We can use principles 1 and 2 given above to prove two properties of derivatives, the Constant Multiple Rule and the Sum Rule, stated in Section 7.4 and worked with in the Exploratory Pr
Trang 1Suppose limx→af (x) = L1and limx→ag(x) = L2, where L1and L2are finite Then: (1) lim
x→a[f (x) ± g(x)] = L1± L2 The limit of a sum/difference is the
sum/difference of the limits
(2) lim
x→a[f (x) · g(x)] = L1· L2 The limit of a product is the product of the
limits (in particular, g(x) may be constant: limx→akf (x) = kL1)
We can use principles (1) and (2) given above to prove two properties of derivatives, the Constant Multiple Rule and the Sum Rule, stated in Section 7.4 and worked with in the Exploratory Problem for Chapter 7
Properties of Derivatives
C o n s t a n t M u l t i p l e R u l e
d
dxkf (x) = k d
dxf (x) or (kf )(x) = kf(x), where k is any constant Multiplying f by a constant k multiplies its derivative by k
S u m R u l e
d
dx[f (x) + g(x)] = dxd f (x) + dxd g(x) or (f + g)(x) = f(x) + g(x) The derivative of a sum is the sum of the derivatives
These properties are natural from a real-world point of view
Sum Rule:. Suppose a juice manufacturer is packaging cranapple juice by pouring cran-berry and apple juice simultaneously into juice containers
Let c(t) be the number of liters of cranberry juice in a container at time t
Let a(t) be the number of liters of apple juice in a container at time t
Then c(t) + a(t) is the amount of juice in the container at time t
It follows thatda
dt anddc
dt are the rates at which apple and cranberry juice, respectively, are entering the container dtd[a(t) + c(t)] is the rate at which juice is entering
d
dt[a(t) + c(t)] =da
dt +dc dt
Constant Multiple Rule:. An athlete and a Sunday jogger start at the same place and run down a straight trail Let s(t) be the position of the Sunday jogger at time t If the position
of the athlete is always k times s(t), then her velocity is k times that of the jogger
d
dtks(t ) = kdsdt
Trang 2The Sum and Constant Multiple Rules are very important results and were the topic
of the exploratory problems for the previous chapter Proofs are provided at the end of the section We will assume in the chapters that follow that you have completed the exercises below asking you to prove these rules and make sense out of them
EXERCISE 8.5 Using the limit definition of derivative prove the two statements made above
EXERCISE 8.6 Argue that the two properties above make sense from a graphical perspective
EXERCISE 8.7 Differentiate the following
(a) y =2 + 3x
3+√5x
√x
√π x 7
Answers to these Exercises are provided at the end of the section.
The Product Rule: Differentiating f (x) · g(x)
From the simplicity of the Sum Rule and the Constant Multiple Rule one might initially hope that the derivative of a product is the product of the derivatives and similarly for quotients But this hope is quickly dashed by working through an example.2
Suppose, for instance, f (x) = x and g(x) = x2 Then f(x) = 1 and g(x) = 2x, so
f(x) · g(x) = 1 · 2x = 2x On the other hand, f (x) · g(x) = x · x2= x3; we know that (f (x) · g(x))should be 3x2, not 2x.
CAVEATThe derivative of a product is not the product of the derivatives.
However, if we know the derivatives of f and g, we can find the derivative of their product
in terms of f , g, f, and g We’ll figure out how to do this by working through the following example
EXAMPLE 8.4 A farmer growing a crop of tomatoes wants to know when he should harvest and sell them
in order to collect the most revenue Both the weight of the crop and the price of tomatoes are changing over time Let W (t) be the weight (in pounds) of his crop of tomatoes t days after the beginning of the harvest season, and let P (t) be the going price (in dollars per pound) for tomatoes on day t Then the revenue the farmer will receive for selling his crop
on day t will be given by (weight in pounds) · ($/lb):
R(t ) = W (t) · P (t)
Suppose that W (t) and P (t) are both differentiable functions As long as R(t ) >0, waiting
to harvest the crop will increase revenue We want to compute R(t )
Let W , P , and R be the change in weight, price, and revenue over the time interval
t There are two components contributing to the change in revenue:
W · P (the change in revenue due to increase in the weight of the tomatoes), and
W · P (the change in revenue due to increase in the price)
2 Leibniz himself wondered about this in an unpublished manuscript dated November 1675 and concluded that the rules were
Trang 3We are interested indRdt, the rate at which revenue is changing with respect to time.
dR
dt = lim
t →0
R
t The broad brushstroke of our argument looks like this
R = W · P + W · P Divide by t
R
t = W t · P + W · P t Take the limit as t → 0
lim
t →0
R
t = lim
t →0
W
t · P + lim
t →0W · P
t dR
dt =dW
dt · P + W ·dP
dt
We will fill in more of the details while looking at the geometry of the problem For simplicity’s sake, suppose that W and P are both increasing with time
W = W (t + t) − W (t) and P = P (t + t) − P (t)
R = R(t + t) − R(t)
R = W (t + t)P (t + t) − W (t)P (t) The drawings below will help us express R in terms of W and P
Revenue is represented by the area of a rectangle the length of whose sides represent the weight and price of tomatoes at a fixed time
W (t + ∆t)
R (t + ∆t) = W(t + ∆t) P(t + ∆t) R (t) = W(t) P(t)
P (t + ∆t)
∆ P
∆ P
}
P (t)
P (t)
W (t)
P (t)
Figure 8.7
R = R(t + t) − R(t)
=area of the entire large rectangle
−
area of the hatched rectangle
Rcan be represented by the three shaded rectangles shown in Figure 8.7(c)
Trang 4R = ( W )P (t) + ( P )W (t) + ( W )( P ) R
t =( W )P (t )
t +( P )W (t )
t +( W )( P )
t dR
dt = lim
t →0
R
t = lim
t →0
( W )P (t)
t +( P )W (t )
t +( W )( P )
t
= lim
t →0
( W )P (t )
t →0
( P )W (t )
t →0
( W )( P ) t Note that P (t) and W (t) are unaffected by letting t approach zero, so we obtain
dR
dt = P (t) lim
t →0
W
t + W (t) lim
t →0
P
t + lim
t →0
W
t P
By definition
lim
t →0
W
t =dWdt and lim
t →0
P
t =dPdt Therefore,
dR
dt = P (t)dWdt + W (t)dPdt +dWdt lim
t →0 P
As t → 0, we know that P → 0,3so the last term is actuallydWdt · 0 Thus,
dR
dt = P (t)dWdt + W (t)dPdt Another way of writing this is
R(t ) = P (t)W(t ) + W (t)P(t )
We have constructed an argument that for increasing functions f and g,
f (x)g(x)= f (x)g(x) + g(x)f(x)
This is known as the Product Rule We will show below that this holds for the product of
anytwo differentiable functions f and g Before doing this, let’s try an example
EXAMPLE 8.5 Find the derivative of each of the following functions using the Product Rule
i q(x) = x3= x · x2
ii h(x) =√7x(x2+ 3x + 2)
SOLUTION
i q(x) = x ·dxd x2 + x2
· dxd [x] = x · 2x + x2· 1 = 2x2+ x2= 3x2
ii h(x) =√7 ·√x(x2+ 3x + 2), so
h(x) =√7 ·2√1
x(x2+ 3x + 2) +√x(2x + 3)
3 We know that the change in price approaches 0 as t approaches 0, because P is continuous P must be continuous, because
Trang 5Proof of the Product Rule
Let j (x) = f (x)g(x), where f and g are differentiable functions
f (x)g(x) = j(x) = lim
h→0
j (x + h) − j (x) h
= lim h→0
f (x + h)g(x + h) − f (x)g(x)
h
We’d love to see a f (x + h)g(x + h) − g(x) in the numerator, but we’re missing the
−f (x + h)g(x) term However, −f (x + h)g(x) + f (x + h)g(x) = 0, so we can write
f (x)g(x) = lim
h→0
f (x + h)g(x + h) − f (x + h)g(x) + f (x + h)g(x) − f (x)g(x)
h
= lim h→0
f (x + h)g(x + h) − f (x + h)g(x)
h
= lim h→0
f (x + h)g(x + h) − g(x)
h + g(x)f (x + h) − f (x)h
= f (x)g(x) + g(x)f(x)
We’ve now proven the Product Rule
The Product Rule:f (x)g(x)= f (x)g(x) + g(x)f(x)
EXERCISE 8.8 Show that computing the derivative of kf (x) is a special case of the Product Rule In other
words, apply the Product Rule to kf (x) and show dxdkf (x) = kdfdx
The answer to this Exercise is provided at the end of the section.
The usefulness of the Product Rule will become increasingly apparent as the number
of types of functions we can differentiate increases Now we will see how the Product Rule enables us to differentiate a whole family of power functions in the blink of an eye, without having to return to the limit definition of the derivative
The Derivative of xn for n a Positive Integer
Mathematicians, like other scientists, look for patterns Upon observing a pattern, a math-ematician tries to generalize his observations and prove that the pattern holds true Let’s gather our results from applying the limit definition of derivative to power functions, functions of the form xnfor n a constant When looking for patterns, one must often play around with the form in which an expression is written Results from Examples and Exercises are gathered in the table on the following page and written in a form so that the pattern is readily accessible You can add results from other problems you have completed
Trang 6f (x) f(x)
x1 12x−1
x−12 −1
2 x−3
x−1 −x−2
Looking at this table we might conjecture that the derivative of xnis nxn−1 Below we
prove this conjecture for n any positive integer The method we will use is called mathe-matical induction; it works on the domino principle Suppose we arrange a collection of dominos in a line so that if any one domino falls, we are sure that the domino next to it will also fall Then, if we knock over the first domino in the line, all the rest of the dominos will also be knocked down
Proof by mathematical induction works similarly except that the dominos are the positive integers, so there are infinitely many of them
i First show that the statementdxdxn= nxn−1holds for n = 1
We show that the first domino will fall.
ii Then we show that if dxdxk
= kxk−1is true for some arbitrary k, where k is a positive
integer, then it must also be true for k + 1 In other words, we must show that if
d
dxxk= kxk−1, thendxdxk+1= (k + 1)xk This is called the inductive step.
We show that if any one domino falls, it will knock down the domino after it.
Once this is shown our proof is complete We will have shown that dxd xn= nxn−1holds true for n = 1, and that it therefore must hold true for n = 2; it holds true for n = 2, and therefore must hold true for n = 3, and so on, ad infinitum It must hold true for all positive integers
For more work with proof by induction, refer to Appendix D: Proof by Induction.
Proof Our hypothesis: d
dxxn= nxn−1
i Our hypothesis holds for n = 1
If f (x) = x, then f(x) = 1 = 1 · x0
ii Let’s suppose that our hypothesis holds for n = k
(This is known as the induction hypothesis.) We will show that it must then hold for
n = k + 1 In other words, we show thatdxdxk+1= (k + 1)xkassuming dxdxk= kxk−1 Let h(x) = xk+1= x · xk
By the Product Rule we know that
h(x) = 1 · xk+ x ·dxd xk But dxk= kxk−1by the induction hypothesis
Trang 7h(x) = xk+ x · kxk−1
= xk+ kxk
= (k + 1)xk Our proof is complete The derivative of xnis nxn−1for n = 1, 2, 3, , that is, for n any positive integer
EXERCISE 8.9 Differentiate f (x) = πx19− 5x5+3x26 + 7(π)3
Answer
f(x) = 19πx18− 25x4+ 9x5 (Why is the derivative of 7(π )3zero and not 21(π )2?)
The Quotient Rule
Although it is tempting to try to differentiate f (x)g(x) by dividing fby g, this is incorrect Below we use the Product Rule to arrive at a rule for differentiating a quotient.4
Let h(x) =f (x)g(x), where f and g are differentiable functions We are looking for h(x)
f (x) = g(x) · h(x) Differentiate both sides of the equation
f(x) = g(x) · h(x) + g(x) · h(x) Solve for h(x)
g(x) · h(x) = f(x) − g(x) · h(x)
h(x) =f
(x) − g(x) · h(x) g(x) But h(x) =f (x)g(x)
=
f(x) − g(x) ·f (x)g(x) g(x)
=
f(x) ·g(x) g(x) −g(x) ·f (x)g(x) g(x)
= f
(x) · g(x) − g(x) · f (x)
g(x)
· g(x)1
=f
(x) · g(x) − g(x) · f (x) [g(x)]2
f g
(x) =f
(x) · g(x) − g(x) · f (x) [g(x)]2
This is the Quotient Rule for taking the derivative of the quotient of two functions
d dx
f (x) g(x)
=g(x) · f
(x) − f (x) · g(x) [g(x)]2 or, without the x’s, f
g
=g · f
− f · g
g2
4 This is the argument that Leibniz used in the 1670s after realizing that the derivative of a quotient is not the quotient of the
Trang 8Notice that the order in which you do the subtraction matters Here’s a mnemonic to remember which term in the numerator comes first; take it or leave it “Don’t get fouled up” translates to gf(g for “get,” f for “fouled,” andfor “up”) Now that you have started, you’re all set
EXERCISE 8.10 Differentiate f (x) =2x2x+17+3
Answer
f(x) = 24x(72x+1)+14x62−6
Knowing the Quotient Rule allows us to differentiate xn, where n is a negative integer We’ll differentiate h(x) =x1m, where m is positive x1m = x−m, so the exponent of x is negative
If the exponent algebra in this proof is hard for you to follow, skip to Section 9.2 or the Algebra Appendix and, after studying that section, try this proof again.
