2 Marco teórico y conceptual
2.1 Tipologías y funciones de los museos
The previous section gave a rather vague definition of the tangent to a curve at a point. All we said is that it is a straight line which just touches the curve at that point. We now give a more formal definition of the same concept.
y x P T Q y x P T Q Figure 1 Figure 2
The geometrical idea behind the definition is easy to understand. Consider a point P on a curve in the xy-plane (see Fig. 1). Take another point Q on the curve. The entire straight line through P and Q is called a secant. If we keep P fixed, but let Q move along the curve toward P , then the secant will rotate around P , as indicated in Fig. 2. The limiting straight line P T toward which the secant tends is called the tangent (line) to the curve at P . Suppose that the curve in Figs. 1 and 2 is the graph of a function f . The approach illustrated in Fig. 2 allows us to find the slope of the tangent P T to the graph of f at the point P .
y x T Q ! (a " h, f (a " h)) P ! (a, f (a)) f(a " h) # f (a) h f Figure 3
Figure 3 reproduces the curve, the points P and Q, and the tangent P T in Fig. 2. Point P in Fig. 3 has the coordinates!a, f (a)". Point Q lies close to P and is also on the graph of f. Suppose that the x-coordinate of Q is a+ h, where h is a small number ̸= 0. Then the x-coordinate of Q is not a (because Q̸= P ), but a number close to a. Because Q lies on the graph of f , the y-coordinate of Q is equal to f (a+ h). Hence, the point Q has coordinates !
a+ h, f (a + h)". The slope mPQof the secant P Q is therefore
mPQ=
f (a+ h) − f (a) h
This fraction is often called a Newton quotient of f . Note that when h= 0, the fraction becomes 0/0 and so is undefined. But choosing h= 0 corresponds to letting Q = P . When Qmoves toward P along the graph of f , the x-coordinate of Q, which is a+ h, must tend to a, and so h tends to 0. Simultaneously, the secant P Q tends to the tangent to the graph at P . This suggests that we ought to define the slope of the tangent at P as the number that mPQapproaches as h tends to 0. In the previous section we called this number f′(a). So
we propose the following definition of f′(a):
f′(a)= # the limit as h tends to 0 of $ f (a+ h) − f (a) h It is common to use the abbreviated notation limh→0, or lim
h→0, for “the limit as h tends to
zero” of an expression involving h. We therefore have the following definition:
D E F I N I T I O N O F T H E D E R I V A T I V E
The derivative of the function f at point a, denoted by f′(a), is given by the
formula f′(a)= lim h→0 f (a+ h) − f (a) h (1)
The number f′(a)gives the slope of the tangent to the curve y= f (x) at the point!a, f (a)".
The equation for a straight line passing through (x1, y1)and having a slope b is given by
y− y1 = b(x − x1). Hence, we obtain:
D E F I N I T I O N O F T H E T A N G E N T
The equation for the tangent to the graph of y= f (x) at the point!a, f (a)"is y− f (a) = f′(a)(x− a)
(2)
So far the concept of a limit in the definition of f′(a)is mathematically somewhat imprecise.
Section 6.5 discusses the concept of limit in more detail. Because it is relatively complicated, we rely on intuition for the time being. Consider a simple example.
E X A M P L E 1 Use (1) to compute f′(a)when f (x)= x2.Find in particular f′(1/2) and f′(−1). Give
geometric interpretations, and find the equation for the tangent at the point (1/2, 1/4).
Solution: For f (x) = x2, we have f (a + h) = (a + h)2 = a2 + 2ah + h2, and so
f (a+ h) − f (a) = (a2+ 2ah + h2)− a2= 2ah + h2. Hence, for all h̸= 0, we obtain f (a+ h) − f (a) h = 2ah+ h2 h = h(2a+ h) h = 2a + h (∗)
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because we can cancel h whenever h̸= 0. But as h tends to 0, so 2a + h obviously tends to 2a. Thus, we can write
f′(a)= lim
h→0
f (a+ h) − f (a)
h = limh→0(2a+ h) = 2a
This shows that when f (x) = x2, then f′(a) = 2a. For a = 1/2, we obtain f′(1/2)=
2· 1/2 = 1. Similarly, f′(−1) = 2 · (−1) = −2.
Figure 4 provides a geometric interpretation of (∗). In Fig. 5, we have drawn the tangents to the curve y = x2 corresponding to a = 1/2 and a = −1. At a = 1/2, we have
f (a)= (1/2)2 = 1/4 and f′(1/2)= 1. According to (2), the equation of the tangent is y− 1/4 = 1 · (x − 1/2) or y = x − 1/4. (Show that the other tangent drawn in Fig. 5 has the equation y= −2x − 1.) Note that the formula f′(a)= 2a shows that f′(a) <0 when
a <0, and f′(a) >0 when a > 0. Does this agree with the graph?
a h (a " h)2 # a2 ! 2ah " h2 a " h P Q y x f(x) ! x2 4 3 2 1 #1 #2 #1 1 2 f(x) ! x2 y x Figure 4 f (x)= x2 Figure 5 f (x)= x2
If f is a relatively simple function, we can find f′(a)by using the following recipe:
R E C I P E F O R C O M P U T I N G f ’ ( a )
(A) Add h to a and compute f (a+ h).
(B) Compute the corresponding change in the function value: f (a+ h) − f (a). (C) For h̸= 0, form the Newton quotient f (a+ h) − f (a)
h .
(D) Simplify the fraction in step (C) as much as possible. Wherever possible, cancel h from the numerator and denominator.
(E) Then f′(a)is the limit of f (a+ h) − f (a)
h as h tends to 0.
(3)
Let us apply this recipe to another example.
Solution: We follow the recipe in (3).
(A) f (a+ h) = (a + h)3 = a3+ 3a2h+ 3ah2+ h3
(B) f (a+ h) − f (a) = (a3+ 3a2h+ 3ah2+ h3)− a3 = 3a2h+ 3ah2+ h3
(C)–(D) f (a+ h) − f (a)
h =
3a2h+ 3ah2+ h3
h = 3a
2+ 3ah + h2
(E) As h tends to 0, so 3ah+h2also tends to 0; hence, the entire expression 3a2+3ah+h2
tends to 3a2. Therefore, f′(a)= 3a2.
We have thus shown that the graph of the function f (x) = x3 at the point x = a has a
tangent with slope 3a2. Note that f′(a) = 3a2 > 0 when a ̸= 0, and f′(0) = 0. The
tangent points upwards to the right for all a̸= 0, and is horizontal at the origin. You should look at the graph of f (x)= x3in Fig. 4.3.7 to confirm this behaviour.
The recipe in (3) works well for simple functions like those in Examples 1 and 2. But for more complicated functions such as f (x)=√3x2+ x + 1 it is unnecessarily cumber-
some. The powerful rules explained in Sections 6.6–6.8 allow the derivatives of even quite complicated functions to be found quite easily. Understanding these rules, however, relies on the more precise concept of limits that we will provide in Section 6.5.
On Notation
We showed in Example 1 that if f (x) = x2, then for every a we have f′(a) = 2a. We
frequently use x as the symbol for a quantity that can take any value, so we write f′(x)= 2x.
Using this notation, our results from the two last examples are as follows:
f (x)= x2 '⇒ f′(x)= 2x (4)
f (x)= x3 '⇒ f′(x)= 3x2 (5)
The result in (4) is a special case of the following rule, which you are asked to show in Problem 6.
f (x)= ax2+ bx + c '⇒ f′(x)= 2ax + b (a, b, and c are constants) (6) For a= 1, b = c = 0, this reduces to (4). Here are some other special cases of (6):
f (x)= 3x2+ 2x + 5 '⇒ f′(x)= 2 · 3x + 2 = 6x + 2 f (x)= −16 +12x−161x2 '⇒ f′(x)= 12 −18x
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If we use y to denote the typical value of the function y = f (x), we often denote the derivative simply by y′. We can then write y= x3 '⇒ y′= 3x2.
