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Chapter 2
The Derivative
2.1 The Derivative and Slope of the Tangent Line
- The Slope of the Tangent Line • The Derivative of a Function • Cases When the Derivative is Undefined • Differentiable Functions are Continuous Func- tions
The moving power of mathematical invention is not reasoning but imagination.
The Slope of the Tangent Line Mathematics advances whenever a solution to a long standing mathematical problem is found. Such solutions are groundbreaking that they bring forth new ideas and techniques. In the 18th century, Newton and Leibniz invented the theory of differential calculus in solving the tangent line problem. In this section we describe the slope of a tangent line.
P
Q
c c x x
f c
f c x
y
Figure 2 The secant line through P (c, f (c)) and Q(c + ∆x, f (c + ∆x)).
An example of a tangent line to the graph of a function f at a point P (c, f (c)) is shown in Figure 1. In order to define the slope of a tangent line, we consider the slopes of secant lines.
P
c x
f c
y
Figure 1 Tangent line at point P (c, f (c)).
Let Q(c + ∆x, f (c + ∆x)) be another point on the graph of y = f (x) where ∆x = 0, see Figure 2. Since the slope of the line through the points (x 1 , y 1 ) and (x 2 , y 2 ) is given by y 2 − y (^1) x 2 − x (^1)
72 CHAPTER 2. THE DERIVATIVE
we obtain by substitution that the slope of the secant line joining P (c, f (c)) and Q(c + ∆x, f (c + ∆x)) is
msecant line =
f (c + ∆x) − f (c) (c + ∆x) − c
=
f (c + ∆x) − f (c) ∆x
Note, when Q(c + ∆x, f (c + ∆x)) is very near P (c, f (c)), the position of the secant line is almost in the same position as the tangent line. In other words, when ∆x is close to but not equal 0, the slope of the secant line joining P (c, f (c)) and Q(c + ∆x, f (c + ∆x)) is close to the slope of the tangent line at P (c, f (c)). This idea motivates the following definition.
Definition 1 The Slope of the Tangent Line
The slope of the tangent line to the graph of y = f (x) at the point (c, f (c)) is
lim ∆x→ 0
f (c + ∆x) − f (c) ∆x
provided the limit exists.
The slope of the tangent line at (c, f (c)) is also called the slope of the graph of y = f (x) at (c, f (c)).
Example 1 The Slope of the Graph of a Linear Function
Find the slope of the graph of f (x) = 4x − 2 at the point (3, 10).
Solution Using Definition 1 with c = 3, the slope of the graph at (3, 10) is
lim ∆x→ 0
f (3 + ∆x) − f (3) ∆x
= lim ∆x→ 0
(4(3 + ∆x) − 2) − (4(3) − 2) ∆x
= (^) ∆limx→ 0
12 + 4∆x − 2 − 10 ∆x = lim ∆x→ 0
4 ∆x ∆x
Then the slope of the graph at (3, 10) is 4. ✷
Try This 1 Find the slope of the graph of g(x) = − 3 x + 12 at the point (2, 6).
Note, the slope of the graph of a linear function is the same at every point in the line. For nonlinear functions, this is not the case, i.e., the slope of the tangent line may vary from one point to another point. This is shown in the next example.
74 CHAPTER 2. THE DERIVATIVE
Try This 3 Find the slope of the tangent line to the graph of the given function at the indicated point.
a) g(x) =
x
b) h(x) =
x + 1
In the previous three examples, all the tangent lines had slopes. However, it is possible that a tangent line may not have a slope 1. Such is the case when we have a vertical tangent line.
Definition 2 Definition of a Vertical Tangent Line
If f (x) is continuous at x = c and either
∆limx→ 0
f (c + ∆x) − f (c) ∆x
= ∞ or (^) ∆limx→ 0
f (c + ∆x) − f (c) ∆x
we say the graph of y = f (x) has a vertical tangent line at the point (c, f (c)).
