Section 33–4
◆
981
Derivatives of Logarithmic Functions
◆◆◆
Example 14:
(a) If y Cot1 (x2 1), then
1
y = (2x)
1 (x2 1)2
2x
2 2x2 x4
(b) If y Cos1 1 x , then
1
1
y = 1
(1
x)
2
1 x
1
= 2
2x x
Exercise 3
◆
◆◆◆
Derivatives of the Inverse Trigonometric Functions
Find the derivative.
1. y x Sin1 x
x
2. y Sin1 a
x
3. y Cos1 a
4. y Tan1(sec x tan x)
sin x cos x
5. y Sin1 2
2ax x2
6. y 2ax x2 a Cos1 a
7. y t2 Cos1 t
9. y Arctan(1 2x)
8. y Arcsin 2x
10. y Arccot(2x 5)2
x
11. y Arccot a
1
12. y Arcsec x
13. y Arccsc 2x
14. y Arcsin x
t
15. y t2 Arcsin 2
a
1
17. y Sec 2
a x2
x
16. y Sin1 2
1 x
Find the slope of the tangent to each curve.
1
18. y x Arcsin x at x 2
20. y x2 Arccsc x at x 2
Arctan x
19. y at x 1
x
x
x Arccot at x 4
21. y 4
22. Find the equations of the tangents to the curve y Arctan x having a slope of 14 .
33–4 Derivatives of Logarithmic Functions
Derivative of logb u
Let us now use the delta method again to find the derivative of the logarithmic function y logb u. We first let u take on an increment u and y an increment y.
982
Chapter 33
◆
Derivatives of Trigonometric, Logarithmic, and Exponential Functions
y logb u
y y logb (u u)
Subtracting gives us
u u
y logb (u u) logb u logb u
by the law of logarithms for quotients. Now dividing by u yields
y
u u
1
logb u
u u
We now do some manipulation to get our expression into a form that will be easier to evaluate.
We start by multiplying the right side by u/u.
y u 1
u u
logb u
u u u
u u
1 u
logb u u
u
Then, using the law of logarithms for powers (Eq. 189), we have
u/u
y 1
p u u q
logb u
u u
We now let u approach zero.
u/u
y dy
1
p u u q
lim lim logb u
u → 0 u
du u l 0 u
[
]
u q u/u
1
p
logb lim 1 u
u
u l 0
Let us simplify the expression inside the brackets by making the substitution
u
k u
Then k will approach infinity as u approaches zero, and the expression inside the brackets
becomes
Glance back at Sec. 20–2 where
we developed this expression.
p
u q
1 u
1
1 q
k→
u → 0
k
This limit defines the number e, the familiar base of natural logarithms.
lim
Definition
of e
u/u
lim
e lim
k→
p
p
k
1
1 q
k
k
185
Our derivative thus becomes
dy
1
du u
We are nearly finished now. Using the chain rule, we get dy/dx by multiplying dy/du by du/dx.
Thus,
logb e
Derivative of
logb u
d(logb u)
1
u
du
dx
logb e dx
362a
Or, since logb e 1/ln b,
This form is more useful when
the base b is a number other
than 10.
Derivative of
logb u
d(logb u)
1 du
u ln b dx
dx
362b
Section 33–4
◆
983
Derivatives of Logarithmic Functions
Example 15: Take the derivative of y log(x2 3x).
◆◆◆
Solution: The base b, when not otherwise indicated, is taken as 10. By Eq. 362a,
dy
1
(log e) (2x 3)
dx x2 3x
2x 3
log e
x2 3x
Or, since log e 0.4343,
dy
p 2x 3 q
0.4343 dx
x2 3x
◆◆◆
Example 16: Find the derivative of y x log3 x2.
◆◆◆
Solution: By the product rule,
1
y x p q (2x) log3 x2
2
x ln 3
2
log3 x2 1.82 log3 x2
ln 3
◆◆◆
Derivative of ln u
Our efforts in deriving Eqs. 362a and 362b will now pay off, because we can use that result
to find the derivative of the natural logarithm of a function, as well as the derivatives of exponential functions in the following sections. To find the derivative of y ln u we use Eq. 362a.
y ln u
dy
du
dx
1
u
ln e dx
But, by Eq. 192, ln e 1. Thus,
d(ln u)
Derivative
of ln u
1 du
u dx
dx
363
The derivative of the natural logarithm of a function is the reciprocal of that function multiplied by its derivative.
