Exponentials and logarithms The exponential

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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 $\begin{align*} \frac{de^x}{dx} &=\frac{e^{x+dx}-e^x}{dx} \\ &=\frac{e^xe^{dx}-e^x}{dx} \\ &= e^x\frac{e^{dx}-1}{dx} \end{align*}$

The second factor, $(e^{dx}-1)/dx$ , doesn't have 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,
$\frac{de^x}{dx}=ce^x$

A rough check by graphing at, say x=0 , shows that the slope is close to 1, so 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 is equal to 1 except for negligible terms involving higher powers of . A rigorous proof is given in Derivative of ex.

Example

The concentration of a foreign substance in the bloodstream generally falls off exponentially with time as $c=c_0e^{-t/a}$ , where c_0 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 2.11. Differentiate the concentration with respect to time, and interpret the result. Check that the units of the result make sense.
Using the chain rule, $\begin{align*} \frac{dc}{dt} &=c_0e^{-t/a}\cdot \left ( -\frac{1}{a} \right ) \\ &= -\frac{c_0}{a}e^{-t/a} \end{align*}$ 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 , 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 on top to compensate for the large 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_0/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.

Figure 2.11

A typical graph of the concentration of caffeine in the blood, in units of milligrams per liter, as a function of time, in hours.

Example

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: $\begin{align*} y &=10^x \\ \textrm{In} y &=\textrm{In} (10^x) \\ \textrm{In} y &=x \textrm{In} 10 \\ y&=e^{x \textrm{ In} 10} \end{align*}$

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:

$\frac{dy}{dx}=e^{x \textrm{In} 10}\cdot \textrm{In} 10$ In other words, the “c” referred to in the discussion of the derivative of e^x becomes $c= \textrm{In} 10$ in the case of the base-10 exponential.

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Exponentials and logarithms The exponential

Example