In Figure 4, we see a vertical tangent line at the point (c, f (c)). Note, for ∆x = 0, the slope of the secant line joining the points (c, f (c)) and Q(c + ∆x, f (c + ∆x)) is positive. As ∆x approaches 0, the position of the secant line is moving towards the vertical position
. Thus, the slope of the secant line is increasing without bound as ∆x approaches 0, i.e.,
∆limx→ 0
f (c + ∆x) − f (c) ∆x
P
Q
c x
f c
y
Figure 4 A vertical tangent line at (c, f (c)).
(^1) It is also possible that the tangent line may not exist at all.
2.1. THE DERIVATIVE AND SLOPE OF THE TANGENT LINE 75
P Q
x
y
Figure 5
The graph of f (x) = 1 − 3
√ x − 1
has a vertical tangent line at P (1, 1).
Example 4 Verifying the Existence of a Vertical Tangent Line
Prove x = 1 is a vertical tangent line of the graph of
f (x) = 1 − 3
x − 1.
Solution We apply Definition 2 with c = 1:
lim ∆x→ 0
f (1 + ∆x) − f (1) ∆x
= lim ∆x→ 0
1 + ∆x − 1
∆x
= lim ∆x→ 0
∆x ∆x
= lim ∆x→ 0
(^3) (∆x) 2
Note, f is continuous at x = 1 by Theorem 1.2, page 24. Thus, by Definition 2, x = 1 is vertical tangent line. Finally, in Figure 5 we see that the slope of a secant line joining P (1, 1) and Q(1 + ∆x, f (1 + ∆x)) is negative. This explains why the above limit is −∞, and not ∞. ✷
Try This 4 Show that the graph of f (x) = 3
x has a vertical tangent line at the point (0, 0) using Definition 2.
The Derivative of a Function
We begin an important phase in studying of calculus. The limit that we used to define the slope of a tangent line will now be used to define the derivative of a function. The process of finding the derivative of a function is called differentiation.
Definition 3 The Derivative of a Function
The derivative of a function f is a function f ^ whose value at x is given by
f ^ (x) = lim ∆x→ 0
f (x + ∆x) − f (x) ∆x
provided the limit exists.
2.1. THE DERIVATIVE AND SLOPE OF THE TANGENT LINE 77
Example 6 Derivatives and Tangent Lines
Given y =
x, find the following:
a)
dy dx b) the slope of the graph of y =
x at the point (4, 2), and
c) an equation of the tangent line at (4, 2).
Problem-Solving Tips
To simplify a difference quotient ∆y/∆x involving rad- icals, try rationalizing.
x
y
Figure 6.
Since dy/dx = 1/(
√ x), the slope at (4, 2) is m = 1/(
√
- = 1/4.
Solution
a) Applying Definition 3, we find
dy dx
= lim x→ 0
x + x −
x x
= lim x→ 0
x + x −
x x
x + x +
√ x x + x +
x
Rationalize
= lim x→ 0
(x + x) − x x
x + x +
x
= (^) limx→ 0
√^1
x + x +
x
Simplify
dy dx
x
. Substitute x = 0
b) The slope m at (4, 2) is obtained by substituting x = 4 into the derivative.
Slope = m = 1 2
=^1
c) Using the slope-intercept equation of the tangent line, we find
y = mx + b Slope-intercept equation
(4) + b x = 4, y = 2, m =
1 = b.
Thus, the tangent line is given by
y =
x + 1.
✷
Try This 6 If f (x) =
2 x, find f ^ (x) and the slope-intercept equation of the tangent line to the graph of f at (8, 4).
78 CHAPTER 2. THE DERIVATIVE
An alternative definition of the derivative is given by
f ^ (c) = lim x→c
f (x) − f (c) x − c
provided the limit exists. The verification that the alternate definition (2) agrees with Definition 3 is left as an exercise at the end of the section. The one-sided limits
lim x→c +
f (x) − f (c) x − c
and lim x→c −
f (x) − f (c) x − c
are called the derivative from the right and derivative from the left at c, respec- tively, if they exist. Moreover, we say f (x) is differerentiable on a closed interval [a, b] if f (x) is differentiable on (a, b), f (x) has a derivative from the right at a, and f (x) has a derivative from the left at b.
Problem-Solving Tips To simplify an improper fraction, multiply and divide by the least common denominator.
