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Instantaneous Rate of Change:
The Derivative
2.1 The slope of a fun tion
Suppose that y is a function of x, say y = f (x). It is often necessary to know how sensitive the value of y is to small changes in x.
EXAMPLE 2.1.1 Take, for example, y = f (x) =
625 − x^2 (the upper semicircle of radius 25 centered at the origin). When x = 7, we find that y =
625 − 49 = 24. Suppose we want to know how much y changes when x increases a little, say to 7.1 or 7.01. In the case of a straight line y = mx+b, the slope m = ∆y/∆x measures the change in y per unit change in x. This can be interpreted as a measure of “sensitivity”; for example, if y = 100x + 5, a small change in x corresponds to a change one hundred times as large in y, so y is quite sensitive to changes in x. Let us look at the same ratio ∆y/∆x for our function y = f (x) =
625 − x^2 when x changes from 7 to 7.1. Here ∆x = 7. 1 − 7 = 0.1 is the change in x, and
∆y = f (x + ∆x) − f (x) = f (7.1) − f (7)
Thus, ∆y/∆x ≈ − 0. 0294 / 0 .1 = − 0 .294. This means that y changes by less than one third the change in x, so apparently y is not very sensitive to changes in x at x = 7. We say “apparently” here because we don’t really know what happens between 7 and 7.1. Perhaps y changes dramatically as x runs through the values from 7 to 7.1, but at 7. 1 y just happens to be close to its value at 7. This is not in fact the case for this particular function, but we don’t yet know why.
29
30 Chapter 2 Instantaneous Rate of Change: The Derivative
One way to interpret the above calculation is by reference to a line. We have computed the slope of the line through (7, 24) and (7. 1 , 23 .9706), called a chord of the circle. In general, if we draw the chord from the point (7, 24) to a nearby point on the semicircle (7 + ∆x, f (7 + ∆x)), the slope of this chord is the so-called difference quotient
slope of chord = f^ (7 + ∆ ∆xx)^ −^ f^ (7)=
625 − (7 + ∆x)^2 − 24 ∆x.
For example, if x changes only from 7 to 7.01, then the difference quotient (slope of the chord) is approximately equal to (23. 997081 − 24)/ 0 .01 = − 0 .2919. This is slightly less steep than the chord from (7, 24) to (7. 1 , 23 .9706). As the second value 7 + ∆x moves in towards 7, the chord joining (7, f (7)) to (7 + ∆x, f (7 + ∆x)) shifts slightly. As indicated in figure 2.1.1, as ∆x gets smaller and smaller, the chord joining (7, 24) to (7 + ∆x, f (7 + ∆x)) gets closer and closer to the tangent line to the circle at the point (7, 24). (Recall that 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 ∆x gets smaller and smaller, the slope ∆y/∆x of the chord gets closer and closer to the slope of the tangent line. This is actually quite difficult to see when ∆x is small, because of the scale of the graph. The values of ∆x used for the figure are 1, 5, 10 and 15, not really very small values. The tangent line is the one that is uppermost at the right hand endpoint.
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Figure 2.1.1 Chords approximating the tangent line. (AP)
So far we have found the slopes of two chords 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 chords as they get “closer and closer” to the tangent line.
