P A R TV Adding Sophistication to Your Differentiation 16 Taking the Derivative of Composite Functions 16.1 THE CHAIN RULE We can construct conglomerate functions in two different ways..
Trang 1Exploratory Problems for Chapter 15 511
(a) Let N = N(t) be the number of moles of substance A at time t Translate the
statement above into mathematical language (Note: The number of moles of
substance B should be expressed in terms of the number of moles of substance A.)
(b) N (t) is a decreasing function The rate at which N is changing is a function of N , the number of moles of substance A When the rate at which A is being converted
to B is highest, how many moles are there of substance A?
Trang 3P A R T
V Adding Sophistication to Your Differentiation
16
Taking the Derivative
of Composite Functions
16.1 THE CHAIN RULE
We can construct conglomerate functions in two different ways One way is to combine the functions’ outputs by taking, for example, their sum or product We can differentiate a sum
by summing the derivatives and differentiate a product by applying the Product Rule Another way to construct a conglomerate is to have functions operate in an assembly-line manner In this configuration, the output of one function becomes the input of the next, creating a composite function In this section we will look at the derivatives of composite functions The work we do will have plentiful rewards, as the results have extensive application
Our goal is to express the derivative of the composite function f (g(x)) in terms of
f, g, and their derivatives Our assumption throughout is that f and g are differentiable functions
Many functions we can’t yet differentiate can be decomposed and expressed as the composite of simpler functions whose derivatives we know For example:
513
Trang 4514 CHAPTER 16 Taking the Derivative of Composite Functions
h(x) =√ln x can be decomposed into f (g(x)), where f (x) =√xand g(x) = ln x h(x) = (x2+ 1)30 can be decomposed into f (g(x)), where f (x) = x30and g(x) = x2+ 1 h(x) = ex2 can be decomposed into f (g(x)), where f (x) = exand g(x) = x2 h(x) = ln(x + 8x2) can be decomposed into f (g(x)), where f (x) = ln x and g(x) = x + 8x2 We’ll look at the problem of differentiating a composite function in the context of the next example
EXAMPLE 16.1 The number of fish a lake can support varies with the water quality The water quality is
affected by industry around the lake; the level of grime in the lake varies with time For the purposes of this example, we’ll assume that the level of grime in the lake is always increasing The number of fish decreases as the level of grime goes up
Let g(t) give the level of grime in the lake as a function of time t, measured in years Let f (g ) be the fish population as a function of the amount of grime Then the population
of fish as a function of time is the composite function f (g(t)) Assume that f and g are differentiable functions Find the rate at which the fish population is changing over time
SOLUTION We must computedtdf (g(t )), ordfdt Essentially, we argue as follows
"f
"t ="f"g ·"g"t From a purely algebraic standpoint, this must be true, provided "g and "t are not zero Let "g = g(t + "t) − g(t) and "f = f (g + "g) − f (g ) As "t → 0, we know
"g → 0 because g is a continuous function
lim
"t →0
"f
"t = lim
"t →0
"f
"g ·"g"t
lim
"t →0
"f
"t =
lim
"t →0
"f
"g
·
lim
"t →0
"g
"t
lim
"t →0
"f
"t =
lim
"g→0
"f
"g
·
lim
"t →0
"g
"t
df
dt =df
dg · dg dt For the purposes of this problem, since we’ve asserted that g is increasing, we haven’t run into trouble In terms of generalizing, we have trouble only if "g = 0 infinitely many times
as "t → 0 But if this is the case, it can be shown that both dgdt = 0 and dfdt = 0, so the equation dfdt =dfdg ·dgdt still holds
Let’s return to the fish and make things more concrete Suppose we want to find the rate
of change of the fish population with respect to time at t = 3, and suppose that at that time the grime level in the lake is 700 units Let’s look at"f"t, the rate of change of the number
of fish over the time interval [3, 3 + "t], where "t is very small
Over a given time interval, the level of grime varies, and this causes a fluctuation in the number of fish in the lake We’re interested in finding out how a change in time, "t, affects the fish population The change in time affects the fish population indirectly via the change
in the grime level of the lake There is a chain reaction; time affects grime, and grime affects
Trang 516.1 The Chain Rule 515 fish Therefore, we first look at the change in the level of grime produced by a "t change
in time
How does g(t) change during this small interval? We know thatdgdt ≈"g"t for "t very small, so we can solve for "g
"g ≈dgdt"t Looking at the time interval [3, 3 + "t] gives us the following
"g = g(3 + "t) − g(3)
"g ≈ dg
dt
t =3
"t
≈ g (3)"t
g
g (t)
g (3 + ∆t)
3 + ∆t
∆t
3
700 = g(3) ∆g
t
Figure 16.1
Now we need to determine how this change in the level of grime, "g, will affect the
t = 3, the grime level is 700 As the grime level changes from 700 to 700 + "g, how does the fish population change?