h(x) =x1m
h(x) = x
m
· 0 − mxm−1 (xm)2
m−1
x2m (because (xa)b
= xab)
= −mx(m−1)−2m (because xa
x b = xa−b)
= −mx−m−1 This result fits our established pattern
d
dxx
−m= −mx−m−1 Therefore,
d
dxx
n
= nxn−1for any integer n
We’ve proven this result for n a positive integer using induction and for n a negative integer using the Quotient Rule Verify that the result is true for n = 0 on your own In fact, this
result is true for any exponent, but we will not be able to prove this until we know more
about exponential functions, their inverse functions, and taking the derivative of composite functions
Answers to Selected Exercises Answers to Exercise 8.5
i If k is a constant, the derivative of kf (x) is kf(x)
Trang 9Let g(x) = kf (x)
g(x) = lim h→0
g(x + h) − g(x) h
= lim h→0
kf (x + h) − kf (x)
h
= lim h→0k ·f (x + h) − f (x)
h
= k · lim h→0
f (x + h) − f (x)
= kf(x)
ii d
dx[f (x) + g(x)] = d
dxf (x) +dxd g(x) Let j (x) = f (x) + g(x)
j(x) = lim
h→0
j (x + h) − j (x) h
= lim
h→0
f (x + h) + g(x + h) − [f (x) + g(x)]
h
= lim
h→0
f (x + h) − f (x)
h
= lim
h→0
f (x + h) − f (x)
h→0
g(x + h) − g(x)
= f(x) + g(x)
Partial Answers to Exercise 8.6
i This makes graphical sense because multiplying f by k rescales the height of the graph
by a factor of k; therefore the ratio riserun is rescaled by a factor of k as well (See Figure 8.8.)
ii The height of the function f + g is the height of f plus the height of g Therefore,
rise run=rise due to f + rise due to g
run
= f
x + g
x
x
y
2 f
f
the slope of 2 f is twice the slope of f.
Figure 8.8
Trang 10Answers to Exercise 8.7
(a) y =2x+ 3x2+√5, so y= −x22 + 6x (b) y =π4 ·√x +3
√ π
7 · x, so y=π4 ·2√1
x+3
√ π
7 · 1 =π√2
x+3
√ π 7
Answer to Exercise 8.8
Let y = kf (x)
Then y= 0 · f (x) + k · f(x) = kf(x)
P R O B L E M S F O R S E C T I O N 8 3
For Problems 1 through 8, find f(x) Strategize to minimize your work For example,
x2+3 3x does not require the Quotient Rule.x23x+3=x3 +x1=13x + x−1 This is simpler
to differentiate.
1 f (x) = 3x2+ 3x + 3 + 3x−1+ 3x−2
2 f (x) = x−2x5 2
3 f (x) = π(3x2+ 7x + 1)(x − 2)
4 f (x) = x21
+4
5 f (x) = x+2x
6 f (x) = x+2x
7 f (x) = (5x22 + 7x5− 5x)x
8 f (x) = ax+ bx(c − dx), where a, b, c, and d are constants
For Problems 9 through 11, rewrite each of these functions in a form that allows you to differentiate using the tools you have now, but with as little exertion as possible Then differentiate Work on strategy; none of these problems require the Quotient Rule.
9 (a) f (x) = (x2+2x)x3 (b) f (x) =(x2+1)x 2
10 (a) f (x) =
1
+1
x+1 x2+2x
(b) f (x) =(x32x+3x)4
11 (a) f (x) = (x + 1)(x − 1)x (b) f (x) = x+
1
x
For each function in Problems 12 through 14:
(a) Sketch the graph of f (b) Find f(x) Are there any values of x for which fis undefined?
12 f (x) = x + |x|
... product of< /b>anytwo differentiable functions f and g Before doing this, let’s try an example
EXAMPLE 8.5 Find the derivative of each of the following functions. .. enables us to differentiate a whole family of power functions in the blink of an eye, without having to return to the limit definition of the derivative
The Derivative of xn... true for any exponent, but we will not be able to prove this until we know more
about exponential functions, their inverse functions, and taking the derivative of composite functions