Page 164 - Calculus Demystified

CHAPTER 6
SOLUTION Transcendental Functions 151
For the first problem, we let u = 4 + x and du/dx = 1. Therefore we have
d 1 d 1
ln(4 + x) = · (4 + x) = .
dx 4 + x dx 4 + x
Similarly,

2
d 3 1 d 3 3x − 1
ln(x − x) = · (x − x) =
3
3
dx x − x dx x − x
d 1 d − sin x
ln(cos x) = · (cos x) =
dx cos x dx cos x
d 5 4 d 4 1 5(ln x) 4
[(ln x) ]= 5(ln x) · (ln x) = 5(ln x) · =
dx dx x x

d d d
[(ln x) · (cot x)]= ln x · (cot x) + (ln x) · cot x
dx dx dx
1
2
= · cot x + (ln x) · (− csc x).
x
2
3
You Try It: What is the derivative of the function ln(x + x ) ?
Now we examine the graph of y = ln x. Since
d 1
(i) (ln x) = > 0,
dx x
1
d 2 d 1
(ii) (ln x) = =− < 0,
dx dx x x 2
(iii) ln(1) = 0,

we know that ln x is an increasing, concave down function whose graph passes
through (1, 0). There are no relative maxima or minima (since the derivative is
n
never 0). Certainly ln .9 < 0; the formula ln(.9 ) = n ln .9 therefore tells us that
+
ln x is negative without bound as x → 0 . Since ln x =− ln(1/x), we may also
conclude that ln x is positive without bound as x →+∞. A sketch of the graph of
y = ln x appears in Fig. 6.2.
We learned in the last paragraph that the function ln x takes negative values which
are arbitrarily large in absolute value when x is small and positive. In particular,
the negative y axis is a vertical asymptote. Since ln(1/x) =− ln x, we then find
that ln x takes arbitrarily large positive values when x is large and positive. The
graph exhibits these features.
Since we have only defined the function ln x when x> 0, the graph is only
sketched in Fig. 6.2 to the right of the y-axis. However it certainly makes sense to