CHAPTER
9

Laying Down the Law for Derivatives

In This Chapter

• When can you find a derivative?
• Calculating rates of change
• Simple derivative techniques
• Derivatives of trigonometric functions
• Multiple derivatives
• Using calculators to evaluate derivatives

One of my most memorable college professors was a kindly Korean man named Dr. Oh. One of the reasons his class sticks out in my mind is the way he was able to illustrate things with bizarre but poignant imagery. The day we first discussed the Fundamental Theorem of Calculus, he described it in his usual understated way. “Today’s topic is like the day the world was created. Yesterday, not interesting. Today, interesting!”

One of the classes I took from Dr. Oh was Differential Equations. Dr. Oh constantly (but jokingly) harassed the young lady who sat next to me, because she would always do things the long way. No matter what shortcuts we learned, she wouldn’t use them. I never understood why, and she simply explained to me, “This is the way I do things. I can’t change it now!” I remember Dr. Oh repeatedly asking her, “If you want potatoes, do you buy a farm, till the field, plant the seed, nurture the plants, and then harvest the potatoes? If I were you, I would just go to the grocery store.”

In the land of derivatives, the difference quotient (see Chapter 8) is the equivalent of growing your own potatoes. Sure, the process works, but I gave you very specific examples so that it would work for you without any trouble or heartache. I was shielding you against the harsh weather of complicated derivatives to come. However, I have to let you grow up sometime and stare in the face of an ugly, complicated derivative. The good news, though, is that you can buy all your solutions from the grocery store.

When Does a Derivative Exist?

Before you run around finding derivatives willy-nilly, you should know that there are three specific instances in which the derivative to a function fails to exist. Even if you get a numerical answer when calculating a derivative, it’s possible that the answer is invalid, because there actually is no derivative! Be extra cautious if the graph of your function contains any of the following things.

Discontinuity

A derivative cannot exist at a point of discontinuity. It doesn’t matter if the discontinuity is removable or not. If a function is discontinuous at a specific x value, there cannot be a derivative there. For example, take a look at this function:

Without doing a bit of work, you can conclude that f(x) has no derivative at x = –2 and x = 6. In other words, f(x) is not differentiable at those values of x.

Sharp Point in the Graph

If a graph contains a sharp point (also known as a cusp), then the function has no derivative at that point. Not many functions have cusps; in fact, they are pretty rare. You’re most likely to see them in functions containing absolute values and in piecewise-defined functions whose pieces meet, but not smoothly. In Figure 9.1, you’ll find the graphs of the function and this piecewise-defined function:

Both f(x) and g(x) contain nondifferentiable cusps at x = 1.

Figure 9.1

Both graphs have a sharp point at x + 1 and, therefore, are not differentiable there.

Vertical Tangent Line

Remember that the derivative is defined as the slope of the tangent line. What if the tangent line is vertical? Keep in mind that vertical lines don’t have a slope, so a derivative cannot exist there. It’s pretty tough to spot when this happens using only a graph, but luckily, the mathematics of derivatives is quick to expose it when it happens, as shown in the next example.

Example 1: Show that no derivative exists for the function f(x) = x^1/3 when x = 0.

Solution: You don’t know how to find the derivative of f(x) = x^1/3 yet (but you will soon), so I’ll tell you that it’s . If you try to evaluate f ′(0), you get:

The slope of the tangent line is a nonexistent number, because you can’t divide by 0.

Basic Derivative Techniques

Learning how to find derivatives using the difference quotient can be long, tedious work, but once you’ve mastered it, you’ve “paid your dues,” so to speak. Now, you’ll learn some really handy techniques. Here are three derivative shortcuts that will make things a whole lot faster and easier.

The Power Rule

Even though the Power Rule can only find very basic derivatives, you’ll definitely use it more than any other of the rules we’ll learn. In fact, it often pops up in the final steps of other rules, but let’s not get ahead of ourselves. Any term in the form axⁿ can be differentiated using the Power Rule.

The Power Rule: The derivative of the term axⁿ (with respect to x), where a and n are real numbers, is (a ∙ n) x^n – 1.

Here are the steps you’ll use to find a derivative with the Power Rule:

1. Multiply the coefficient by the variable’s exponent. If no coefficient is stated—in other words, the coefficient equals 1—the exponent becomes the new coefficient.

2. Subtract 1 from the exponent.

Some examples will shed some light on the matter.

Example 2: Use the Power Rule to find the derivative of .

Solution: Even though there are a number of terms here, you can find the derivative of each one separately using the Power Rule. Before you start, I’ll tell you that the derivative of the constant term (–5) is 0. (More on this in a minute.) For the other terms, multiply the coefficient of each one by the exponent and then subtract 1 from the exponent:

Remember, the exponent of 6x is understood to be 1 because it’s not written explicitly, and a variable to the zero power equals 1: 6x⁰ = 6 ∙ 1 = 6.

