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Jim Lamb ers Math 2B Fall Quarter 2004- Le ture 13 Notes
These notes orresp ond to Se tion 7.2* in the text.
The Natural Logarithmi Fun tion
De nition of the Natural Logarithmi Fun tion
Re all that the p ower rule for integration, Z xn^ dx = x
n+ n + 1
+ C ; (1)
do es not apply if n = 1, sin e then there is a division by zero. However, sin e 1 =t is ontinuous for all t > 0, then its de nite integral exists over any interval (a; b), where b > a > 0. We an therefore de ne a fun tion of x to b e the integral of 1 =t over an interval whose right endp oint dep ends on x.
De nition 1 (Natural Logarithmi Fun tion) The natural logarithmi fun tion ln x is de ned by
ln x =
Z (^) x
1
t
dt; x > 0 : (2)
Note that this fun tion an easily b e di erentiated using Part 1 of the Fundamental Theorem of Cal ulus: d dx
ln x = d dx
Z (^) x
1
t
dt = 1 x
From the hain rule, we obtain the following more general di erentiation rule.
Theorem 1 If g (x) is a positive, di erentiable fun tion, then
d dx
[ln g (x)℄ = g^
(^0) (x) g (x)
We now illustrate the usage of this rule with some examples.
Example 1 Compute the derivative of f (x) = ln(sin x) sin(ln x):
Solution Using the Pro du t Rule and Chain Rule, we obtain
f 0 (x) = sin(ln x)[ln(sin x)℄^0 + ln(sin x)[sin(ln x)℄^0 = sin(ln x) 1 sin x
os x + ln(sin x) os (ln x) 1 x = sin(ln x) ot x + ln(sin^ x)^ os^ (ln^ x) x
Example 2 Di erentiate f (x) = ln(xe^ + ln x):
Solution We have
f 0 (x) = 1 xe^ + ln x
(xe^ + ln x)^0 = 1 xe^ + ln x
exe ^1 + 1 x
Example 3 Di erentiate f (x) = ln(1 + x^2 ):
Solution From the Chain Rule, we have
f 0 (x) = (^1) +^1 x 2 (2x) = (^1) +^2 x x 2 : (7)
2
Prop erties of the Natural Logarithmi Fun tion
The natural logarithmi fun tion has some interesting prop erties, all of whi h an b e proven by applying the hain rule for di erentiation:
ln(xy ) = ln x + ln y (8)
ln
x y
= ln x ln y (9) ln(xr^ ) = r ln x (10)
In addition, ln x has the prop erties
xlim!1 ln^ x^ =^1 ;^ lim x! 0 +^
Furthermore, b e ause its derivative is always p ositive for x > 0, it an b e seen that ln x is one-to-one and therefore has an inverse fun tion.
This rule an b e used to address a onspi uous void in the list of integration rule we have develop ed. While we have simple rules for anti-di erentiating sin x and os x, we do not have su h rules for the other four basi trigonometri fun tions. We an now derive su h rules, as the following example illustrates.
Example 5 Compute (^) Z
tan x dx: (19)
Solution Writing tan x = sin x= os x and letting u = os x, we have du = sin x dx. It follows that Z tan x dx =
Z (^) sin x os x
dx
=
Z (^) du u = ln juj + C = ln j os xj + C = ln j os xj ^1 + C = ln 1 os x
+ C
= ln j se xj + C : (20)
As will b e seen in later le tures, similar integration rules an b e obtained for ot x, se x, and s x. 2
Logarithmi Di erentiation
The natural logarithmi fun tion an b e used to more easily di erentiate fun tions that are om- pli ated pro du ts or quotients, sin e logarithms of pro du ts and quotients are equal to sums and di eren es. For example, if f (x) = g (x)h(x), then
ln f (x) = ln g (x) + ln h(x): (21)
Di erentiating b oth sides, we obtain
f 0 (x) f (x)
= g^
(^0) (x) g (x)
(^0) (x) h(x)
whi h an then b e solved for f 0 (x).
Example 6 Given y = (sin x)x^2 +3x^ ; use logarithmi di erentiation to ompute y 0.
Solution Taking the natural logarithm of b oth sides of the ab ove equation, we have
ln y = ln(sin x)x^2 +3x^ = (x^2 + 3 x) ln(sin x): (23)
Di erentiating b oth sides impli itly with resp e t to x (i.e., treating y as a fun tion of x) yields, via the Pro du t Rule and the Chain Rule,
1 y
y 0 = ln(sin x)(x^2 + 3 x)^0 + (x^2 + 3 x)(ln(sin x))^0 = ln(sin x)(2x + 3) + (x^2 + 3 x) 1 sin x
os x: (24)
Multiplying through by y , we obtain
y 0 = y
h ln(sin x)(2x + 3) + (x^2 + 3 x) (^) sinos xx
i = (sin x)x^2 +3x^
ln(sin x)(2x + 3) + (x^2 + 3 x) ot x
Example 7 Compute the slop e of the tangent line to the graph of y = xx^ at x = e.
Solution To obtain the slop e of the tangent line, we need to ompute the derivative of xx^ at x = e. Taking the natural logarithm of b oth sides of the equation y = xx^ we obtain
ln y = ln xx^ = x ln x: (26)
Di erentiating b oth sides with resp e t to x (and rememb ering that y dep ends on x) yields, via the Pro du t Rule, 1 y y^
(^0) = ln x(x) (^0) + x(ln x) (^0) = ln x + x 1 x =^ ln^ x^ +^1 ;^ (27) and therefore y 0 = y (ln x + 1) = xx^ (ln x + 1): (28)
Substituting x = 1 into this equation yields y 0 = 11 (ln 1 + 1) = 1, and therefore the slop e of the tangent line at the p oint x = 1 is equal to 1. 2
The Inverse of The Natural Logarithmi Fun tion
It is interesting to note that the derivative of ln x is 1 =x, while the derivative of an inverse fun tion f ^1 (x) is equal to 1 =f 0 (f ^1 (x)). Supp ose that g (x) is the inverse fun tion of f (x) = ln x. Then, b e ause f 0 (x) = 1 =x, it follows that
g 0 (x) = (^11) g (x)
= g (x): (29)
In other words, the derivative of the inverse of ln x is equal to itself. This will b e dis ussed further in the next le ture.
