The derivative of logbxis just a constant times1x.. EXAMPLE 14.1 Find the equation of the line tangent to f x = log327x2at the point 1, 3.. Graph f x =√x − ln x, indicating all local max
Trang 1y = logbx ⇐⇒ by= x
ln by
= ln x
yln b = ln x
y =ln x
ln b logbx = ln x
ln b = 1
ln bln x 1
ln bis simply a constant We know that the derivative of a constant times f (x) is that constant times f (x), so
d dx
1
ln bln x
ln b
d
dx(ln x) =ln b1 1x
Conclusion:
d
dx logbx = 1
ln b
1
x The derivative of logbxis just a constant times1x For example, d
dx log x =ln 101 x1, because log x =ln 101 ln x Notice that there is no escaping the natural logarithm It pops its head3 right up into the derivative of the log base b of x no matter what the value of b!
EXAMPLE 14.1 Find the equation of the line tangent to f (x) = log327x2at the point (1, 3)
The equation of the line is y − y1= m (x − x1), or y − 3 = f (1) (x − 1)
f (x) = log327 + log3x2= 3 + log3x2 For positive x, f (x) = 3 + 2 log3x We need
to differentiate this and evaluate the derivative at x = 1
Recall: d
dx(3 + 2 log3x)
x=1
says “take the derivative of 3 + 2 log3x and evaluate it at x = 1.”
d
dx 3 + 2 log3x
x=1
=
0 + 2
1
ln 3
1 x
x=1
=ln 32 Therefore, the equation of the tangent line is
y − 3 = 2
ln 3(x − 1) or y = 2
ln 3x − 2
ln 3+ 3
This is a linear equation;ln 32 is a constant
P R O B L E M S F O R S E C T I O N 1 4 1
In Problems 1 through 7, find y .
3 Of course we could writelog blog ein place of ln b In particular, ln 10 =log 10log e = 1
log e So d
dx log x = 1
ln 101= (log e) 1 ; but e
Trang 21 y = 2 ln 5x
2 y = π ln√x
3 y = ln 3x5
4 y = x ln x
5 y = ln√x2x
6 y = 3 log x
7 y = log2 x
3
8 Show that f (x) =ln√23x + 3 is invertible Find f−1(x)
9 Show that g(x) = π log2(π x) − π2is invertible Find g−1(x)
10 Find and classify the critical points of f (x) = x ln x
11 What is the lowest value taken on by the function g(x) = x2ln x? Is there a highest value? Explain
12 Use a tangent line approximation of ln x at x = 1 to approximate:
(a) ln(0.9)
(b) ln(1.1)
13 Graph f (x) =√x − ln x, indicating all local maxima, minima, and points of inflection
Do this without your graphing calculator (You can use your calculator to check your answer.) To aid in doing the graphing, do the following
(a) On a number line, indicate the sign of f Above this number line draw arrows indicating whether f is increasing or decreasing
(b) On a number line indicate the sign of f Above this number line indicate the concavity of f
(c) Find limx→0+f (x)and limx→∞f (x)using all tools available to you You should
be able to give a strong argument supporting your answer to the former The latter requires a bit more ingenuity, but you can do it
14 Let f (x) = ln x − x
(a) What is the domain of this function?
(b) Find all the critical points of f (The critical points must be in the domain of f ) (c) By looking at the sign of f , find all local maxima and minima Give both the x-and y-coordinates of the extrema
(d) Find f Where is f concave up and where is f concave down?
(e) Sketch the graph of ln x − x without using a calculator (except possibly to check your work)
Trang 314.2 THE DERIVATIVE OF bx REVISITED
EXERCISE 14.3 Label the y-coordinates of the three points indicated in Figure 14.4 below Use numerical
approximations, rounding off your answers to three decimal places Do these numbers look familiar?
y
1 2 3 4 5 6 7 8 9 10 x
y = ln x
(2, ) (3, )
(10, )
Figure 14.4
When we first investigated the derivative of bx, we showed that d
dxbx= kbx, where k
is a constant that depends on the base b We know that k is the slope of bxat x = 0, and we approximated k for various values of b We found
d
dx2x≈ (0.693)2x dxd 3x≈ (1.099)3x dxd 10x≈ (2.303)10x, and as a result of work on Exploratory Problems you may also know that
d
dx2.7x≈ (0.993)2.7x and d
dx2.8x≈ (1.030)2.8x None of this is completely satisfying These results drive us to search for some structure to these constants That search, or the exercise at the beginning of this section, might lead us
to conjecture thatdxd bx
= ln b · bx This conjecture we have is wonderful and delightful If it is true, those constants that you’ve been memorizing or looking up in order to approximate the derivatives of 2x,
3x, and 10x(the proportionality constants for the rates of growth of these functions), are just ln 2, ln 3, and ln 10, respectively The natural logarithm function just pops up—oh so naturally!