The derivative of any constant is 0. Here is a quick justification if you are interested. Consider the constant function g(x) = 7. If you wanted to, you could write this function with a variable term: g(x) = 7x⁰. I am not changing the value of the function because a variable to the 0 power has a value of 1, and 7 ∙ 1 = 7. So now that you’ve rewritten the function, use the Power Rule:

The Product Rule

If a function contains two variable expressions multiplied together, you cannot simply find the derivative of each and multiply the results. For example, the derivative of x² × (x³ – 3) is not (2x)(3x²). Instead, you have to use a very simple formula, which (by the way) you should memorize.

The Product Rule: If a function h(x) = f(x) × g(x) is the product of two differentiable functions f(x) and g(x), then

h′(x) = f(x) ∙ g′(x) + f ′(x) ∙ g(x)

Here’s what that means. If a function is created by multiplying two other functions together, then the derivative of the overall function is the first one times the derivative of the second plus the second one times the derivative of the first.

Example 3: Differentiate f(x) = (x² + 6) (2x – 5) using (1) the Product Rule, and (2) the Power Rule, and show that the results are equal. Hint: to use the Power Rule, you’ll first have to multiply the terms together.

Solution: (1) Apply the Product Rule:

(2) As the hint indicates, you need to multiply those binomials together before you can apply the Power Rule: f(x) = 2x³ – 5x² + 12x – 30. Now, apply the Power Rule to get f ′(x) = 6x² – 10x + 12, which matches the answer from part (1).

The Quotient Rule

Just as the Product Rule prevents you from simply taking individual derivatives when you’re multiplying, the Quotient Rule prevents the same for division. Every year on my first derivatives exam, one of the problems is to find the derivative of something like , and half of my students always answer , no matter how many times I warn them to use the Quotient Rule. You must use the Quotient Rule any time two variable expressions are divided.

The Quotient Rule: If , where f(x) and g(x) are differentiable functions and g(x) ≠ 0, then:

In other words, to find the derivative of a fraction, take the bottom times the derivative of the top and subtract the top times the derivative of the bottom; divide all of that by the bottom squared. Of course, by top and bottom, I mean numerator and denominator, respectively.

Example 4: Find the derivative of using the Quotient Rule.

Solution: The numerator is f(x) in the Quotient Rule, and the denominator is g(x): f(x) = 3x + 7 and g(x) = x² – 1. Therefore, f ′(x) = 3 and g′(x) = 2x. Plug all of these values into the appropriate spots in the Quotient Rule:

The Chain Rule

Consider, for a moment, the functions and g(x) = 3x + 1. With the skills you now possess, you could find the derivative of each using the Power Rule. You could even find the derivatives of their product or their quotient , using the Product and Quotient Rules, respectively (no big surprise there).

However, you don’t know how to find the derivative of two functions plugged into (or “composed with”) one another. In other words, the derivative of requires a technique you’ve not yet learned, a technique called the Chain Rule.

If this function were simpler, such as , there would be no need for the Chain Rule, but the inner function (in this case 3x + 1) is too complicated. Here’s a good rule of thumb: if a function contains something other than a single variable, like x, then you should use the Chain Rule to find its derivative.

The Chain Rule: Given the composite function h(x) = f(g(x)), where f(x) and g(x) are differentiable functions, then

In other words, to take the derivative of an expression where one function is “trapped inside” another function, you follow these steps:

1. Take the derivative of the “outer” function, leaving the trapped, “inner” function alone.

2. Multiply the result by the derivative of the “inner” function.

Example 5: Use the Chain Rule to find the derivative of .

Solution: Rewrite the function so that it’s clear what’s actually plugged into what. In this case, 3x + 1 is plugged into . In other words, if (the outer function) and g(x) = 3x + 1 (the inner function, because it’s trapped inside the square root symbol in f(x)), then . Rewriting the function like this helps you plug everything into the right spots in the Chain Rule formula.

Your first step is to take the derivative of f(x), leaving g(x) alone. This just means you should find the derivative of f(x) and, once you’re done, plug g(x) into all of its x spots. According to the Power Rule, if f(x) = x^1/2:

Now plug g(x) in for x:

You’re almost done. The final step is to multiply this ugly monstrosity of a fraction by the derivative of g(x). A quick nod to the Power Rule tells you that if g(x) = 3x + 1, then g ′(x) = 3:

On the other hand, the graph of a line always changes at the exact same rate. For example, g(x) = 4x – 3 will always increase at a rate of 4, because that is the slope of the line and also the derivative. Curves, however, do not have the same slope everywhere, so we rely on the slopes of their tangent lines. Sometimes a curve is increasing quickly and the tangent line is steep (causing a high-valued derivative). At other places the curve may be increasing shallowly or even decreasing, causing the derivative to be small or negative, respectively. Look at the graph of f(x) in Figure 9.2.

At x = a, f is increasing ever so slightly, causing the tangent line there to be shallow. Because a shallow line has a slope close to 0, the derivative here will be very small. In other words, the rate of change of the graph is very small at the instant that x = a. However, at the instant that x = b, the graph is climbing more rapidly, causing a steeper tangent line, which in turn causes a larger derivative. Finally, at x = c, the graph is decreasing, so the instantaneous rate of change there is negative (because the slope of the tangent line is negative).

You can also use the slope of a secant line to determine rates of change on a graph. However, the slope of a secant line describes something different: the average rate of change over some portion of the graph. Finding the slope of a secant line is very easy, as you’ll see in the next example.

Example 6: Poteet, Inc. has just introduced a new, revolutionary brand of athletic sock into the market. The new innovation is a special sweat-absorbing “cotton-esque” material that supposedly prevents foot odor. On their fourth day of sales, the snappy slogan, “If you smell feet, they ain’t wrapped in Poteet’s,” was released, and sales immediately increased. Figure 9.3 is a graph of the number of units sold during the first six days of sales.

Figure 9.3

The rise of a new sweat sock empire.

What was the average rate of units sold per day between day one and day six?

Solution: The problem asks us to find an average rate of change, which translates to finding the slope, m, of the secant line connecting the points (1,6.5) and (6,8.5). To do that, use our tried-and-true method from algebra:

That means Poteet’s socks sold at a rate of two-fifths of a thousand units per day, or units/day on average. So even through the moderate decreases they experienced, the new slogan probably helped.

It’s not as bad as you think—half of the inverse trig derivatives are different from the other half by only a negative sign.

You’ll notice that I have included the inverse trig functions in this list, but you may not recognize them. Instead of using the notation y = sin^–1x to indicate the inverse sine, I use the notation y = arcsin x. I am a huge fan of the latter notation, because sin^–1x looks a lot like (sin x)^–1, which is equal to csc x (which is not the inverse of sin x).

You’ll have to be able to use these formulas with the Product, Quotient, and Chain Rules, so here are a couple of examples to get you used to them. Remember, if a trig function contains anything except a single variable (like x), you have to use the Chain Rule to find the derivative.

Example 7: If f(x) = cos x sin 2x, identify f ′(x) and evaluate .

Solution: Because this function is the product of two variable expressions, you’ll have to use the Product Rule. In addition, you’ll have to use the Chain Rule to differentiate sin 2x, because it contains more than just x inside the sine function. According to the Chain Rule, Here’s the Product Rule in action:

Example 8: Assume functions f(x) and g(x) are continuous and differentiable for all real numbers. The following table lists values of the functions and their derivatives for specific x values.

Based on this information, calculate the following:

(a)

(b)

Solution: Although you don’t have specific functions that define f(x) and g(x), you can still apply the derivative techniques you learned in this chapter.

(a) The function h(x) is defined as the product of functions f(x) and g(x). That means you need to use the Product Rule to calculate h ′(x):

What now? You don’t know what f(x) or g(x) are equal to, so how are you supposed to know what their derivatives are? No need to worry. You aren’t asked to find the general derivative, just the value of the derivative when x = 1.

h′(1) = f(1) ∙ g′(1) + g(1) ∙ f ′(1)

Find f(1), f ′(1), g(1), and g′(1) in the table and substitute them into the formula.

(b) Function j(x) is a composition of functions: f(x) is substituted into g(x). You need to apply the Chain Rule to calculate j′(x):

You’re asked to calculate j′(–3), so replace x with –3:

According to the chart, f(–3) = 0.

j′(–3) = g′(0) ∙ f ′(–3)

Once again, refer to the chart to identify the values of g′(0) and f ′(–3):

Example 9: Given the graph of f(x) in Figure 9.4, estimate the following:

(a) f ′(3)

(b) f ′(–2)

Figure 9.4

The graph of a function f(x).

Solution: As Chapter 8 explains, the derivative of a function is the slope of the tangent line at a specific x value. Therefore, to estimate the derivatives of f(x) in this example, you should draw tangent lines at the given x values and calculate their slopes.

(a) The graph of f(x) seems to be changing direction at x = 3. Before it passes x = 3, as you travel from left to right, it is decreasing. However, after x = 3, it increases. The tangent line at x = 3, therefore, will be horizontal, as illustrated in Figure 9.5. Remember, this is an estimation, so even if the tangent isn’t perfectly horizontal, this is a good enough guess.