Example 17: Differentiate y ln(2x3 5x).
◆◆◆
Solution: By Eq. 363,
dy
1
dx 2x 5x
6x2 5
◆◆◆
2x3 5x
The rule for derivatives of the logarithmic function is often used with other rules for
derivatives.
(6x2 5)
3
Example 18: Take the derivative of y x3 ln(5x 2).
◆◆◆
Solution: Using the product rule together with our rule for logarithms gives us
5x3
1 q
◆◆◆
dx
5x 2
5x 2
Our work is sometimes made easier if we first use the laws of logarithms to simplify a given
expression.
dy
p
x3 5 [ln (5x 2)] 3x2 3x2 ln (5x 2)
984
Chapter 33
◆
Derivatives of Trigonometric, Logarithmic, and Exponential Functions
x2x 3
Example 19: Take the derivative of y ln .
3 4x 1
◆◆◆
Solution: We first rewrite the given expression using the laws of logarithms.
1
1
y ln x ln(2x 3) ln(4x 1)
2
3
We now take the derivative term by term.
dy 1 1 p 1 q
1p 1 q
2 4
x
2 2x 3
3 4x 1
dx
1
1
4
x 2x 3 3(4x 1)
◆◆◆
Logarithmic Differentiation
Here we use logarithms to aid in differentiating nonlogarithmic expressions. Derivatives of
some complicated expressions can be found more easily if we first take the logarithm of both
sides of the given expression and simplify by means of the laws of logarithms. (These operations will not change the meaning of the original expression.) We then take the derivative.
◆◆◆
3
x 2 x 3
Example 20: Differentiate y .
4
x1
We could instead take the
common log of both sides, but
the natural log has a simpler
derivative.
Solution: We will use logarithmic differentiation here. Instead of proceeding in the usual way,
we first take the natural log of both sides of the equation and apply the laws of logarithms.
1
1
1
ln y ln (x 2) ln (x 3) ln (x 1)
2
3
4
Taking the derivative, we have
1 dy 1 p 1 q 1 p 1 q 1 p 1 q
y dx 2 x 2
3 x3
4 x1
We could now use the original
expression to replace the y in the
answer.
Finally, multiplying by y to solve for dy/dx gives us
dy
[
1
2 (x 2)
1
3 (x 3)
1
4 (x 1)
y dx
]
◆◆◆
In other cases, this method will allow us to take derivatives not possible with our other
rules.
Example 21: Find the derivative of y x2x.
◆◆◆
Solution: This is not a power function because the exponent is not a constant. Nor is it an exponential function because the base is not a constant. So neither Rule 345 nor 364 applies. But let
us use logarithmic differentiation by first taking the natural logarithm of both sides.
ln y ln x2x 2x ln x
Now taking the derivative by means of the product rule yields
1 dy
y dx
p
1q
x
2 2 ln x
2x (ln x)2
Multiplying by y, we get
dy
2 (1 ln x)y
dx
Replacing y by x2x gives us
dy
2 (1 ln x)x2x
dx
◆◆◆
Section 33–4
◆
Exercise 4
◆
985
Derivatives of Logarithmic Functions
Derivatives of Logarithmic Functions
Derivative of logb u
Differentiate.
1. y log 7x
2. y log x2
3. y logb x3
4. y loga(x2 3x)
5. y log (x5 6x )
2
7. y x log x
1
6. y loga p q
2x 5
(1 3x)
8. y log x2
Derivative of ln u
Differentiate.
9. y ln 3x
10. y ln x3
11. y ln(x2 3x)
12. y ln(4x x3)
13. y 2.75x ln 1.02x3
14. w z2 ln(1 z2)
ln (x 5)
15. y x2
ln x2
16. y 3 ln (x 4)
17. s ln t 5
18. y 5.06 ln x2 3.25x
With Trigonometric Functions
Differentiate.
19. y ln sin x
20. y ln sec x
21. y sin x ln sin x
22. y ln(sec x tan x)
Implicit Relations
Find dy/dx.
