Exponential and Log, Logarithmic Differentiation

Background
We always assume the base, a, is greater than 1. a0 = 1; a1 = a; a2 = aa; ...
·
ax1+x2 = ax1ax2 (ax1 )x2 = ax1x2
p qq
a = ?ap (where p and q are integers)
r
To define afor real numbers r, fill in by continuity.
d
Today’s main task: find ax
dx
We can write
dax+?xx x
a = lim ? a
dx ?x0 ?x
?
We can factor out the ax:
x+?xx ?x ?x
lim a? a= lim ax a? 1= ax lim a? 1
?x0 ?x ?x0 ?x ?x0 ?x
???
Let’s call M(a) ? lim a?x ? 1
?x0 ?x
?
We don’t yet know what M(a) is, but we can say d
ax = M(a)ax
dx Here are two ways to describe M(a): d
1. Analytically M(a)= ax at x = 0.
dx
Indeed, M(a) = lim a0+?x ? a0 = dax
?x0 ?x dx
?x=0
1
Lecture 6 18.01 Fall 2006
M(a)
(slope of ax at x=0)
ax
Figure 1: Geometric definition of M(a)
2. Geometrically, M(a) is the slope of the graph y = ax at x = 0.
The trick to figuring out what M(a) is is to beg the question and define e as the number such
that M(e) = 1. Now can we be sure there is such a number e? First notice that as the base a
increases, the graph ax gets steeper. Next, we will estimate the slope M(a) for a = 2 and a = 4
geometrically. Look at the graph of 2x in Fig. 2. The secant line from (0, 1) to (1, 2) of the graph
y = 2x has slope 1. Therefore, the slope of y = 2x at x = 0 is less: M(2) < 1 (see Fig. 2).
1 1
Next, look at the graph of 4x in Fig. 3. The secant line from (?2, 2
) to (1, 0) on the graph of
y = 4x has slope 1. Therefore, the slope of y = 4x at x = 0 is greater than M(4) > 1 (see Fig. 3).
Somewhere in between 2 and 4 there is a base whose slope at x = 0 is 1.