Page 169 - Calculus Demystified

CHAPTER 6
156
For all real numbers a and b we have Transcendental Functions
(a) exp(a + b) =[exp(a)]·[exp(b)].
exp(a)
(b) For any a and b we have exp(a − b) = .
exp(b)
These properties are verified just by exploiting the fact that the exponential is
the inverse of the logarithm, as we saw in Example 6.7.

EXAMPLE 6.8
Use the basic properties to simplify the expression
2 3
[exp(a)] ·[exp(b)]
.
4
[exp(c)]
SOLUTION
We calculate that

2 3
[exp(a)] ·[exp(b)] [exp(a)]·[exp(a)]·[exp(b)]·[exp(b)]·[exp(b)]
=
4
[exp(c)] [exp(c)]·[exp(c)]·[exp(c)]·[exp(c)]
exp(a + a + b + b + b)
=
exp(c + c + c + c)
= exp(a + a + b + b + b − c − c − c − c)
= exp(2a + 3b − 4c).

2
You Try It: Simplify the expression (exp a) −3 · (exp b) / exp(5c).

6.2.2 CALCULUS PROPERTIES OF THE EXPONENTIAL
Now we want to learn some “calculus properties” of our new function exp(x).
These are derived from the standard formula for the derivative of an inverse, as in
Section 2.5.1.
For all x we have
d
(exp(x)) = exp(x).
dx
In other words,

exp(x) dx = exp(x).

More generally,

d du
exp(u) = exp(u)
dx dx