23. y ln y cos x 0
24. ln x2 2x sin y 0
25. x y ln(x y)
x
26. xy = a2 ln a
27. ln y x 10
Logarithmic Differentiation
Differentiate.
x 2
28. y = 3 2 x
30. y xx
31. y xsin x
32. y (cot x)sin x
33. y (Cos1 x)x
a2 x2
29. y = x
Remember to start these by
taking the logarithm of both
sides.
986
Chapter 33
◆
Derivatives of Trigonometric, Logarithmic, and Exponential Functions
Tangent to a Curve
Find the slope of the tangent at the point indicated.
34. y log x at x 1
35. y ln x where y 0
2
37. y log(4x 3) at x 2
36. y ln(x 2) at x 4
38. Find the equation of the tangent to the curve y ln x at y 0.
Angle of Intersection
Find the angle of intersection of each pair of curves.
39. y ln(x 1) and y ln(7 2x)
40. y x ln x and y x ln(1 x)
at x 2
1
at x 2
Extreme Values and Points of Inflection
Find the maximum, minimum, and inflection points for each curve.
41. y x ln x
x
43. y ln x
42. y = x3 ln x
44. y ln(8x x2)
Newton’s Method
Find the smallest positive root between x 0 and 10 to two decimal places.
45. x 10 log x 0
46. tan x log x 0
Applications
r1
r2
FIGURE 33–16 Insulated
pipe.
47. A certain underwater cable has a core of copper wires covered by insulation. The speed of
transmission of a signal along the cable is
1
S x2 ln x
where x is the ratio of the radius of the core to the thickness of the insulation. What value
of x gives the greatest signal speed?
48. The heat loss q per foot of cylindrical pipe insulation (Fig. 33–16) having an inside radius
r1 and outside radius r2 is given by the logarithmic equation
2k(t1t2)
q kJh2
ln(r2/r1)
Hint: Treat B2 as a constant in
problem 50.
where t1 and t2 are the inside and outside temperatures (°C) and k is the conductivity of the
insulation. A 10-cm-thick insulation having a conductivity of 0.04 is wrapped around a
30-cm-diameter pipe at 282°C, and the surroundings are at 32°C. Find the rate of change
of heat loss q if the insulation thickness is decreasing at the rate of 0.3 cmh.
49. The pH value of a solution having a concentration C of hydrogen ions is pH log10 C.
Find the rate at which the pH is dropping when the concentration is 2.0 105 and decreasing at the rate of 5.5 105 per minute.
50. The difference in elevation, in metres, between two locations having barometric readings
of B1 and B2 mm of mercury is given by h 18 413 log B2B1, where B1 is the pressure at
the upper location. At what rate is an airplane changing in elevation when the barometric
pressure outside the airplane is 546 mm of mercury and decreasing at the rate of 12.7 mm
per minute?
_
51. The power input to a certain amplifier is 2.0 W; the power output is 400 W but because of
a defective component is dropping at the rate of 0.50 W per day. Use Eq. A104 to find the
rate (decibels per day) at which the decibels are decreasing.
Section 33–5
◆
987
Derivatives of the Exponential Function
33–5 Derivatives of the Exponential Function
Derivative of bu
We seek the derivative of the exponential function y bu, where b is a constant and u is a function of x. We can get the derivative without having to use the delta method by using the rule we
derived for the logarithmic function (Eq. 363). We first take the natural logarithm of both sides.
y bu
ln y ln bu u ln b
by Eq. 189. We now take the derivative of both sides, remembering that ln b is a constant.
1 dy
y dx
du
dx
dy
du
dx
ln b
Multiplying by y, we have
y ln b
dx
Finally, replacing y by bu gives us the following:
d(bu)
Derivative
of bu
du
dx
bu ln b
dx
364
The derivative of a base b raised to a function u is the product of bu, the derivative of the
function, and the natural log of the base.
◆◆◆
Example 22: Find the derivative of y 10x 2.
2
Solution: By Eq. 364,
dy
10x 2(2x) ln 10 2x(ln 10)10x 2
2
2
◆◆◆
dx
Common
Error
Do not confuse the exponential function y bx with the
power function y xn. The derivative of bx is not xbx1!
The derivative of bx is bx ln b.
Derivative of eu
We will use Eq. 364 mostly when the base b is the base of natural logarithms, e.
y eu
Taking the derivative by Eq. 364 gives us
dy
du
dx
= eu ln e
dx
But since ln e 1, we get the following:
Derivative
of eu
d
dx
du
dx
eu eu 365