32 Chapter 2 Instantaneous Rate of Change: The Derivative
example, the “24” in the calculation came from
625 − 72 , so we’ll need to fix that too. √ 625 − (x + ∆x)^2 −
625 − x^2 ∆x
625 − (x + ∆x)^2 −
625 − x^2 ∆x
625 − (x + ∆x)^2 +
√^625 −^ x^2 625 − (x + ∆x)^2 +
625 − x^2
= 625 −^ (x^ + ∆x)
(^2) − 625 + x 2 ∆x(
625 − (x + ∆x)^2 +
625 − x^2 )
= 625 −^ x
(^2) − 2 x∆x − ∆x (^2) − 625 + x 2 ∆x(
625 − (x + ∆x)^2 +
625 − x^2 )
= ∆x(−^2 x^ −^ ∆x) ∆x(
625 − (x + ∆x)^2 +
625 − x^2 )
= √ −^2 x^ −^ ∆x 625 − (x + ∆x)^2 +
625 − x^2
Now what happens when ∆x is very close to zero? Again it seems apparent that the quotient will be very close to
√^ −^2 x 625 − x^2 +
625 − x^2
= −^2 x 2
625 − x^2
= √ −x 625 − x^2
Replacing x by 7 gives − 7 /24, 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, simple formula, −x/
625 − x^2 , that tells us the slope of the tangent line for any value of x. This slope, in turn, tells us how sensitive the value of y is to changes in the value of x. What do we call such a formula? That is, a formula with one variable, so that substi- tuting an “input” value for the variable produces a new “output” value? This is a function.
Starting with one function,
625 − x^2 , we have derived, by means of some slightly nasty algebra, a new function, −x/
625 − x^2 , that gives us important information about the original function. This new function in fact is called the derivative of the original func- tion. If the original is referred to as f or y then the derivative is often written f ′^ or y′^ and pronounced “f prime” or “y prime”, so in this case we might write f ′(x) = −x/
625 − x^2. At a particular point, say x = 7, we say that f ′(7) = − 7 /24 or “f prime of 7 is − 7 /24” or “the derivative of f at 7 is − 7 /24.” To summarize, we compute the derivative of f (x) by forming the difference quotient f (x + ∆x) − f (x) ∆x ,^ (2.^1 .1) which is the slope of a line, then we figure out what happens when ∆x gets very close to
2.1 The slope of a function 33
We should note that 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. In
general, a radius to the point (x,
625 − x^2 ) has slope
625 − x^2 /x, so the slope of the
tangent line is −x/
625 − x^2 , 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. As above, and as you might expect, for different values of x we generally get different values of the derivative f ′(x). Could it be that the derivative always has the same value? This would mean that the slope of f , or the slope of its tangent line, is the same everywhere. One curve that always has the same slope is a line; it seems odd to talk about the tangent line to a line, but if it makes sense at all the tangent line must be the line itself. It is not hard to see that the derivative of f (x) = mx + b is f ′(x) = m; see exercise 6.
Exercises 2.1.
- Draw the graph of the function y = f (x) =
√ 169 − x^2 between x = 0 and x = 13. Find the slope ∆y/∆x of the chord between the points of the circle lying over (a) x = 12 and x = 13, (b) x = 12 and x = 12.1, (c) x = 12 and x = 12.01, (d) x = 12 and x = 12.001. Now use the geometry of tangent lines on a circle to find (e) the exact value of the derivative f ′(12). Your answers to (a)–(d) should be getting closer and closer to your answer to (e). ⇒
- Use geometry to find the derivative f ′(x) of the function f (x) =
√ 625 − x^2 in the text for each of the following x: (a) 20, (b) 24, (c) −7, (d) −15. Draw a graph of the upper semicircle, and draw the tangent line at each of these four points. ⇒
- Draw the graph of the function y = f (x) = 1/x between x = 1/2 and x = 4. Find the slope of the chord between (a) x = 3 and x = 3.1, (b) x = 3 and x = 3.01, (c) x = 3 and x = 3.001. Now use algebra to find a simple formula for the slope of the chord between (3, f (3)) and (3 + ∆x, f (3 + ∆x)). Determine what happens when ∆x approaches 0. In your graph of y = 1/x, draw the straight line through the point (3, 1 /3) whose slope is this limiting value of the difference quotient as ∆x approaches 0. ⇒
- Find an algebraic expression for the difference quotient
( f (1 + ∆x) − f (1)
) /∆x when f (x) = x^2 − (1/x). Simplify the expression as much as possible. Then determine what happens as ∆x approaches 0. That value is f ′(1). ⇒
- Draw the graph of y = f (x) = x^3 between x = 0 and x = 1.5. Find the slope of the chord between (a) x = 1 and x = 1.1, (b) x = 1 and x = 1.001, (c) x = 1 and x = 1.00001. Then use algebra to find a simple formula for the slope of the chord between 1 and 1 + ∆x. (Use the expansion (A + B)^3 = A^3 + 3A^2 B + 3AB^2 + B^3 .) Determine what happens as ∆x approaches 0, and in your graph of y = x^3 draw the straight line through the point (1, 1) whose slope is equal to the value you just found. ⇒
- Find an algebraic expression for the difference quotient (f (x + ∆x) − f (x))/∆x when f (x) = mx + b. Simplify the expression as much as possible. Then determine what happens as ∆x approaches 0. That value is f ′(x). ⇒
2.2 An example 35
a bit of algebra:
h(2) − h(2 + ∆t) ∆t =
- 4 − (100 − 4 .9(2 + ∆t)^2 ) ∆t =^80.^4 −^ 100 + 19.6 + 19.6∆t^ + 4.9∆t
2 ∆t =^19 .6∆t^ + 4.9∆t
2 ∆t = 19.6 + 4.9∆t
When ∆t is very small, this is very close to 19.6, and indeed it seems clear that as ∆t goes to zero, the average speed goes to 19.6, so the exact speed at t = 2 is 19.6 meters per second. This calculation should look very familiar. In the language of the previous section, we might have started with f (x) = 100 − 4. 9 x^2 and asked for the slope of the tangent line at x = 2. We would have answered that question by computing
f (2 + ∆x) − f (2) ∆x =^
− 19 .6∆x − 4 .9∆x^2 ∆x =^ −^19.^6 −^4 .9∆x
The algebra is the same, except that following the pattern of the previous section the subtraction would be reversed, and we would say that the slope of the tangent line is − 19 .6. Indeed, in hindsight, perhaps we should have subtracted the other way even for the dropping ball. At t = 2 the height is 80.4; one second later the height is 55.9. The usual way to compute a “distance traveled” is to subtract the earlier position from the later one, or 55. 9 − 80 .4 = − 24 .5. This tells us that the distance traveled is 24.5 meters, and the negative sign tells us that the height went down during the second. If we continue the original calculation we then get − 19 .6 meters per second as the exact speed at t = 2. If we interpret the negative sign as meaning that the motion is downward, which seems reasonable, then in fact this is the same answer as before, but with even more information, since the numerical answer contains the direction of motion as well as the speed. Thus, the speed of the ball is the value of the derivative of a certain function, namely, of the function that gives the position of the ball. (More properly, this is the velocity of the ball; velocity is signed speed, that is, speed with a direction indicated by the sign.) The upshot is that this problem, finding the speed of the ball, is exactly the same problem mathematically as finding the slope of a curve. This may already be enough evidence to convince you that whenever some quantity is changing (the height of a curve or the height of a ball or the size of the economy or the distance of a space probe from earth or the population of the world) the rate at which the quantity is changing can, in principle, be computed in exactly the same way, by finding a derivative.
36 Chapter 2 Instantaneous Rate of Change: The Derivative
Exercises 2.2.
