}{\Delta x}\text{.} \end{equation*}

For example, if \(x\) changes only from 7 to 7.01, then the difference quotient (slope of the secant line) is approximately equal to \((23.997081-24)/0.01=-0.2919\text{.}\) This is slightly different than for the secant line from \((7,24)\) to \((7.1,23.9706)\text{.}\)

As \(\Delta x\) is made smaller (closer to 0), \(7+\Delta x\) gets closer to 7 and the secant line joining \((7,f(7))\) to \((7+\Delta x,f(7+\Delta x))\) shifts slightly, as shown in Figure 4.1. The secant line gets closer and closer to the tangent line to the circle at the point \((7,24)\text{.}\) (The tangent line is the line that just grazes the circle at that point, i.e., it doesn't meet the circle at any second point.) Thus, as \(\Delta x\) gets smaller and smaller, the slope \(\Delta y/\Delta x\) of the secant line gets closer and closer to the slope of the tangent line. This is actually quite difficult to see when \(\Delta x\) is small, because of the scale of the graph. The values of \(\Delta x\) used for the figure are \(1\text{,}\) \(5\text{,}\) \(10\) and \(15\text{,}\) not really very small values. The tangent line is the one that is uppermost at the right hand endpoint.

So far we have found the slopes of two secant lines that should be close to the slope of the tangent line, but what is the slope of the tangent line exactly? Since the tangent line touches the circle at just one point, we will never be able to calculate its slope directly, using two “known” points on the line. What we need is a way to capture what happens to the slopes of the secant lines as they get “closer and closer” to the tangent line.

Instead of looking at more particular values of \(\Delta x\text{,}\) let's see what happens if we do some algebra with the difference quotient using just \(\Delta x\text{.}\) The slope of a secant line from \((7,24)\) to a nearby point \(\left(7+\Delta x,f(7+\Delta x)\right)\) is given by

\begin{align*} \frac{f(7+\Delta x)-f(7)}{\Delta x}\amp =\frac{\sqrt{625-(7+\Delta x)^2} - 24}{\Delta x}\\ \amp = \left( \ds\frac{\sqrt{625-(7+\Delta x)^2} - 24}{\Delta x} \right) \left(\frac{\sqrt{625-(7+\Delta x)^2}+24}{\sqrt{625-(7+\Delta x)^2}+24} \right)\\ \amp =\frac{625-(7+\Delta x)^2-24^2}{\Delta x(\sqrt{625-(7+\Delta x)^2}+24)}\\ \amp =\frac{49-49-14\Delta x-\Delta x^2}{\Delta x(\sqrt{625-(7+\Delta x)^2}+24)}\\ \amp =\frac{625-(7+\Delta x)^2-24^2}{\Delta x(\sqrt{625-(7+\Delta x)^2}+24)}\\ \amp =\frac{49-49-14\Delta x-\Delta x^2}{\Delta x(\sqrt{625-(7+\Delta x)^2}+24)}\\ \amp =\frac{\Delta x(-14-\Delta x)}{\Delta x(\sqrt{625-(7+\Delta x)^2}+24)}\\ \amp =\frac{-14-\Delta x}{\sqrt{625-(7+\Delta x)^2}+24} \end{align*}

Now, can we tell by looking at this last formula what happens when \(\Delta x\) gets very close to zero? The numerator clearly gets very close to \(-14\) while the denominator gets very close to \(\ds \sqrt{625-7^2}+24=48\text{.}\) The fraction is therefore very close to \(-14/48 = -7/24 \cong -0.29167\text{.}\) In fact, the slope of the tangent line is exactly \(-7/24\text{.}\)

What about the slope of the tangent line at \(x=12\text{?}\) Well, 12 can't be all that different from 7; we just have to redo the calculation with 12 instead of 7. This won't be hard, but it will be a bit tedious. What if we try to do all the algebra without using a specific value for \(x\text{?}\) Let's copy from above, replacing 7 by \(x\text{.}\)

\begin{align*} \frac{f(x+\Delta x)-f(x)}{\Delta x}\amp =\frac{\sqrt{625-(x+\Delta x)^2} - \sqrt{625-x^2}}{\Delta x}\\ \amp =\frac{\sqrt{625-(x+\Delta x)^2} - \sqrt{625-x^2}}{\Delta x} \frac{\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2}}{\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2}}\\ \amp =\frac{625-(x+\Delta x)^2-625+x^2}{\Delta x(\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2})}\\ \amp =\frac{625-x^2-2x\Delta x-\Delta x^2-625+x^2}{\Delta x(\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2})}\\ \amp =\frac{\Delta x(-2x-\Delta x)}{\Delta x(\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2})}\\ \amp =\frac{-2x-\Delta x}{\sqrt{625-(x+\Delta x)^2}+\sqrt{625-x^2}} \end{align*}