Example 7 The Derivative of a Rational Function
Find the derivative f ^ (x) of f (x) =^1 x
using the alternative definition (2).
x
y
Figure 7
The graph of f (x) =
1 x
.
x
y
Figure 8
The graph of f ^ (x) = − 1 x 2
.
Solution Applying the alternative definition (2), we find
f ^ (c) = (^) xlim→c
f (x) − f (c) x − c
= lim x→c
x
c x − c
= (^) xlim→c
x
c x − c
xc xc
Multiply and divide by xc
= lim x→c
c − x xc(x − c)
= (^) xlim→c
−(x − c) xc(x − c)
Rewrite
= lim x→c
xc
Simplify
f ^ (c) = −
c 2
Substitute c into x
Rewriting, the derivative is f ^ (x) = − 1 x 2
See the graphs of y = f (x) and y = f ^ (x) in Figures 7 and 8. ✷
Try This 7
Find the derivative of g(x) = −
x
using the alternative definition (2).
Cases When the Derivative is Undefined If the slopes of the graph abruptly change as the values of x passes through c, we say the graph of y = f (x) has a ‘sharp turn’ or ‘cusp’ at (c, f (c)). In such a case , f ^ (c) is undefined. For instance, see Figure 9. Another case when f ^ (c) fails to exist is when the line x = c is a vertical tangent line to the graph of y = f (x), see Figure 10.
80 CHAPTER 2. THE DERIVATIVE
Try This 9
Verify f (x) = 3
x + 1 is not differentiable at x = −1.
Differentiable Functions are Continuous Functions
If a function is continuous at a point, the function is not necessarily differentiable at the point. Examples of such functions is seen in Figure 11, and the previous two examples. However, if a function is differentiable at a point then it is continuous at the point.
Theorem 2.1 Differentiability Implies Continuity
If a function f (x) is differentiable at x = c, then f (x) is continuous at x = c
1 1 2 3 x
1
3
2
y
Figure 11 f (x) = [[x]] is not continuous at
x = 0, ± 1 , ± 2 , ...
and thus not differentiable at the same values of x.
Proof To prove f (x) is continuous at x = c, we must verify
lim x→c
f (x) = f (c)
or equivalently to show hlim→ 0 f^ (c^ +^ h) =^ f^ (c). Let h = 0 and consider the identity
f (c + h) = f (c) +
f (c + h) − f (c) h
· h
Applying limits and the definition of the derivative, we find
hlim→ 0 f^ (c^ +^ h)^ =^ hlim→ 0 f^ (c) + lim h→ 0
f (c + h) − f (c) h
· (^) hlim→ 0 h
= f (c) + f ^ (c) · 0 lim h→ 0
f (c + h) = f (c).
Thus, f (x) is continuous at x = c. ✷
Theorem 2.1 implies that if a function is not continuous at x = c, then it not differen- tiable at x = c. For instance, the greatest integer function f (x) = [[x]] is not continuous at integer values of x, see Figure 11. Then f ^ (x) is undefined if x is an integer.
2.1 Check-It Out
- Find the slope of the graph of f (x) = x 2 at the point (3, 9).
- Find the derivative f (x) = 5x 2.
2.1. THE DERIVATIVE AND SLOPE OF THE TANGENT LINE 81
True or False. If false, explain or rewrite the statement to make it true.
- If f (x) = x 2 , then f ^ (x) = lim ∆x→ 0
(x + ∆x) 2 − x 2 ∆x
- If f (x) = 3
x, then f ^ (4) = (^) ∆limx→ 0
4 + ∆x − 6 ∆x
- If f (x) = x 2 and g(x) = x 2 − 1, then f ^ (x) = g ^ (x).
- The tangent line to the graph of f (x) = x 2 at (0, 0) has slope 0.
- At every point on the graph of f (x) = mx + b the slope of the tangent line is m.
- If f ^ (c) exists, then lim x→c f (x) = f (c).
- The graph of f (x) = 1/x has a vertical tangent line.
- The tangent line to the graph of f (x) = x 3 at (2, 8) is y = 12x − 16.
- If f (x) =
x, then f ^ (0) = 0 10. If g(x) = |x|, then g ^ (x) = 1.
Exercises for Section 2.
In Exercises 1-16, find the slope of the tangent line at the given point. Apply the limit process of Definition 1.
- f (x) = 10x − 4, (2, 16) 2. f (x) = 4 +
3 x 7
, (− 7 , 1) 3. f (x) = 2x 2 + 3, (− 2 , 11)
- f (x) = 4 − x 2 , (− 3 , −5) 5. f (x) = 4x − x 2 , (2, 4) 6. f (x) = 3x 2 − 12 x + 1, (2, −11)
- g(t) =
t + 3, (1, 2) 8. g(t) =
t − 5, (9, 2) 9. g(t) =
4 t + 1, (2, 3)
- g(t) =
4 t − 3 − 1, (1, 0) 11. g(t) = t(2 − t 2 ), (− 2 , 4) 12. h(t) = t(t 2 − 3), (2, 2)
- h(x) =
x
, (1, 3) 14. h(x) =
7 x
- h(x) = √−^4 x
, (4, −2) 16. h(x) = 3
x − 5, (− 3 , −2)
In Exercises 17-20, estimate the slope of the tangent line at the indicated point (x, y).
x,y
x,y x,y
x,y
Figure for No. 17 Figure for No. 18 Figure for No. 19 Figure for No. 20
In Exercises 21-34, find the derivative by applying the limit process in Definition 3.
- f (x) = x 22. f (x) = 5x 23. y = 3x − 4 24. y =
x + 8
- f (x) = 3x 2 + 5 26. f (x) = x 2 − 6 x 27. g(t) =
t + 4
- g(t) =
t + 1
- m(x) =
x + 4 30. m(x) = 2
x − 1 31. f (x) =
3 x 32. f (x) = 4
2 x
- f (x) = 2x 3 34. f (x) = x 4
2.1. THE DERIVATIVE AND SLOPE OF THE TANGENT LINE 83
- Identify or sketch the following.
a) f (1) and f (3) b) f (1) − f (3) c) y = f^ (1)^ −^ f^ (3) 1 − 3
(x − 3) + f (3)
- Which is greater?
a)
f (2) − f (1) 2 − 1
or
f (1) − f (0) 1 − 0
b)
f (1) − f (3) 1 − 3
or
f (2) − f (3) 2 − 3
- Sketch the graph of y = f (x) in each of the following cases: a) f ^ (x) is always positive b) f ^ (x) is always negative
- Sketch the graph of y = g(x) in each of the following cases: a) g ^ (x) is positive on (−∞, 0), and g ^ (x) is negative on (0, ∞) b) g ^ (x) is positive on (0, ∞), and g ^ (x) is negative on (−∞, 0)
- Use the given graph of y = g(x) to answer the following:
B
A
C D
E
x
y
Figure for No. 57
a) Between which two consecutive points is the average rate of change of g(x) the least? b) Is the derivative of g(x) at C greater than or less than the average rate of change of g(x) between B and C? c) Sketch a tangent line to the graph of y = g(x) between A and B such that the slope of the line is the same as the slope of the line segment AB.
- If lim ∆x→ 0
f (c + ∆x) − f (c) ∆x
exists, use the ε-δ definition to prove that lim x→c
f (x) − f (c) x − c exists and equals the former limit. This exercise shows that the alternative definition of the derivative in (2) agrees with Definition 3.
Odd Ball Problems
- Determine if f (x) = x|x| is differentiable at x = 0.
- Determine if f (x) = sin |x| is differentiable at x = 0.
- Find the values of a, b, c if the graph of f (x) = ax 3 + bx + c passes through (− 1 , 0), and y = x − 1 is the tangent to the graph at (1, 0).
- Show f ^ (0) exists if f (x) =
1 − cos x x
if x = 0
0 otherwise
- Consider the function given below.
f (x) =
3 + x 2 if x is a rational number
3 + x 4 otherwise
a) Prove f ^ (0) exists. b) Prove f continuous but not differentiable at x = ±1.