- An object is traveling in a straight line so that its position (that is, distance from some fixed point) is given by this table:
time (seconds) 0 1 2 3 distance (meters) 0 10 25 60 Find the average speed of the object during the following time intervals: [0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]. If you had to guess the speed at t = 2 just on the basis of these, what would you guess? ⇒
- Let y = f (t) = t^2 , where t is the time in seconds and y is the distance in meters that an object falls on a certain airless planet. Draw a graph of this function between t = 0 and t = 3. Make a table of the average speed of the falling object between (a) 2 seconds and 3 seconds, (b) 2 seconds and 2.1 seconds, (c) 2 seconds and 2.01 seconds, (d) 2 seconds and 2.001 seconds. Then use algebra to find a simple formula for the average speed between time 2 and time 2+ ∆t. (If you substitute ∆t = 1, 0. 1 , 0. 01 , 0 .001 in this formula you should again get the answers to parts (a)–(d).) Next, in your formula for average speed (which should be in simplified form) determine what happens as ∆t approaches zero. This is the instantaneous speed. Finally, in your graph of y = t^2 draw the straight line through the point (2, 4) whose slope is the instantaneous velocity you just computed; it should of course be the tangent line. ⇒
- If an object is dropped from an 80-meter high window, its height y above the ground at time t seconds is given by the formula y = f (t) = 80 − 4. 9 t^2. (Here we are neglecting air resistance; the graph of this function was shown in figure 1.0.1.) Find the average velocity of the falling object between (a) 1 seconds and 1.1 seconds, (b) 1 seconds and 1.01 seconds, (c) 1 seconds and 1.001 seconds. Now use algebra to find a simple formula for the average velocity of the falling object between 1 seconds and 1 + ∆t seconds. Determine what happens to this average velocity as ∆t approaches 0. That is the instantaneous velocity at time t = 1 second (it will be negative, because the object is falling). ⇒
2.3 Limits
In the previous two sections we computed some quantities of interest (slope, velocity) by seeing that some expression “goes to” or “approaches” or “gets really close to” a particular value. In the examples we saw, this idea may have been clear enough, but it is too fuzzy to rely on in more difficult circumstances. In this section we will see how to make the idea more precise. There is an important feature of the examples we have seen. Consider again the formula − 19 .6∆x − 4 .9∆x^2 ∆x.
We wanted to know what happens to this fraction as “∆x goes to zero.” Because we were able to simplify the fraction, it was easy to see the answer, but it was not quite as simple
38 Chapter 2 Instantaneous Rate of Change: The Derivative
that − 19. 6 − 4 .9∆x > − 19 .600001. This is something we can manipulate with a little algebra: − 19. 6 − 4 .9∆x > − 19. 600001 − 4 .9∆x > − 0. 000001 ∆x < − 0. 000001 / − 4. 9 ∆x < 0. 0000002040816327...
Thus, we can say with certainty that if ∆x is positive and less than 0.0000002, then ∆x < 0. 0000002040816327... and so − 19. 6 − 4 .9∆x > − 19 .600001. We could do a similar calculation if ∆x is negative. So now we know that we can make − 19. 6 − 4 .9∆x within one millionth of − 19 .6. But can we make it “as close as we want”? In this case, it is quite simple to see that the answer is yes, by modifying the calculation we’ve just done. It may be helpful to think of this as a game. I claim that I can make − 19. 6 − 4 .9∆x as close as you desire to − 19 .6 by making ∆x “close enough” to zero. So the game is: you give me a number, like 10−^6 , and I have to come up with a number representing how close ∆x must be to zero to guarantee that − 19. 6 − 4 .9∆x is at least as close to − 19 .6 as you have requested. Now if we actually play this game, I could redo the calculation above for each new number you provide. What I’d like to do is somehow see that I will always succeed, and even more, I’d like to have a simple strategy so that I don’t have to do all that algebra every time. A strategy in this case would be a formula that gives me a correct answer no matter what you specify. So suppose the number you give me is ǫ. How close does ∆x have to be to zero to guarantee that − 19. 6 − 4 .9∆x is in (− 19. 6 − ǫ, − 19 .6 + ǫ)? If ∆x is positive, we need: − 19. 6 − 4 .9∆x > − 19. 6 − ǫ − 4 .9∆x > −ǫ ∆x < −ǫ/ − 4. 9 ∆x < ǫ/ 4. 9
So if I pick any number δ that is less than ǫ/ 4 .9, the algebra tells me that whenever ∆x < δ then ∆x < ǫ/ 4 .9 and so − 19. 6 − 4 .9∆x is within ǫ of − 19 .6. (This is exactly what I did in the example: I picked δ = 0. 0000002 < 0. 0000002040816327.. ..) A similar calculation again works for negative ∆x. The important fact is that this is now a completely general result—it shows that I can always win, no matter what “move” you make. Now we can codify this by giving a precise definition to replace the fuzzy, “gets closer and closer” language we have used so far. Henceforward, we will say something like “the limit of (− 19 .6∆x− 4 .9∆x^2 )/∆x as ∆x goes to zero is − 19 .6,” and abbreviate this mouthful
2.3 Limits 39
as
∆limx→ 0 −^19 .6∆x^ −^4 .9∆x
2 ∆x
Here is the actual, official definition of “limit”.
DEFINITION 2.3.2 Limit Suppose f is a function. We say that lim x→a f (x) = L if
for every ǫ > 0 there is a δ > 0 so that whenever 0 < |x − a| < δ, |f (x) − L| < ǫ.
The ǫ and δ here play exactly the role they did in the preceding discussion. The definition says, in a very precise way, that f (x) can be made as close as desired to L (that’s the |f (x) − L| < ǫ part) by making x close enough to a (the 0 < |x − a| < δ part). Note that we specifically make no mention of what must happen if x = a, that is, if |x − a| = 0. This is because in the cases we are most interested in, substituting a for x doesn’t even make sense. Make sure you are not confused by the names of important quantities. The generic definition talks about f (x), but the function and the variable might have other names. In the discussion above, the function we analyzed was
− 19 .6∆x − 4 .9∆x^2 ∆x
and the variable of the limit was not x but ∆x. The x was the variable of the original function; when we were trying to compute a slope or a velocity, x was essentially a fixed quantity, telling us at what point we wanted the slope. (In the velocity problem, it was literally a fixed quantity, as we focused on the time 2.) The quantity a of the definition in all the examples was zero: we were always interested in what happened as ∆x became very close to zero. Armed with a precise definition, we can now prove that certain quantities behave in a particular way. The bad news is that even proofs for simple quantities can be quite tedious and complicated; the good news is that we rarely need to do such proofs, because most expressions act the way you would expect, and this can be proved once and for all.
EXAMPLE 2.3.3 Let’s show carefully that lim x→ 2 x + 4 = 6. This is not something we
“need” to prove, since it is “obviously” true. But if we couldn’t prove it using our official definition there would be something very wrong with the definition. As is often the case in mathematical proofs, it helps to work backwards. We want to end up showing that under certain circumstances x + 4 is close to 6; precisely, we want to show that |x + 4 − 6 | < ǫ, or |x − 2 | < ǫ. Under what circumstances? We want this to be true whenever 0 < |x − 2 | < δ. So the question becomes: can we choose a value for δ that
2.3 Limits 41
we don’t have to worry about it ever again. When we say that x might be “complicated” we really mean that in practice it might be a function. Here is then what we want to know:
THEOREM 2.3.5 Suppose lim x→a f (x) = L and lim x→a g(x) = M. Then
limx→a f (x)g(x) = LM.
Proof. We have to use the official definition of limit to make sense of this. So given any ǫ we need to find a δ so that 0 < |x − a| < δ implies |f (x)g(x) − LM | < ǫ. What do we have to work with? We know that we can make f (x) close to L and g(x) close to M , and we have to somehow connect these facts to make f (x)g(x) close to LM. We use, as is so often the case, a little algebraic trick: |f (x)g(x) − LM | = |f (x)g(x) − f (x)M + f (x)M − LM | = |f (x)(g(x) − M ) + (f (x) − L)M | ≤ |f (x)(g(x) − M )| + |(f (x) − L)M | = |f (x)||g(x) − M | + |f (x) − L||M |.