Now what happens when \(\Delta x\) is very close to zero? Again it seems apparent that the quotient will be very close to

\begin{equation*} \frac{-2x}{\sqrt{625-x^2}+\sqrt{625-x^2}} =\frac{-2x}{2\sqrt{625-x^2}}=\frac{-x}{\sqrt{625-x^2}}\text{.} \end{equation*}

Replacing \(x\) by 7 gives \(-7/24\text{,}\) as before, and now we can easily do the computation for 12 or any other value of \(x\) between \(-25\) and 25.

So now we have a single expression, \(\ds \dfrac{-x}{\sqrt{625-x^2}}\text{,}\) that tells us the slope of the tangent line for any value of \(x\text{.}\) This slope, in turn, tells us how sensitive the value of \(y\) is to small changes in the value of \(x\text{.}\)

To summarize, we computed the slope of the tangent line at a point \(P = \left(x, f(x)\right)\) on the curve of a function \(y = f(x)\) by forming the difference quotient and figuring out what happens when \(\Delta x\) gets very close to \(0\text{.}\) At this point, we should note that the idea of letting get closer and closer to \(0\) is precisely the idea of a limit that we discussed in the last chapter. This leads us to the following definition.

Definition 4.3. Slope of Tangent Line—Instantaneous Rate of Change.

The slope of the tangent line to the graph of a function \(y = f(x)\) at the point \(P = \left(x,f(x)\right)\) is given by

\begin{equation*} m = \lim_{\Delta x \to 0} \dfrac{f(x + \Delta x)-f(x)}{\Delta x}\text{,} \end{equation*}

provided this limit exists.

Note: The slope of the tangent line is also referred to as the insantaneous rate of change of \(f\) at \(x\text{.}\)

The expression\(\ds \dfrac{-x}{\sqrt{625-x^2}}\) defines a new function called the derivative (see Section 4.2) of the original function (since it is derived from the original function). If the original is referred to as \(f\) or \(y\) then the derivative is often written as \(f'\) or \(y'\text{,}\) read “f prime” or “y prime”. We also write

\begin{equation*} f'(x)=\dfrac{-x}{\sqrt{625-x^2}} \text{ or } y'=\ds \dfrac{-x}{\sqrt{625-x^2}}\text{.} \end{equation*}

At a particular point, say \(x=7\text{,}\) we write \(f'(7)=-7/24\) and we say that “\(f\) prime of 7 is \(-7/24\)” or “the derivative of \(f\) at 7 is \(-7/24\text{.}\)”

In the particular case of a circle, there's a simple way to find the derivative. Since the tangent to a circle at a point is perpendicular to the radius drawn to the point of contact, its slope is the negative reciprocal of the slope of the radius. The radius joining \((0,0)\) to \((7,24)\) has slope 24/7. Hence, the tangent line has slope \(-7/24\text{.}\) In general, a radius to the point \(\ds (x,\sqrt{625-x^2})\) has slope \(\ds \sqrt{625-x^2}/x\text{,}\) so the slope of the tangent line is \(\ds {-x/ \sqrt{625-x^2}}\text{,}\) as before. It is NOT always true that a tangent line is perpendicular to a line from the origin—don't use this shortcut in any other circumstance.

We now summarize our findings.

From Tangent Line Slope to Derivative.

Given a function \(f\) and a point \(x\) we can compute the derivative of \(f(x)\) at \(x\) as follows:

1. Form the difference quotient \(\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\text{,}\) which is the slope of a general secant line of the curve \(f\) throught the points \(P = \left(x,f(x)\right)\) and \(Q = \left(x+\Delta x, f(x+\Delta x)\right)\text{.}\)

2. Take the limits as \(\Delta x\) goes to zero: \(\lim\limits_{\Delta x \to 0} \dfrac{f(x+\Delta x)-f(x)}{\Delta x}\text{,}\) which is the slope of the tangent line of the curve \(f\) at the point \(P = \left(x,f(x)\right)\text{.}\)

3. If this limit exists, then the derivative exists and is equal to this limit.

In other words,

\begin{equation*} f'(x)=\lim\limits_{\Delta x \to 0} \dfrac{f(x+\Delta x)-f(x)}{\Delta x}\text{,} \end{equation*}

provided the limit exists.

Interactive Demonstration. Use the sliders below to investigate the limit of secant lines (blue and green) as they approach the tangent line (red) to any point \(P \) on the graph: