Calculus 2nd Edition Rogawski

SECTION 7.1 Derivative of f (x) = b x and the Number e 341
(c) Since 1 7 = 7−1 , the right-hand side of the equation is ( 1
7
) 2t = (7 −1 ) 2t = 7 −2t . The
equation becomes 7 t+1 = 7 −2t . Therefore, t + 1 = −2t, or t = − 1 3 .
We shall take for granted that f (x) = b x is
differentiable. Although the proof of this
fact is somewhat technical, it is plausible
because the graph of y = b x appears
smooth and without corners.
Derivative of f (x) = b x
At this point, it is natural to ask: What is the derivative of f (x) = b x ? Our rules of
differentiation are of no help because b x is neither a product, quotient, nor composite of
functions with known derivatives. We must go back to the limit definition of the derivative.
The difference quotient (for h ̸= 0) is
f (x + h) − f (x)
h
= bx+h − b x
h
= bx b h − b x
h
= bx (b h − 1)
h
Now take the limit as h → 0. The factor b x does not depend on h, so it may be taken
outside the limit:
d b x+h − b x b x (b h (
− 1)
b
dx bx = lim
= lim
= b x h )
− 1
lim
h→0 h h→0 h
h→0 h
This last limit (which exists because b x is differentiable) does not depend on x. We denote
its value by m(b):
( b h )
− 1
m(b) = lim
h→0 h
What we have shown, then, is that the derivative of b x is proportional to b x :
d
dx bx = m(b) b x 1
What is the factor m(b)? We cannot determine its exact value at this point (in Section 7.3,
we will learn that m(b) is equal to ln b, the natural logarithm of b). To proceed further,
let’s investigate m(b) numerically.
EXAMPLE 3 Estimate m(b) numerically for b = 2, 2.5, 3, and 10.
Solution We create a table of values of difference quotients to estimate m(b):
h
2 h − 1
h
(2.5) h − 1
h
3 h − 1
h
10 h − 1
h
0.01 0.69556 0.92050 1.10467 2.32930
0.001 0.69339 0.91671 1.09921 2.30524
0.0001 0.69317 0.91633 1.09867 2.30285
0.00001 0.69315 0.916295 1.09861 2.30261
m(2) ≈ 0.69 m(2.5) ≈ 0.92 m(3) ≈ 1.10 m(10) ≈ 2.30
These computations suggest that m(b) is an increasing function of b. In fact, it can
be shown that m(b) is both increasing and continuous as a function of b (we shall take
these facts for granted). Then, since m(2.5) ≈ 0.92 and m(3) ≈ 1.10, there exists a unique
number b between 2.5 and 3 such that m(b) = 1. This is the number e, whose value is
approximately 2.718.
Using infinite series (Exercise 87 in Section 10.7), we can show that e is irrational
and we can compute its value to any desired accuracy.