This is all straightforward except perhaps for the “≤”. That is an example of the triangle inequality, which says that if a and b are any real numbers then |a + b| ≤ |a| + |b|. If you look at a few examples, using positive and negative numbers in various combinations for a and b, you should quickly understand why this is true; we will not prove it formally. Since lim x→a f (x) = L, there is a value δ 1 so that 0 < |x − a| < δ 1 implies |f (x) − L| <
|ǫ/(2M )|, This means that 0 < |x − a| < δ 1 implies |f (x) − L||M | < ǫ/2. You can see where this is going: if we can make |f (x)||g(x) − M | < ǫ/2 also, then we’ll be done. We can make |g(x) − M | smaller than any fixed number by making x close enough to a; unfortunately, ǫ/(2f (x)) is not a fixed number, since x is a variable. Here we need another little trick, just like the one we used in analyzing x^2. We can find a δ 2 so that |x − a| < δ 2 implies that |f (x) − L| < 1, meaning that L − 1 < f (x) < L + 1. This means that |f (x)| < N , where N is either |L − 1 | or |L + 1|, depending on whether L is negative or positive. The important point is that N doesn’t depend on x. Finally, we know that there is a δ 3 so that 0 < |x − a| < δ 3 implies |g(x) − M | < ǫ/(2N ). Now we’re ready to put everything together. Let δ be the smallest of δ 1 , δ 2 , and δ 3. Then |x − a| < δ implies that |f (x) − L| < |ǫ/(2M )|, |f (x)| < N , and |g(x) − M | < ǫ/(2N ). Then
|f (x)g(x) − LM | ≤ |f (x)||g(x) − M | + |f (x) − L||M | < N 2 ǫN +
ǫ 2 M
∣ |M^ |
= ǫ 2 + 2 ǫ= ǫ.
This is just what we needed, so by the official definition, lim x→a f (x)g(x) = LM.
42 Chapter 2 Instantaneous Rate of Change: The Derivative
A handful of such theorems give us the tools to compute many limits without explicitly working with the definition of limit.
THEOREM 2.3.6 Suppose that (^) xlim→a f (x) = L and (^) xlim→a g(x) = M and k is some
constant. Then
xlim→a kf^ (x) =^ k^ xlim→a f^ (x) =^ kL
x^ lim→a(f^ (x) +^ g(x)) = lim x→a f^ (x) + lim x→a g(x) =^ L^ +^ M
x^ lim→a(f^ (x)^ −^ g(x)) = lim x→a f^ (x)^ −^ xlim→a g(x) =^ L^ −^ M x^ lim→a(f^ (x)g(x)) = lim x→a f^ (x)^ ·^ xlim→a g(x) =^ LM
x^ lim→a^ f g^ ((xx) )= lim limx→a^ f^ (x) x→a g(x)
= (^) M L, if M is not 0
Roughly speaking, these rules say that to compute the limit of an algebraic expression, it is enough to compute the limits of the “innermost bits” and then combine these limits. This often means that it is possible to simply plug in a value for the variable, since
xlim→a x^ =^ a.
EXAMPLE 2.3.7 Compute lim x→ 1 x
(^2) − 3 x + 5 x − 2. If we apply the theorem in all its gory detail, we get
xlim→ 1
x^2 − 3 x + 5 x − 2 =
limx→ 1 (x^2 − 3 x + 5) limx→ 1 (x − 2)
= (limx→^1 x
(^2) ) − (limx→ 1 3 x) + (limx→ 1 5) (limx→ 1 x) − (limx→ 1 2)
= (limx→^1 x)
(^2) − 3(limx→ 1 x) + 5 (limx→ 1 x) − 2
=^1
=^1 − −^ 3 + 5 1 = − 3
It is worth commenting on the trivial limit lim x→ 1 5. From one point of view this might
seem meaningless, as the number 5 can’t “approach” any value, since it is simply a fixed
44 Chapter 2 Instantaneous Rate of Change: The Derivative
This theorem is not too difficult to prove from the definition of limit. Another of the most common algebraic tricks was used in section 2.1. Here’s another example:
EXAMPLE 2.3.11 Compute (^) xlim→− 1
x + 5 − 2 x + 1.
xlim→− 1
x + 5 − 2 x + 1 =^ xlim→− 1
x + 5 − 2 x + 1
√x^ + 5 + 2 x + 5 + 2 = (^) xlim→− 1 x^ + 5^ −^4 (x + 1)(
x + 5 + 2)
= (^) xlim→− 1 x^ + 1 (x + 1)(
x + 5 + 2)
= (^) xlim→− 1 √^1 x + 5 + 2
=^1
At the very last step we have used theorems 2.3.9 and 2.3.10.