As above, we can write dfdg ≈"f"g for "g very small Because "t is very small (and will approach zero), "g is very small as well.1Solving for "f gives
"f ≈dfdg"g
Focusing on what is happening at t = 3 (when g = 700) leads to this equation
"f = f (g (3 + "t)) − f (g (3))
= f (700 + "g) − f (700)
"f ≈
df dg
g=700
· "g
≈ f (700)"g
1 This is because g is a continuous function We know g is continuous because we are working under the assumption that g is differentiable.
Trang 6516 CHAPTER 16 Taking the Derivative of Composite Functions
f
g
f (g)
f(700)
|∆f |
700
700
700 + ∆g
700 + ∆g
f (700 + ∆g)
∆g
g
t
g (t)
3 + ∆t
3
∆t
∆g
Figure 16.2
We can use the expression for "g on the preceding page
"f ≈ f (700)"g
≈ f (700)g (3)"t
≈ f (g(3)) · g (3)"t Now we can find an expression for"f"t, the rate of change of the population with respect to time
"f
"t ≈f
(g(3))g (3)"t
"t
≈ f (g(3))g (3) Let’s look at the approximations we made in this discussion As "t gets closer and closer to zero, the approximationdgdt ≈"g"t gets better and better, because
lim
"t →0
"g
"t =dgdt
As "t approaches zero, "g = g(t + "t) − g(t) also approaches zero As "g gets closer and closer to zero, the approximation dfdg ≈"f"g gets better and better, because
lim
"g→0
"f
"g =dfdg This leads us to conclude that
d
dtf (g(t ))
t =3
= f (g(3)) · g (3)
In fact, there was nothing special about the time t = 3, so this relation should hold for all values of t:
= f (g(t )) · g (t )
A unit analysis gives fish
time = grime ·fish
grime time , which makes sense The crucial charac-teristic of f and g is that they are differentiable In our discussion, we need the assumption
Trang 716.1 The Chain Rule 517
The result can be stated for any two differentiable functions f and g; it goes by the name “the Chain Rule.”
The Chain Rule:
d
dtf (g(t )) = f (g(t )) · g (t ) or df
dt =df
dg · dg dt
Interpreting the Chain Rule
Thoughts on the formdtdf (g(t )) = f (g(t )) · g (t ):
More informally, we can state the Chain Rule as
d
dt (f (mess)) = f (mess) · (mess) ,
where mess, of course, is just a function of t; it’s g(t).
Notice that the derivative of f is evaluated at g(t); f (g(t )), not f (t ) We obtain
f (g(t ))by calculating f (x)and replacing x by g(t) Think about our example; f is
a function of grime, so f is also a function of grime, not of time We evaluated f at
g = 700 (or g(3)), not at g = 3
= f (g(t )) · g (t ) = (derivative of f evaluated at g(t)) · (derivative of g) Thoughts on the form dfdt =dfdg ·dgdt: Leibniz’s notation gives us a very nice way of expressing the Chain Rule Although dfdg anddgdt are not fractions, the notation works
well; that is the genius of it
Let’s take a few more quick passes on interpreting the Chain Rule in this form
If g changes three times as fast as t and f changes twice as fast as g, then f changes
6 times as fast as t.dfdt =dfdg· dgdt
Or, think of gears—either interlocking or connected by a chain as in bicycle gears Suppose the little gear spins 12 times per minute and the big gear spins once for every three turns of the little gear Then the big gear spins 4 times per minute
rotations of big gear
rotations of little gear·rotations of little gear
minute =rotations of big gear
Applying the Chain Rule
Let’s apply the Chain Rule to the functions h(x) = f (x + k) and j (x) = f (kx), correspond-ing to a horizontal shift of f and a horizontal compression of f , respectively If you haven’t previously done so, first spend a minute trying to determine h (x)and j (x)by graphical means
Trang 8518 CHAPTER 16 Taking the Derivative of Composite Functions
EXAMPLE 16.2 f (x + k) can be thought of as the composite f (g(x)), where g(x) = x + k.