We will show that the conjecturedxdbx= ln b · bxholds Our argument will rest on the fact that logbxand bxare inverse functions For each point (c, d) on the graph of bx, the point (d, c) lies on the graph of logbx Notice that if (c, d) lies on the graph of bx, then
d = bc
Trang 4L1
L1
L2
L2
y = logb x
y = b x
(c, b c ) = (c, d)
(d, c)
Figure 14.5
The slope of L1, the line tangent to the graph of bxat x = c, isthe slope of1 L
2, where L2is the line tangent to the graph of logbxat the point (d, c)
The slope of L2at the point (d, c) = dxd logbx
x=d
ln b
1 x
x=d
ln b
1
d =d 1
ln b. Therefore,
the slope of L1= 11
d ln b
= d ln b = ln b · d
But (c, d) lies on the graph of bxso d = bc Consequently,
the slope of L1= ln b · bc
We have shown that the derivative of bxat x = c is ln b · bc
Conclusion:
d
dxb
x
= ln b · bx
An Alternative Approach to Finding the Derivative of bx
We know that the derivative of bxis
bx· (the slope of the tangent to bxat x = 0)
We can concentrate our energy simply on finding the slope of the tangent line to bxat (0,1)
It is the reciprocal of the slope of log xat the point (1,0)
Trang 5(1, 0)
(0, 1)
x
y = logb x
y = b x
Figure 14.6
The slope of the line tangent to logbxat x = 1 is dxd logbx
x=1= ln b1 x1
x=1=ln b1 Therefore, the slope of the tangent line to bxat x = 0 is ln b
We conclude that dxdbx= ln b · bx, as before
EXERCISE 14.4 Differentiate f (x) = 3 · 2x
Answer
f (x) = 3 ln 2 · 2x
P R O B L E M S F O R S E C T I O N 1 4 2
For Problems 1 through 3, finddydx.
1 y = x2· 2x
2 y =5·23x
3 y =x55 x
5
4 Find the absolute minimum value of f (x) = x · 2x
5 Find f (x)if (a) f (x) = x2+ ex+ xe+ e2 (b) f (x) = πex
−√6ex
29 (c) f (x) = 3ex+3 (Hint: Break this up into the product of exand a constant.)
6 Using the definition of derivative, show that the derivative of axis axtimes the slope
of the tangent to axat x = 0 (We’ve done this, but refresh your memory.)
7 If your last name begins with A–J: Approximate the derivative of (2.7)xusing the results
of Problem 6 and h = 0.000001
If your last name begins with K–Z: Approximate the derivative of (2.8)xusing the results of Problem 6 and h = 0.000001
Trang 68 Graph f (x) = ex
− x
Onlyuse a calculator to check your work after working on your own
(a) Find f (x) Draw a number line and indicate where f is positive, zero, and negative
(b) Label the x- and y-coordinates of any local extrema (local maxima or minima) (c) Using your picture, determine how many solutions there are to the following equations
i f (x) = 5
ii f (x) = 0.5 Notice that these equations are “intractable”— try to solve ex
− x = 5 alge-braically to see what this means If we want to estimate the solutions, we can do
so using a graphing calculator At this point, we should know how many solutions
to expect
9 Find the quantity indicated
(a) y = ln 5x + ln x5+ ln 5x
i Find y
ii Find the slope of the graph of the function at x = 1
(b) f (x) = log10x
i Find f (x)
ii Find f (100)
(c) P (x) = 7x
i Find P (x)
ii Find the instantaneous rate of change of P with respect to x when x = 0 (d) y = e3x
i Find y
ii Find the slope of the graph of the function at x = 0
(e) f (x) = 14ex/2
i Find f
ii Find f (ln 9) and simplify your answer
10 Differentiate y = e3xln(√1
5x)
14.3 WORKED EXAMPLES INVOLVING DIFFERENTIATION
In Sections 14.1 and 14.2 we found some important derivative formulas We’ve shown the following
i.dxd ln x =1x
ii.dxd logbx =dxd ln xln b=dxd ln b1 ln x =ln b1 ·1x=xln b1 iii dxd bx= ln b · bx
In this section we will work examples using a combination of these differentiation formulas and the laws of logs and exponents The main piece of advice is to spend time
Trang 7putting the expression to be differentiated into a form that makes it simple to apply the differentiation formulas given above Be sure to distinguish between constants and variables
Worked Examples
In each of the following problems, find dydx
EXAMPLE 14.2 y = π ln7x 6
8
y = π(ln 7x6− ln 8)
= π ln 7 + π ln x6− π ln 8
= π ln 7 + 6π ln x − π ln 8 (First and third terms are constant.) Differentiate:
dy
dx =6π
EXAMPLE 14.3 y =10 log(bx3)
√ π
y =√10π logbx3
=√10
π[log b + 3 log x]
=√10π log b + √30πln 10ln x Notice that√10
π log b is constant, as is √30
π
1
ln 10. Differentiate:
dy
dx =√30
π
1
xln 10
EXAMPLE 14.4 y = 17(3)t /5
y = 1731/5t We’ll treat 31/5as the base of y = bx Differentiate:
dy
dx = 17ln 31/5 31/5t=17
5 (ln 3)
3t /5
Trang 8EXAMPLE 14.5 y = 52x+3
y = 52x· 53=53 52x= 5325x Notice that 53is a constant
We’ll treat 52, or 25, as the base of y = bx Differentiate:
dy
dx = 53ln52· 25x
= 125 ln 52· 52x
= 250 ln 5 · 52x
In the following examples, we generalize what we have done Try them on your own before looking at the answers
EXAMPLE 14.6 y = A lnBxC, where A, B, and C are constants and B > 0
y = A[ln B + C ln x]
= A ln B + AC ln x Differentiate:
dy
dx = AC1x =ACx
EXAMPLE 14.7 y = Abkx, where A, b, and k are constants and b > 0
y = Abkx Differentiate:
dy
dx = A lnbk bkx
= Ak ln b · bkx
In Chapter 16 we will acquire alternative means of finding these derivatives
EXAMPLE 14.8 Approximate ln 1.1 with the help of the first derivative
SOLUTION This problem is like the tangent-line approximations we discussed in Chapter 8 We begin
by sketching ln x and getting an off-the-cuff approximation of ln 1.1
Trang 9y
y = lnx
y = lnx
1
1
1.1
1.1 enlarged
tangent line at x = 1
slope = 1
tangent line
∆y
∆x
Figure 14.7
ln 1.1 is a bit larger than ln 1; how much larger can be approximated by looking at the rate at
which ln x is increasing near x = 1 The derivative of ln x at x = 1 gives the rate of increase
d
dx ln x|x=1= x1
x=1
= 1
Knowing the rate of increase of ln x at x = 1 enables us to judge how to adjust our approximation (in this case, 0) to fit a nearby value of x dydx ≈'y'x for 'x small (because dy
dx= lim'x→0 'y'x) In this problem, 'x = 0.1
y = lnx
1 1.1
tangent line approx.
actual y-value
Figure 14.8
dy
dx = 1 ≈ 'y
.1 ⇒ 'y ≈ 0.1 Therefore, 'y ≈ 0.1 We obtain the approximation ln 1.1 ≈ 0 + 0.1 = 0.1
Is the approximation ln 1.1 ≈ 0.1 an overestimate, or an underestimate? We know that
ln x is concave down
d2
dx2ln x =dxd 1x
= −x12<0 Therefore the tangent line lies above the graph of the function, and this approximation is a bit too big
Comparing our approximation, ln 1.1 ≈ 0.1, with the calculator approximation,
ln 1.1 ≈ 0.0953, we see that the approximation is fairly good
Trang 10REMARKIf we wanted to further refine our approximation, we could use information supplied by the second derivative to make the required adjustment.4This is equivalent to approximating the curve y = ln x at x = 1 by a parabola instead of a line Further successive
refinements involve higher order derivatives and give us what is called a Taylor polynomial approximationto the function at x = 1
Answer
ln 0.9 ≈ −0.1
Logarithms: A Summary
logbx = y is equivalent to x = by
This equivalence follows from the definition of logbx logbxis the number we must raise
bto in order to obtain x
logbxand bxare inverse functions, so
logbb(= ( and blogb (
= (
log x is the way we write log10x(log base 10 of x, or the common log)
ln x is the way we write logex (log base e of x, or the natural log)
Thus,
ln x = y is equivalent to x = ey
It follows that
eln (= ( and ln e(= (
y
1
1
x
y = ln x
y = x
y = e x
Figure 14.9
4 The larger the value of the second derivative at x = 1 the larger the rate of change of the slope near x = 1.