Occasionally we will need a slightly modified version of the limit definition. Consider the function f (x) =
1 − x^2 , the upper half of the unit circle. What can we say about
xlim→ 1 f^ (x)? It is apparent from the graph of this familiar function that as^ x^ gets close to 1 from the left, the value of f (x) gets close to zero. It does not even make sense to ask what happens as x approaches 1 from the right, since f (x) is not defined there. The definition of the limit, however, demands that f (1 + ∆x) be close to f (1) whether ∆x is positive or negative. Sometimes the limit of a function exists from one side or the other (or both) even though the limit does not exist. Since it is useful to be able to talk about this situation, we introduce the concept of one sided limit:
DEFINITION 2.3.12 One-sided limit Suppose that f (x) is a function. We say that (^) xlim→a− f (x) = L if for every ǫ > 0 there is a δ > 0 so that whenever 0 < a − x < δ,
|f (x) − L| < ǫ. We say that limx→a+ f (x) = L if for every ǫ > 0 there is a δ > 0 so that whenever 0 < x − a < δ, |f (x) − L| < ǫ.
Usually (^) xlim→a− f (x) is read “the limit of f (x) from the left” and (^) xlim→a+ f (x) is read “the
limit of f (x) from the right”.
EXAMPLE 2.3.13 Discuss lim x→ (^0) |^ xx| , (^) xlim→ 0 − |^ xx| , and (^) xlim→ 0 + |^ xx|.
The function f (x) = x/|x| is undefined at 0; when x > 0, |x| = x and so f (x) = 1;
when x < 0, |x| = −x and f (x) = −1. Thus lim x→ 0 −
x |x|
= lim x→ 0 −^
−1 = −1 while lim x→ 0 +
x |x|
2.3 Limits 45
x^ lim→ 0 +^ 1 = 1. The limit of^ f^ (x) must be equal to both the left and right limits; since they are different, the limit lim x→ 0 x |x|
does not exist.
Exercises 2.3.
Compute the limits. If a limit does not exist, explain why.
- (^) xlim→ 3 x
(^2) + x − 12 x − 3 ⇒^ 2.^ xlim→^1
x^2 + x − 12 x − 3 ⇒
- (^) xlim→− 4 x
(^2) + x − 12 x − 3 ⇒^ 4.^ xlim→^2
x^2 + x − 12 x − 2 ⇒
- (^) xlim→ 1
√ x + 8 − 3 x − 1 ⇒^ 6.^ xlim→ 0 +
√ 1 x + 2^ −
√ 1 x ⇒
- (^) xlim→ 2 3 ⇒ 8. (^) xlim→ 4 3 x^3 − 5 x ⇒
- (^) xlim→ 04 x^ −^5 x
2 x − 1 ⇒^ 10.^ xlim→^1
x^2 − 1 x − 1 ⇒
- (^) xlim→ 0 +
√ 2 − x^2 x ⇒^ 12.^ xlim→ 0 +
√ 2 − x^2 x + 1 ⇒
- (^) xlim→a^ x
(^3) − a 3 x − a ⇒^ 14.^ xlim→^2 (x
(^2) + 4) (^3) ⇒
- (^) xlim→ 1
{ (^) x − 5 x 6 = 1, 7 x = 1. ⇒
- (^) xlim→ 0 x sin
( (^1) x
) (Hint: Use the fact that | sin a| < 1 for any real number a. You should probably use the definition of a limit here.) ⇒
- Give an ǫ–δ proof, similar to example 2.3.3, of the fact that lim x→ 4 (2x − 5) = 3.