Figure 9.5

The line tangent to f(x) when x = 3 is horizontal.

What is the derivative when x = 3? The slope of a horizontal line is 0, so you can confidently assert that f ′(3) ≈ 0.

(b) Draw the line tangent to the graph at x = –2.

Figure 9.6

The line tangent to the graph of f(x) when x = –2.

Your line may vary a little from mine, but remember that you’re only asked for an estimate. My dotted tangent line passes through points (–2,3) and (1,0). Use the slope formula to calculate the slope of the tangent:

You conclude that f ′(–2) ≈ –1.

Technology Focus: Calculating Derivatives

Different calculators have different capabilities when it comes to calculating derivatives. Some calculators, like the TI-89, can differentiate symbolic expressions. By accessing the Calculus menu on the home screen, you can select the differentiate function, as illustrated in Figure 9.7. If you own a symbolic calculator, you can check most of your homework in a snap. (However, don’t be tempted to depend on your calculator to do your work for you.)

Figure 9.7

The Calc menu contains common calculus functions, including “differentiate.”

For example, you can check your work for Example 3 in this chapter by entering the expression, as shown in Figure 9.8. Notice that the expression is followed by “,x” indicating that the expression you typed contains the variable x, and that is the variable you are differentiating with respect to. (More about what “with respect to” means in Chapter 10.)

Figure 9.8

The calculator verifies the solution to Example 3 in this chapter, from the Product Rule section.

Pretty powerful! The calculator applied the Product Rule for you automatically and even simplified the expression. Symbolic calculators are very smart devices. Check out Figure 9.9 to see that even trigonometric and inverse trigonometric functions are just as easy to differentiate.

Figure 9.9

The TI-89 doesn’t blink an electronic eye at the thought of differentiating trigonometric functions.

Here’s the bad news: symbolic calculators are so powerful that most classes and exams don’t allow them. After all, teachers and exam makers want to make sure you understand calculus; they aren’t interested in measuring your graphing calculator skills. Other calculators are allowed, including the TI-84 family. While they cannot handle symbolic differentiation, they can calculate derivatives at specific x values.

For example, in Example 3 (and in the calculator screens above), you determined that the derivative of f(x) = (x² + 6)(2x – 5) was 6x² – 10x + 12. Your TI-84 calculator would not be able to tell you the derivative expression, but it could evaluate the derivative for a specific value of x. In other words, it can’t tell you what f ′(x) is, but it can tell you what f ′(–2) is … almost.

Press the button and scroll down to the “nDeriv(” option, as illustrated in Figure 9.10. This stands for “numeric derivative,” as opposed to symbolic derivative.

Figure 9.10

Some of your screens may look different from mine if you have MathPrint enabled in the Mode menu. You can always toggle display options in the Mode menu.

Now type the expression from Example 3, followed by “,x,–2)” to indicate that you are taking the derivative of the x variable and evaluating the derivative at x = –2. See Figure 9.11 for the result.

Figure 9.11

This is an estimate of the correct answer. It is not 100 percent accurate.

We can give the calculator a lot of credit for trying, but let’s be honest—it did not get the answer right. If you substitute x = –2 into the derivative you get 56, not 56.000002.

Why does the calculator get it wrong? Remember, it’s not actually calculating the symbolic derivative. It is using a method similar to the difference quotient to calculate a tangent slope, so you cannot count on it for exact measurements. However, 56.000002 is pretty close to 56, so you can still use it to check any numeric derivative answers.

One warning: your calculator may try to calculate a derivative even if it doesn’t exist! For example, the function is not differentiable when x = 0, because it contains a cusp there. The algorithm your calculator uses to approximate the derivative may not test for cusps, so the calculator reports a derivative of 0 (see Figure 9.12).

Figure 9.12

The derivative of does not exist when x = 0, but that’s news to your calculator, which managed to find a derivative anyway.

There are pros and cons to using your graphing calculator. A symbolic calculator is very smart—too smart, you could argue, to use on a quiz or test. A calculator that provides only numeric derivatives is smart enough to be helpful, as long as you interpret the results appropriately.

The Least You Need to Know

• If a function is differentiable, it must also be continuous.
• A function is not differentiable at a point of discontinuity, a sharp point (cusp), or where the tangent line is vertical.
• The slope of a function’s tangent line gives its instantaneous rate of change, and the slope of its secant line gives the average rate of change.
• Products and quotients of variable expressions must be differentiated using the Product and Quotient Rules, respectively.
• You must use the Chain Rule to differentiate any function that contains something other than just x.
• You can use your calculator to check your work when evaluating derivatives.