b Express the rate of change of volume with respect to time in terms of r, r , h, and h.. b ExpressdPdt, the rate of change of the power with respect to time, in terms of the functions P
Trang 1(c) Finddxd fg
x=3 (d) Let y = 2g(x2) What is dxdy
x=2? (e) Let y =f (x2) Find y (2)
9 Below is a graph of f (x) on the interval [−2, 3] j (x) is given by j (x) = ln
the domain [−2, 3]
1
1
–1
2
3
f
x
(a) How many zeros does j (x) have?
(b) Find j (x)
(c) Approximate the critical points of j
(d) Identify the local extrema of j (Estimate the positions.)
(e) Where does j attain its absolute maximum value? Its absolute minimum value? (f ) Which function has a higher absolute maximum value, f or j ? Which function has a lower absolute minimum value, f or j ? Explain
10 g(x) is a continuous function with exactly two zeros, one at x = 1 and the other at
x = 4 g(x) has a local minimum at x = 3 and a local maximum at x = 7 These are
(a) Find f (x)in terms of g and its derivatives
(b) Can we determine (definitively) whether g has an absolute minimum value on (−∞, ∞)? If we can, where is that absolute minimum value attained? Can we determine (definitively) whether g has an absolute maximum value? If we can, where is that absolute maximum value attained?
(c) What are the critical points of f ?
(d) On what intervals is the graph of f increasing? On what intervals is it decreasing? (e) Identify the local maximum and minimum points of f
(f ) Can we determine (definitively) whether f has an absolute minimum value? If so, can we determine what that value is?
If you haven’t already done so, step back, take a good look at the problem (a bird’s-eye view) and make sure your answers make sense
11 Assume that f , g, and h are differentiable Differentiate p(x) where
(a) p(x) = f (x)g(x)h(x) (Hint: Use the Product Rule twice.)
(b) p(x) =√g(x) + ln f (x)
Trang 2532 CHAPTER 16 Taking the Derivative of Composite Functions
12 Let f (x) be the function whose graph is drawn on the axes below
1
1 –1
–1 –2
–2
–3 –4 –5
2 3 4
f (x)
Let a(x) = 2f (x), b(x) = f (x + 2), and c(x) = f (2x)
(a) On three separate sets of axes, draw the graphs of the function a(x) = 2f (x), b(x) = f (x + 2), and c(x) = f (2x), labeling the x-intercepts, the y-intercept, and the x- and y-coordinates of the local extrema
(b) Suppose we know that
f (−4) = 1/2, f (−2) = 3/2, f (0) = 0
Find the following, explaining your reasoning briefly
i a (−2) =
ii b (−2) = iii c (−2) =
In Problems 13 through 18, find h (x) Assume that f and g are differentiable on (−∞, ∞) Your answers may be in terms of f , g, f , and g
13 h(x) = f (ln x) − ln(f (x))
14 h(x) =√f (x)g(x)
f (g(x))
16 h(x) = f (x2)eg(x)
17 h(x) =[f (x)]1 2 + f x12
18 h(x) = [f (x)]3g(2x)
In Problems 19 through 22, find dydx Take the time to prepare the expression so that it
is as simple as possible to differentiate.
19 y = 3 lnx2−1
x+2
Trang 3
20 y =(x2+ 3)5
21 y =5xπ−x3−1
x 2
22 y = 5 lnx3x2 +1
In Problems 23 through 29, differentiate In Problems 23 through 25, assume f is differentiable Your answers may be in terms of f and f .
23 y = lnf (xx2 )
24 y = [f (x)]2+ 2f (x)
25 y = ln
x·f (x)
√
3x 3
+2
26 f (x) = ln(e(x+5)2)
27 f (x) = e(g(x))2
28 f (x) = (x3+ e)π
29 Let f (x) = xx
(a) Use numerical methods to approximate f (2)
that makes it not a power function?
that makes it not an exponential function?