2.4 The Derivative Function 47
We know that f ′^ carries important information about the original function f. In one example we saw that f ′(x) tells us how steep the graph of f (x) is; in another we saw that f ′(x) tells us the velocity of an object if f (x) tells us the position of the object at time x. As we said earlier, this same mathematical idea is useful whenever f (x) represents some changing quantity and we want to know something about how it changes, or roughly, the “rate” at which it changes. Most functions encountered in practice are built up from a small collection of “primitive” functions in a few simple ways, for example, by adding or multiplying functions together to get new, more complicated functions. To make good use of the information provided by f ′(x) we need to be able to compute it for a variety of such functions. We will begin to use different notations for the derivative of a function. While initially confusing, each is often useful so it is worth maintaining multiple versions of the same thing. Consider again the function f (x) =
625 − x^2. We have computed the derivative
f ′(x) = −x/
625 − x^2 , and have already noted that if we use the alternate notation
y =
625 − x^2 then we might write y′^ = −x/
625 − x^2. Another notation is quite different, and in time it will become clear why it is often a useful one. Recall that to compute the the derivative of f we computed
∆limx→ 0
625 − (7 + ∆x)^2 − 24 ∆x.
The denominator here measures a distance in the x direction, sometimes called the “run”, and the numerator measures a distance in the y direction, sometimes called the “rise,” and “rise over run” is the slope of a line. Recall that sometimes such a numerator is abbreviated ∆y, exchanging brevity for a more detailed expression. So in general, a derivative is given by
y′^ = (^) ∆limx→ 0 ∆y ∆x
To recall the form of the limit, we sometimes say instead that
dy dx =^ ∆limx→ 0
∆y ∆x.
In other words, dy/dx is another notation for the derivative, and it reminds us that it is related to an actual slope between two points. This notation is called Leibniz notation, after Gottfried Leibniz, who developed the fundamentals of calculus independently, at about the same time that Isaac Newton did. Again, since we often use f and f (x) to mean the original function, we sometimes use df /dx and df (x)/dx to refer to the derivative. If
48 Chapter 2 Instantaneous Rate of Change: The Derivative
the function f (x) is written out in full we often write the last of these something like this
f ′(x) = (^) dxd
625 − x^2
with the function written to the side, instead of trying to fit it into the numerator.
EXAMPLE 2.4.2 Find the derivative of y = f (t) = t^2. We compute y′^ = (^) ∆limt→ 0 ∆ ∆yt = (^) ∆limt→ 0 (t^ + ∆t)
(^2) − t 2 ∆t
= (^) ∆limt→ 0 t
(^2) + 2t∆t + ∆t (^2) − t 2 ∆t
= (^) ∆limt→ 02 t∆t^ + ∆t
2 ∆t = (^) ∆limt→ 0 2 t + ∆t = 2t.
Remember that ∆t is a single quantity, not a “∆” times a “t”, and so ∆t^2 is (∆t)^2 not ∆(t^2 ).
EXAMPLE 2.4.3 Find the derivative of y = f (x) = 1/x. The computation:
y′^ = (^) ∆limx→ 0 ∆y ∆x
= (^) ∆limx→ 0
1 x+∆x −^
1 x ∆x
= (^) ∆limx→ 0
x x(x+∆x) −^
x+∆x x(x+∆x) ∆x
= (^) ∆limx→ 0
x−(x+∆x) x(x+∆x) ∆x = (^) ∆limx→ 0 x^ x(x^ − + ∆^ x^ −x^ ∆)∆xx
= (^) ∆limx→ (^0) x(x + ∆^ −∆xx)∆x
= (^) ∆limx→ 0 x(x −+ ∆^1 x) = − x^12
Note. If you happen to know some “derivative formulas” from an earlier course, for the time being you should pretend that you do not know them. In examples like the ones above and the exercises below, you are required to know how to find the derivative