84 CHAPTER 2. THE DERIVATIVE
2.2 Basic Differentiation Formulas
- Derivative of a Constant Function • The Power Rule • The Constant Multiple Rule • Sum and Difference Rules • The Derivatives of the Sine and Cosine Functions • Rate of Change
Derivative of a Constant Function In this section, we develop rules for evaluating the derivative of several basic functions. These rules are specially practical when we need to find the derivative without using the limit process of Definition 3.
f x c
x
f x
Figure 1
The derivative or slope of the constant function is 0: d dx
[c] = 0
for each real number c.
Theorem 2.2 The Derivative of a Constant Function
For any constant c, we have d dx
[c] = 0.
Proof Let f (x) = c be a constant function. Then the derivative is
d dx
[c] = f ^ (c)
= lim ∆x→ 0
f (c + ∆x) − f (c) ∆x
= lim ∆x→ 0
c − c ∆x
Since f (c + ∆x) = c
d dx
[c] = (^) ∆limx→ 0 0 = 0
✷
In Figure 1, we see the graph of a constant function. Since the slope of a horizontal line is 0, the derivative of a constant function is the zero function.
Example 1 When the Derivative is the Zero Function
The following is a list of constant functions and their derivatives.
Function Derivative
a) y = 16
dy dx
b) a(t) = − 32 a ^ (t) = 0
c) f (x) = π
d dx
[f (x)] = 0
✷
Try This 1 Find the derivative of each function.
a) p(x) = 2 − π b) a(t) = − 9. 8
86 CHAPTER 2. THE DERIVATIVE
In Example 2b), the derivative f ^ (x) = 1 is the slope of the linear function f (x) = x. We see the graphs of f (x) = x and its derivative in Figure 2. In Examples 2c) and d), we had to rewrite y = 1/x 5 and y =
x before applying the Power Rule. ✷
Try This 2 Find the derivative of the function.
a) f (x) = x 5 b) g(x) =
x 2
c) y =
√^1
x
d) p(t) = t
f x x^2
1 , 1
2 , 4
0 , 0
m 2
m 4
m 0
x
y
Figure 3.
The slope m of the graph at (c, f (c)) is m = f ^ (c).
Example 3 The Slopes of a Graph
Find the slope of the graph of f (x) = x 2 at the given value of x.
a) x = − 2 b) x = 0 c) x = 1
Solution Applying the Power Rule to f (x) = x 2 , we obtain
f ^ (x) = 2x.
a) If x = −2, the slope of the graph is f ^ (−2) = 2(−2) = −4. b) If x = 0, the slope of the graph is f ^ (0) = 2(0) = 0. c) If x = 1, the slope of the graph is f ^ (1) = 2(1) = 2. In Figure 3, we see the tangent lines with their slopes. ✷
Try This 3 Find the slope of the graph of y = g(x) at the indicated value of x.
a) g(x) = x 2 , x = 3 b) g(x) =
x 2
, x = 1 c) g(x) = 3
x, x = 8
x
y
Figure 4
At the point (2, 16) on the graph of f (x) = x 4 the tangent line is y = 32x − 48.
Example 4 Finding an Equation of a Tangent Line
Find the slope-intercept form of the equation of the tangent line to the graph of f (x) = x 4 at the point (2, 16), see Figure 3.
Solution To find the slope of the tangent line at the point (2, 16), we substitute x = 2 into the derivative f ^ (x) = 4x 3. Then the slope is
m = f ^ (2) = 4(2) 3 = 32.
Next, substitute the slope m = 32 and the coordinates of (2, 16) into the slope-intercept form of the equation of the line.
y = mx + b 16 = 32(2) + b − 48 = b
Hence, the slope-intercept equation of the tangent line is
y = 32x − 48.
✷
2.2. BASIC DIFFERENTIATION FORMULAS 87
Try This 4 Find the slope-intercept equation of the tangent line to the graph of f (x) = x 3 at the point (2, 8).
The Constant Multiple Rule
Theorem 2.4 The Constant Multiple Rule
If f (x) is a differentiable function and c is any constant, then cf (x) is differentiable and
d dx
[cf (x)] = c
d dx
[f (x)].