[f (x + k)] = f (g(x)) · g (x)
= f (x + k) · 1
= f (x + k) This is in agreement with our graphical intuition
[f (kx)] = f (g(x)) · g (x)
= f (kx) · k
= kf (kx) This makes sense graphically Suppose k > 0 Because the function’s graph is compressed horizontally, its derivative will be as well This horizontal compression results in steeper slopes; hence f (kx)is multiplied by k
Some functions that we could differentiate by taking advantage of laws of logs and exponentials we can now differentiate using the Chain Rule The next example illustrates these options
EXAMPLE 16.4 Find the derivatives of the following, where k is a constant
(a) q(x) = ekx+2 (b) s(x) = ln(kx)
SOLUTION (a) Using Exponent Laws: q(x) = ekxe2, so q (x) = e2kekx= k · ekx+2
Using the Chain Rule:Let the inside function, g(x), be kx + 2 and the outside function,
f (u), be eu Then g (x) = k, f (u) = eu, and f (g(x)) = eg(x)= ekx+2 So
q (x) = f (g(x)) · g (x)
= ekx+2· k
(b) Using Log Laws: s(x) = ln k + ln x ln k is a constant, so s (x) = 1x
Using the Chain Rule:Let the inside function, g(x), be kx and the outside function,
f (u), be ln u Then g (x) = k, f (u) =1u, and f (g(x)) =g(x)1 =kx1 So
s (x) = f (g(x)) · g (x) =kx1 · k = 1x
Observe that, by using the Chain Rule, we can generalize the three basic derivative rules
d
dx[xn] = nxn−1 can be generalized to dxd n= n n−1·dgdx
d
dx[bx] = ln b · bx can be generalized to dxd g(x) = ln b · bg(x)
·dgdx
d
dx bx = 1
ln b·x1 can be generalized to dxd b ln b1 ·g(x)1 ·dgdx
EXAMPLE 16.5 Decompose each of the following functions into f (g(x)) and then compute the derivative
(a) j (x) = (x6+ 5x3+ x2)8 (b) k(x) = ln(x3+ 3x) (c) h(x) = 3x2
Trang 916.1 The Chain Rule 519
SOLUTION (a) j (x) = (x6+ 5x3+ x2)8 Multiplying out would be terribly tiring and tedious Instead,
use the Chain Rule, where the inside function, g(x), is x6+ 5x3+ x2and the outside function, f (u), is u8 (Check that f (g(x)) = j (x).)
j (x) = f (g(x)) · g (x)
We know f (u) = 8u7, so
f (g(x)) = 8 7= 8(x6+ 5x3+ x2)7
g (x) = 6x5+ 15x2+ 2x
So j (x) = 8(x6+ 5x3+ x2)7· (6x5+ 15x2+ 2x)
This is a lot easier than multiplying out in the beginning!
Basically, this function is of the form (mess)8, so its derivative is 8(mess)7· (mess) (b) k(x) = ln(x3+ 3x) Let the inside function, g(x), be x3+ 3xand the outside function
be f (u) = ln u (Check that f (g(x)) = k(x).)
k (x) = f (g(x)) · g (x)
We know f (u) = 1u, so
f (g(x)) = 1
g(x)= 1
x3+ 3x
g (x) = 3x2+ (ln 3)3x
So k (x) = x3 +31 x · (3x2+ (ln 3)3x)
Basically, this function is of the form ln (mess), so its derivative is mess · (mess)1 (c) h(x) = 3x2 Let the inside function, g(x), be x2and the outside function be f (u) = 3u
h (x) = f (g(x)) · g (x)
We know f (u) = (ln 3)3u, so
f (g(x)) = (ln 3)3g(x)= (ln 3)3x2
g (x) = 2x
So h (x) = (ln 3)3x2(2x) = (2 ln 3)(x)(3x2)
Basically, this function is of the form 3mess, so its derivative is
(ln 3)3mess · (mess)
P R O B L E M S F O R S E C T I O N 1 6 1
1 (a) Which of the following are equal to (ln x)2?
i (ln x)(ln x) ii ln x2 iii ln[(x)(x)] iv 2 ln x (b) Which of the following are equal to 2 ln x?
i (ln x)(ln x) ii ln x2 iii ln[(x)(x)]
(c) Differentiate y = (ln x)2 (Do this twice, first using the product rule and then using the Chain Rule.)
(d) Differentiate y = ln x2 (Do this twice, first using the log rules and the derivative
of ln x and then using the Chain Rule.)
Trang 10520 CHAPTER 16 Taking the Derivative of Composite Functions
In Problems 2 through 20, find f (x) Do these problems without using the Quotient Rule.
2 (a) f (x) = 3(x + 2)−5 (b) f (x) = 2(3x + 7)−8
3 f (x) = ln√π x + 1 +√π x + (πx + π)5+(π x21
+1)3
(Hint: Use log operations to simplify the first term.)
4 f (x) = (x3 +7x)x 4
5 f (x) = 3xe2x
+1
6 f (x) = e5x(1 + 2x)6
7 f (x) = (1 − 1x)e−x
8 f (x) = ln(√x3)e6x
9 f (x) = 5 ln(2x2+ 3x)
10 f (x) = (3x3+ 2x)13
11 f (x) = (x+xeπ x2 ) 3
12 f (x) = 3(xπ3 +2)2 6
13 f (x) = x3 +7x+51
14 f (x) = 23x+1x
15 f (x) = x5x+12
16 f (x) = √4
e x +1
17 f (x) =ex+ ln(x + 1)2
18 f (x) = ln(x12
+2)
19 f (x) = ln(ex+ x2)
20 f (x) = x lnx2x+1
21 Find a formula for dydx if y = f (g(h(x))), where f , g, and h are differentiable every-where
In Problems 22 through 25, graph f (x), labeling the x-coordinates of all local extrema Strategize Is it more convenient to keep expressions factored?