Several other forms of notation for the derivative are often used in mathematics and its applications. One of them, originally due to Leibniz, is called the differential notation. If y= f (x), then in place of f′(x), we write
dy dx = dy/dx or df (x) dx = df (x)/dx or d dxf (x)
For instance, if y= x2, then
dy
dx = 2x or d dx(x
2)= 2x
We can think of the symbol d/dx as an instruction to differentiate what follows with respect to x. Differentiation occurs so often in mathematics that it has become standard to use w.r.t. as an abbreviation for “with respect to”. At this point, we will only think of the symbol dy/dx as meaning f′(x)and will not consider how it might relate to dy divided by dx. Later chapters discuss this notation in greater detail.
When we use letters other than f , x, and y, the notation for the derivative changes accordingly. For example:
P (t )= t2 '⇒ P′(t )= 2t; Y = K3'⇒ Y′= 3K2; and A = r2'⇒ dA/dr = 2r
P R O B L E M S F O R S E C T I O N 6 . 2
1. Let f (x)= 4x2. Show that f (5+ h) − f (5) = 40h + 4h2. Hence, f (5+ h) − f (5)
h = 40 + 4h
Using this result, find f′(5). Compare the answer with (6).
2. (a) Let f (x)= 3x2+ 2x − 1. Show that for h ̸= 0, f (x+ h) − f (x)
h = 6x + 2 + 3h
Use this result to find f′(x).
(b) Find in particular f′(0), f′(−2), and f′(3). Find also the equation of the tangent to the graph at the point (0,−1).
3. (a) The demand function for a commodity with price P is given by the formula D(P )= a−bP . Find dD(P )/dP .
(b) The cost of producing x units of a commodity is given by the formula C(x)= p + qx2. Find C′(x), the marginal cost (see Section 6.4).
4. Show that f (x)= 1 x = x −1 '⇒ f′(x)= −1 x2 = −x −2 (Hint: Show that [f (x+ h) − f (x)]/h = −1/x(x + h).)
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5. In each case, find the slope of the tangent to the graph of f at the specified point:(a) f (x)= 3x + 2 at (0, 2) (b) f (x)= x2− 1 at (1, 0) (c) f (x)= 2 + 3/x at (3, 3) (d) f (x)= x3− 2x at (0, 0) (e) f (x)= x + 1/x at (−1, −2) (f) f (x)= x4 at (1, 1)
6. (a) If f (x)= ax2+ bx + c, show that [f (x + h) − f (x)]/h = 2ax + b + ah. Use this to show that f′(x)= 2ax + b.
(b) For what value of x is f′(x)= 0? Explain this result in the light of (4.6.3) and (4.6.4).
7. The figure shows the graph of a function f . Determine the sign of the derivative f′(x)at each of the four points a, b, c, and d.
y ! f (x) y x a b c d Figure 6
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8. (a) Show that!√x+ h −√x"!√x+ h +√x"= h.(b) If f (x)=√x, show that [f (x+ h) − f (x)]/h = 1/!√x+ h +√x". (c) Use the result in part (b) to show that for x > 0,
f (x)=√x '⇒ f′(x)= 1 2√x =
1 2x−1/2
9. (a) If f (x)= ax3+ bx2+ cx + d, show that f (x+ h) − f (x)
h = 3ax
2+ 2bx + c + 3axh + ah2+ bh and hence find f′(x).
(b) Show that the result in part (a) generalizes Example 2 and Problem 6(a).
HARDER PROBLEM
10. Show that
f (x)= x1/3 '⇒ f′(x)=13x−2/3
(Hint: Prove that [(x+ h)1/3− x1/3][(x+ h)2/3+ (x + h)1/3x1/3+ x2/3]= h. Then follow the argument used to solve Problem 8.)
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