# Untitled Page 10

- Page ID
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## 2.5 Exponentials and logarithms

### The exponential

The exponential function *e*^{x}, where *e*=2.71828… is the base of natural
logarithms, comes constantly up in applications as diverse as credit-card interest, the growth of
animal populations, and electric circuits. For its derivative we have

The second factor, (*e*^{dx}-1)/*dx*, doesn't have *x* in it, so it
must just be a constant. Therefore we know that the derivative of *e*^{x} is simply
*e*^{x}, multiplied by some unknown constant,

A rough check by graphing at, say *x*=0, shows that the slope is close to 1, so *c* is close to
1. Numerical calculation also shows that, for example, (*e*^{0.001}-1)/0.001=1.00050016670838 is very close
to 1. But how do we know it's exactly one when *dx* is really infinitesimal?
We can use Inf:

: [exp(d)-1]/d 1+0.5d+...

(The ... indicates where I've snipped some higher-order terms out of the output.) It seems clear
that *c* is equal to 1 except for negligible terms involving higher powers of *dx*.
A rigorous proof is given on page 151.

##### Example 14

◊ The concentration of a foreign substance in the bloodstream generally falls off exponentially with time as*c*=

*c*

_{o}

*e*

^{-t/a}, where

*c*

_{o}is the initial concentration, and

*a*is a constant. For caffeine in adults,

*a*is typically about 7 hours. An example is shown in figure k. Differentiate the concentration with respect to time, and interpret the result. Check that the units of the result make sense.

◊ Using the chain rule,

This can be interpreted as the rate at which caffeine is being removed from the blood and put into the person's urine. It's
negative because the concentration is decreasing.
According to the original expression for *x*, a substance with a large *a* will take a long time to reduce its concentration,
since *t*/*a* won't be very big unless we have large *t* on top to compensate for the large *a* on the bottom.
In other words, larger values of *a* represent substances that the body has a harder time getting rid of efficiently.
The derivative has *a* on the bottom, and the interpretation of this is that for a drug that is hard to eliminate,
the rate at which it is removed from the blood is low.

It makes sense that *a* has units of time, because the exponential function has to have a unitless argument, so the units
of *t*/*a* have to cancel out. The units of the result come from the factor of *c*_{o}/*a*, and
it makes sense that the units are concentration divided by time, because the result
represents the rate at which the concentration is changing.

##### Example 15

◊ Find the derivative of the function *y*=10^{x}.

◊ In general, one of the tricks to doing calculus is to rewrite functions in forms that
you know how to handle. This one can be rewritten as a base-*e* exponent:

Applying the chain rule, we have the derivative of the exponential, which is just the same exponential, multiplied by the derivative of the inside stuff:

In other words, the “*c*” referred to in the discussion of the derivative of *e*^{x} becomes
*c*=ln 10 in the case of the base-10 exponential.

### The logarithm

The natural logarithm is the function that undoes the exponential. In a situation like this, we have

where on the left we're thinking of *y* as a function of *x*, and on the right we consider
*x* to be a function of *y*. Applying this to the natural logarithm,

This is noteworthy because it shows that there must be an exception to the rule
that the derivative of *x*^{n} is *nx*^{n-1}, and the integral of *x*^{n-1} is
*x*^{n}/*n*. (On page 37 I remarked that this rule could
be proved using the product rule for negative integer values of *k*, but that I would
give a simpler, less tricky, and more general proof later. The proof is example
24 below.) The integral of *x*^{-1} is not *x*^{0}/0, which wouldn't make sense anyway
because it involves division by zero.^{5} Likewise the derivative of
*x*^{0}=1 is 0*x*^{-1}, which is zero. Figure l shows the idea.
The functions *x*^{n} form a kind of ladder, with differentiation taking us down
one rung, and integration taking us up. However, there are two special cases where differentiation
takes us off the ladder entirely.

##### Example 16

◊ Prove d(*x*

^{n})/

*dx*=

*nx*

^{n-1}for any real value of

*n*, not just an integer.

◊

(For *n*=0, the result is zero.)

When I started the discussion of the derivative of the logarithm, I wrote *y*=ln *x* right
off the bat. That meant I was implicitly assuming *x* was positive.
More generally, the derivative of ln|*x*| equals 1/*x*, regardless of the sign (see
problem 29 on page 50).