(d) Challenge: Figure out how to rewrite xxso you can use the Chain Rule to differ-entiate it
30 Consider the function f (x) = e−x2(−x2+ 1) You must give exact answers for all of the following questions Show your work Your work must stand independent of your calculator
(a) Find all the x-intercepts
(b) Identify the local extrema of f (x)
(c) Sketch a graph of f , labeling the x-coordinates of all local and global extrema (d) Now consider the function g(x) = |f (x)|
i What are the critical points of g?
ii Classify the critical points of g(x)
31 A craftsman is making a mobile consisting of hanging circles each with an inscribed triangle of stained glass Each piece of stained glass will be an isosceles triangle Show that if she wants to maximize the amount of stained glass used, the glass triangles should
be equilateral In other words, show that the isosceles triangle of maximum area that can be inscribed in a circle of radius R is an equilateral triangle
Trang 4534 CHAPTER 16 Taking the Derivative of Composite Functions
32 A holiday ornament is being constructed by inscribing a right circular cone of brightly colored material in a transparent spherical ball of radius 2 inches What is the maximum possible volume of such an inscribed cone?
33 A craftsman is making a ribbon ornament by inscribing an open hollow cylinder of colored ribbon in a transparent spherical ball of radius R What is the maximum surface area of such a cylinder?
34 Phone cables are to be run from an island to a town on the shore The island is 5 miles from the shore and 13 miles from the town The cable will be run in a straight line from the island to the shore and then in another straight line along the shoreline If it costs 60% more to run the cable under water than it does to run it under the ground, how far should the cable be run along the shore?
35 The volume of a cylindrical tree trunk varies with time Let r(t) give the radius of the trunk at time t and let h(t) give the height of time t
(a) Express the rate of change of A, the cross-sectional area, with respect to time in terms of r and r
(b) Express the rate of change of volume with respect to time in terms of r, r , h, and
h
36 Suppose the amount of power generated by an energy generating system is a function of
v, the volume of water flowing through the system The function is given by P = P (v) The volume of water in the sytem is determined by r, the radius of an adjustable valve;
v = v(r) The radius varies with time: r = r(t)
(a) ExpressdP
dr, the rate of change of the power with respect to a change in the valve’s radius, in terms of the functions P (v) and v(r) and their derivatives
(b) ExpressdPdt, the rate of change of the power with respect to time, in terms of the functions P (v), v(r), and r(t) and their derivatives
Trang 5Implicit Differentiation and its Applications
In this section you will add sophistication to your differentiation skills by applying two fundamental principles
Principle (i):If two expressions are equal, their derivatives are equal
Principle (ii): Whether y is given implicitly or explicitly in terms of x,1 we can differentiate an expression in y with respect to x by treating y the same way we would treat g(x) and using the Chain Rule
Try Example 17.1 below Do the problem on your own, writing down your solution so you can compare it with the discussion that follows
EXAMPLE 17.1 Let f (x) = xx+1, where x > 0
(a) Approximate f (2) numerically
(b) Using appropriate rules of differentiation, find f (2) exactly
(c) Compare your answers to parts (a) and (b) If they are not very close to one another, identify your error
slope of the secant line through (2, f (2)) and a nearby point on the graph of f , say (2.0001, f (2.0001))
1 For example, xy 3
= 1 is an equation that gives y implicitly in terms of x; y = x2+ 3x is an equation that gives y explicitly
in terms of x.
535
Trang 6536 CHAPTER 17 Implicit Differentiation and its Applications
f (2) ≈f (2.0001) − f (2)
2.0001 − 2 (Equivalently, we could say f (2) = limh→0 f (2+h)−f (2)
h , so f (2) ≈f (2+h)−f (2)h for
hvery small Then we choose a very small value for h, say h = 0.0001.)
f (2) ≈(2.0001)0.00012.0001+1− 22+1
f (2) ≈(2.0001)
3.0001
− 23 0.0001
≈ 17.547
before dividing by 0.0001 To divide by 0.0001 is to multiply by 10,000, so any roundoff error in the numerator will be multiplied ten thousand-fold
(b) There are two popular but fatally flawed approaches to differentiating xx+1 We’ll set them up and knock them down before proceeding
Fatally flawed approach #1:We know that dxdxn= nxn−1for any constant n It is easy
to succumb to temptation and propose (x + 1)xxas the derivative of xx+1 But this
differentiation rule cannot be applied because (x + 1) is not a constant; xx+1is not a power function! (If you apply it anyway, comparing your answers to parts (a) and (b) (17.547 versus 24) ought to alert you to the error.)
Fatally flawed approach #2:We know thatdxdbg(x)
= ln b · bg(x)· g (x)for any positive constant b Trying to apply this rule to xx+1by letting b = x leads to a derivative of
ln x · xx+1 The error is similar to that in the previous approach; b must be a constant,
and x is not constant Again, comparing your answers in parts (a) and (b) (17.547
versus 5.545) ought to alert you to an error if you use this approach
A correct approach:In order to apply a differentiation rule we know, we must alter the
form of f (x) The problem is that x is in both the base and the exponent A flash of
inspiration can lead us to express f (x) as follows
f (x) = xx+1= eln
x x+1
= e(x+1) ln x
dxeg(x)=
eg(x)· g (x)to obtain
f (x) = e(x+1) ln x d
dx[(x + 1) ln x]
= e(x+1) ln x x + 1
= xx+1 x + 1
Then f (2) = 233
2+ ln 2= 12 + 8 ln 2 This is 17.54517 , quite close to the approximation found in part (a)
From what source could this flash of inspiration, expressing xx+1as e(x+1) ln x, have arisen? We know that eln A= A for any positive A, but what could make us think
Trang 7of applying that knowledge in this problem? Recall that in provingdxdxn= nxn−1for any real number n we used the same idea.2We rewrote xnas eln xnand then as enln x (This is why we needed the Chain Rule before we could establish the proof.)
An alternative correct approach:In the approach taken above, to differentiate f (x) =
xx+1we changed its form, converting the expression xx+1to something we could more
easily differentiate Another approach is to deal instead with the entire equation f (x) =
xx+1 If we can bring the (x + 1) down from the exponent before we differentiate, we’ll
be in good shape Taking logs of both sides of the equation f (x) = xx+1accomplishes this while still preserving the balance of the equation Keep in mind that taking the log
of an expression changes it into a new and different expression; we are not proposing
to take the logarithm of an expression, but rather of both sides of an equation.
f (x) = xx+1
ln f (x) = ln(xx+1) Take ln of each side (If A = B and A, B > 0,
then ln A = ln B.)
ln f (x) = (x + 1)(ln x) Use ln ab= b ln a to bring down the exponent Now we can differentiate both sides using principle (i), which says that if g(x) = h(x), then dxdg(x) = dxdh(x) Remember to use the Chain Rule on the left side (where we have ln(f (x))) and the Product Rule on the right
d
1
f (x)f
(x) = (1)(ln x) + (x + 1) 1x
f (x) =
ln x +x + 1
x
f (x) =
ln x +x + 1x
This gives us an expression for f (x), as desired
Expressions Versus Equations
The distinction between an expression and an equation is critical to understanding the two approaches discussed above, so we’ll reiterate the distinction
The governing criterion for working with an expression is that its value remain un-changed; only its form may vary
Zero can be added to an expression; an expression can be multiplied by 1; an expression can be factored An expression A in the domain of f can be written as
f−1(f (A)) For example, g(x) = ln(eg(x))and if g(x) > 0, then g(x) = eln g(x)
2 Sometimes establishing proofs and doing practical problem solving seem to be disparate activities Often, however, they are more intimately intertwined; a proof can provide problem-solving inspiration, and problem solving might suggest a more general
Trang 8538 CHAPTER 17 Implicit Differentiation and its Applications
An equation, on the other hand, establishes an equality of two expressions; we can manipulate an equation in any way that does not alter the equality of the two sides The governing criterion for working with equations is that balance be maintained
Adding/subtracting the same thing to/from both sides of the equation maintains the balance Multiplying/dividing both sides of the equation by the same nonzero quantity maintains the balance
If A = C, where A and C are both positive, then An
= Cn For example, squaring both sides of an equation (n = 2) maintains balance (although it may introduce extrane-ous roots) Similarly, taking reciprocals of both sides of an equation (n = −1) main-tains balance If J + K = C, then (J + K)n
= Cn In particular, if J + K = C, then 1
J +K =C1
If A = C, where A and C are both positive, then ln A = ln C For example, if J + K =
C, then ln(J + K) = ln C
If A = C, then bA
= bCfor any positive constant b For example, if J + K = C then
eJ +K= eC
We have presented two different successful approaches to tackling the problem
pre-sented in Example 17.1(b) The first dealt with the expression xx+1; we used the fact that
eln A= A for any positive A to convert the expression xx+1to the equivalent expression
e(x+1) ln x The second approach dealt with the equation f (x) = xx+1 We took the natural
logarithm of both sides of the equation to obtain an equivalent equation, differentiated both sides of the equation, and solved for f (x) This latter technique of differentiation is called
logarithmic differentiation
P R O B L E M S F O R S E C T I O N 1 7 1
Differentiate the following.
