Page 179 - Calculus Demystified

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SOLUTION CHAPTER 6 Transcendental Functions
We have

1
(log 9) · (log 15) = · log 16 = log 16 = 2.
9
4
4
9
log 4
9
6.4 Calculus with Logs andExponentials to
Arbitrary Bases

6.4.1 DIFFERENTIATION AND INTEGRATION OF
log x AND a x
a
We begin by noting these facts:
If a> 0 then
x
d a
x
x
x
(i) a = a · ln a; equivalently, a dx = + C.
dx ln a
d 1
(ii) (log x) =
a
dx x · ln a
Math Note: As always, we can state these last formulas more generally as
d u u du
a = a · · ln a
dx dx
and
d 1 du 1
log u = · · .
a
dx u dx ln a
EXAMPLE 6.25
Calculate
d x d cos x d d
(5 ), (3 ), (log x), (log (x · cos x)).
4
8
dx dx dx dx
SOLUTION
We see that
d x x
(5 ) = 5 · ln 5.
dx
For the second problem, we apply our general formulation with a = 3, u =
cos x to obtain

d cos x cos x d cos x
(3 ) = 3 · cos x · ln 3 = 3 · (− sin x) · ln 3.
dx dx