Proof Applying Definition 3 in Section 2.1, we find
d dx
[cf (x)] = (^) ∆limx→ 0
cf (x + ∆x) − cf (x) ∆x
= lim ∆x→ 0
c
f (x + ∆x) − f (x) ∆x
= c lim ∆x→ 0
f (x + ∆x) − f (x) ∆x d dx
[cf (x)] = c
d dx
[f (x)]
This proves the theorem. ✷
By combining the Power Rule and the Constant Multiple Rule, we obtain the following theorem.
Theorem 2.5 The Combination Rule
If c is any constant, then d dx
[cx n^ ] = cnx n−^1.
In using the combination rule, we may have to rewrite a function in the form y = cx n^. This way, the coefficient c and exponent n are explicit.
2.2. BASIC DIFFERENTIATION FORMULAS 89
d dx
[f (x) − g(x)] = d dx
[f (x) + (−1)g(x)]
d dx
[f (x)] +
d dx
[(−1)g(x)] Sum Rule
d dx
[f (x)] + (−1)
d dx
[g(x)] Constant Multiple Rule
d dx
[f (x) − g(x)] = f ^ (x) − g ^ (x)
✷
The Sum and Difference Rules extend to sums and differences of any finite number of differentiable functions, by repeated applications of Theorem 2.6.
Example 6 Applying the Sum and Difference Rules
We list certain functions and evaluate their derivatives.
Problem-Solving Tips
Before differentiating, rewrite the function.
Function Derivative
a) f (x) = 8x 3 − 12 f^
(^) (x) = d dx
8 x 3
− d dx
[12]
= 24 x 2
b) y =
x 2
9 x 4
dy dx
= d dx
3 x −^2
x 1
Rewrite
x 3
Combination Rule, simplify
✷
Try This 6 Find the derivative of each function.
a) f (x) = 7x 6 − 32 x + 1 b) g(x) = 4 x 7
+^8 x^
2 3
− 5 x
The Derivatives of the Sine and Cosine Functions
We recall the following limits
lim ∆x→ 0
sin ∆x ∆x
= 1 and lim ∆x→ 0
1 − cos ∆x ∆x
For reference, see Theorem 1.7 and Example 8 in Section 1.3. The limits (3) are essential in the proof of the next theorem.
90 CHAPTER 2. THE DERIVATIVE
Theorem 2.7 The Derivatives of the Sine and Cosine Functions
a)
d dx
[sin x] = cos x b)
d dx
[cos x] = − sin x
Proof Applying Definition 3 in Section 2.1 and the sum formula
sin(α + β) = sin α cos β + cos α sin β
we obtain
d dx
[sin x] = lim ∆x→ 0
sin(x + ∆x) − sin x ∆x = lim ∆x→ 0
sin x cos ∆x + cos x sin ∆x − sin x ∆x
= (^) ∆limx→ 0
cos x sin ∆x − sin x (1 − cos ∆x) ∆x d dx
[sin x] = lim ∆x→ 0
cos x
sin ∆x ∆x
− sin x
1 − cos ∆x ∆x
If we fix the value of x in the limit process as ∆x approaches 0, then sin x and cos x are constants. Since the limits in (3) exist, we rewrite the previous equation as follows:
d dx
[sin x] = cos x lim ∆x→ 0
sin ∆x ∆x
− sin x lim ∆x→ 0
1 − cos ∆x ∆x = cos x · 1 − sin x · 0 d dx
[sin x] = cos x
This proves part a) of the Theorem. The proof of part b) is assigned to Exercise 67 at the end of the section. ✷
Example 7 Using the Derivatives of Sine and Cosine
Evaluate the derivative of each function.
Function Derivative
a) f (x) = 3 sin x f^
(^) (x) = 3 d dx
[sin x] Constant Multiple Rule
f ^ (x) = 3 cos x Theorem 2.
b) y =
4 cos x 5
dy dx
d dx
[cos x] Constant Multiple Rule
sin x Theorem 2.
c) y = 7x − cos x
dy dx
d dx
[7x] −
d dx
[cos x] Difference Rule
= 7 − (− sin x) Theorem 2. dy dx
= 7 + sin x
✷