1 (a) y = 3x (b) y = x3 (c) y = xx, where x > 0
2 y = (x + 1)(x+1), where x > −1
3 y = (3x2+ 2)x
4 y = xx2, where x > 0
17.2 LOGARITHMIC DIFFERENTIATION
Logarithmic differentiation deals with the task of differentiating a positive function f (x)
by working with both sides of the equation y = f (x) as follows
Using Logarithmic Differentiation to Find y
1.Begin with an equation y = f (x), where f (x) > 0 Take the natural logarithm of both
sidesof the equation
Trang 92.Use log rules to bring down exponents and/or simplify expressions.
3.Differentiate both sides of the equation y is a function of x so the Chain Rule must be
applied to differentiate ln y or ln f (x)
d
dx[ln f (x)] =f (x)1 f (x)or, equivalently, d
dx[ln y] = 1ydydx
4.Solve fordydx or f (x)
5.To express y in terms of x, replace y or f (x) with the equivalent expression in terms
of x
When to Use Logarithmic Differentiation
i The technique is useful in differentiating a function that has the variable in both the base and the exponent
ii We may choose to use logarithmic differentiation to make the differentiation of quo-tients or products more palatable, provided that we only take the log of positive quan-tities
EXAMPLE 17.2 Let y = 2xe x
, where x > 0 Finddxdy
SOLUTION The variable is in the base and the exponent; logarithmic differentiation enables us to bring
down the expression in the exponent
y = 2xex
d
dx[ln y] = dxd [ln 2 + exln x] Differentiate each side ln 2 is a constant
1 y
dy
dx = (ex)(ln x) + (ex) 1
x
Use the Chain Rule on left because y is a function of x dy
dx = ex
ln x +1x
dy
dx = ex
ln x +1 x
2xex Replace y by its expression in x
EXAMPLE 17.3 Differentiate y =(x + 3)
5(x2+ 7x)8 x(x2+ 5)3 , where x > 0.3
3 The condition x > 0 assures that (x + 3), (x 2 + 7x), and x are all positive.
Trang 10540 CHAPTER 17 Implicit Differentiation and its Applications
5(x2+ 7x)8 x(x2+ 5)3
ln y = ln(x + 3)
5(x2+ 7x)8
ln y = 5 ln(x + 3) + 8 ln(x2+ 7x) − ln x − 3 ln(x2+ 5) Use ln(ab) = ln a + ln b and ln ab= b ln a d
dx[ln y] =dxd [5 ln(x + 3) + 8 ln(x2+ 7x) − ln x − 3 ln(x2+ 5)] Differentiate each side
1
y
dy
8(2x + 7)
1
x −x3(2x)2
dy
16x + 56
1
x2+ 5
dy
16x + 56
1
x −x26x + 5
(x + 3)5(x2+ 7x)8
Ugly, but not as painful as differentiating this using the Quotient and Product Rules.4
P R O B L E M S F O R S E C T I O N 1 7 2
1 Find f (x)
(a) f (x) = 2xx, where x > 0 (b) f (x) = 5(x2+ 1)x (c) f (x) = (2x4+ 5)3x+1
2 Find f (x)
(a) f (x) = 3 · 2x
+ 2 · x3+ 3 · x2x+3, where x > 0 (b) f (x) = x(2x3+ 1)x+ 5, where x > 0
3 Find g (t )
(a) g(t) = t22tt, where t > 0 (b) g(t) = ln(t + 1)t2+1, where t > −1
4 Find dxdy using logarithmic differentiation You need not simplify.
(a) y = xln√x, where x > 0 (b) y =(x+1)xe25x√
x−2, where x > 0 (c) y = (e2x)(x2+ 3)5(2x2+ 1)3
5 Suppose y = f (x)g(x), where f (x) and g(x) are positive for all x Use logarithmic differentiation to find dxdy Verify that your result is simply the Product Rule
4 If you ever forget the Product or Quotient Rules but remember the derivative of ln x, you can use logarithmic differentiation