Return now to the example of that bacteria that doubled its numbers
every 60 minutes. Remember that after 10 hours, a single bacterium
had multiplied into 1024 (that is `2 ^{10}`).
In the 11th hour, we expected that the
population would increase from 1024 to 2048 (that is to

In fact, in any hour, the net increase is we expect precisely
the amount there were at the beginning of the hour. But what
is the increase we expect in, say, half an hour? If the population
grows in 1 hour to `2 ^{1}` times what it was, then
in half an hour, we expect it to grow to

(1024 × 1.41421356) - 1024 = 1024 × (1.41421356 - 1) = 424.15What I'd like you to notice is not the

And what about the net increase in a quarter-hour period? We expect
that if there are `n` at the beginning of the quarter-hour,
then at the end of the quarter-hour there ought to be
`n × 2 ^{0.25}`

Remember also that we determined that if the bacteria colony is allowed
to grow unchecked for several days, the population would grow so large
that cell divisions would be taking place in even the shortest imaginable
time intervals. Still you would find that even over a millisecond, the
net increase in population would continue to be in direct proportion to
the population at the start of that millisecond. One millisecond is
`0.0000002778` hours. So if there are `n` bacteria at
the beginning of a millisecond, there should be
`n × 2 ^{0.00000002778}`

This is the operative property of all exponentials: *the rate of increase
is always in direct proportion to the exponential itself*. And in place
of the phrase *rate of increase* we can substitute the word
*derivative*. We have demonstrated the truth of this informally with
our discussion of the bacteria colony. We now go on to demonstrate it
mathematically. **Warning: The derivation that follows is likely to be on the
exam.**

Recall from

df f(x + h) - f(x) f'(x) =If you wanted to know the derivative of the function~~= lim~~~~eq. 4.1-2 dx h~~~~> 0 h~~

f(x) = 2for example, you would substitute^{x}eq. 6.2-1

2By taking this limit as^{x+h}- 2^{x}f'(x) = lim~~eq. 6.2-2 h~~~~> 0 h~~

But how do you take that limit? First we can apply the basic rule
of exponents that we developed in the last section. Notice that we
have a term, `2 ^{x+h}`. And recall we had a rule about
the sum of exponents.

2If you substitute this in for^{x+h}= 2^{x}× 2^{h}eq. 6.2-3

(2Notice that the two terms in the numerator have a common factor of^{x}× 2^{h}) - 2^{x}f'(x) = lim~~eq. 6.2-4 h~~~~> 0 h~~

2Remember also the product rule for limits (that is,^{h}- 1 f'(x) = lim~~× 2~~^{x}eq. 6.2-5 h~~> 0 h~~

2Look carefully at the two limits in eq. 6.2-6. The limit on the left has no^{h}- 1 f'(x) = lim~~× lim 2~~^{x}eq. 6.2-6 h~~> 0 h h~~~~> 0~~

2This makes it quite clear that the derivative of^{h}- 1 f'(x) = 2^{x}× lim~~eq. 6.2-7 h~~~~> 0 h~~

2The table seems to indicate that limit exists for the limit on the right-hand side of 6.2-7, at least to 5 places beyond the decimal point. Assuming that it does, we can say that the derivative of^{h}- 1 h~~h 0.1 0.71773... 0.01 0.69555... 0.001 0.69339... 0.0001 0.69317... 0.00001 0.69314... 0.000001 0.69314... Table 6.2-1~~

And we needn't restrict ourselves to
`2 ^{x}`. Pick any positive real number,

bAgain, the derivative of^{x+h}- b^{x}f'(x) = lim~~eq. 6.2-8 h~~~~> 0 h (b~~^{x}× b^{h}) - b^{x}f'(x) = lim~~eq. 6.2-9 h~~~~> 0 h b~~^{h}- 1 f'(x) = lim~~× b~~^{x}eq. 6.2-10 h~~> 0 h b~~^{h}- 1 f'(x) = lim~~× lim b~~^{x}eq. 6.2-11 h~~> 0 h h~~~~> 0 b~~^{h}- 1 f'(x) = b^{x}× lim~~eq. 6.2-12 h~~~~> 0 h~~

This limit of

bis so important that we give it a special name. We call it^{h}- 1 lim~~eq. 6.2-13 h~~~~> 0 h~~

f'(x) = bwhich is simply the the exponential,^{x}× ln(b) eq. 6.2-14

Figure 6.2-1 shows a plot of ` f(x) = 2 ^{x}`

- The derivative of an exponential is
*always*that same exponential times a constant. - The constant is always the
*natural log*of the base (the base,`b`, being the positive real number to which the exponent is applied -- e.g.`b`).^{x} - For the purposes of this tutorial, we define the
*natural log*of`b`according to the limit given in 6.2-13 (and we abrieviate it as`ln(b)`).

**1)** Let `a` and `b` be any positive real numbers.
Apply the product rule to

f(x) = ato determine its derivative. Use the shorthand,^{x}b^{x}

f(x) = (ab)What special property of^{x}

**2)** Use the chain rule to determine
the derivative of

f(x) = bwhere^{cx}

bFind the derivative of the right-hand expression above. What other special property of^{cx}= (b^{c})^{x}

**3)** Take everything you did in the second problem and simply
substitute the expression, `1/ln(b)`, for every occurrence you
find of
`c`. Take whatever cancellations you can. What unusual and
mathematically interesting relationship do you now find between the
resulting function and its derivative? When you see an interesting
property and you think it's the one I'm after,
click here.

**4)** Use the product rule to determine
the derivative of

xwhere^{2}f(x) =~~b~~^{x}2

**5)** Use the chain rule to determine
the derivative of

f(x) = bwhere^{-x2}

**6)** Let `b` be a positive real number and let `n` be
a counting number.
Suppose you didn't know what a particular `f(x)` was, but
you knew that the equation,

balways holds. Use implicit differentiation (which is really just an application of the chain rule) to find the derivative of^{f(x)}= x^{n}

Move on to Be Frugal and Add (logrithms)

email me at *hahn@netsrq.com*