2.4.2 Derivatives of Exponential Functions

Our discussion of interest rates and bank balances indicates that the functions we seek, functions with growth rate proportional to the amount present, might be exponential functions with constant base and variable exponent. To check this, we want to find the derivative of an exponential function of the form b t. The base might be \(2\) (if we want to talk about things doubling) or \(10\) (to invert the common logarithm function discussed in Chapter 1) or \(1.011\) (if we want to talk about bank balances) or any other constant that makes sense as a base for exponentiation. As we did in finding the derivative of k ⁢ t 2, we will approach this task first from a graphical point of view (with the aid of computer tools) and then from a numerical point of view (on the next page with calculations performed by an applet).

In our study of functions \(f\) of the form f ( t ) = k ⁢ t 2, we saw that we could get a very good approximation to the derivative function at every instant by calculating difference quotients Δ ⁢ f / Δ ⁢ t with a small value of \(\Delta t\). We use the same idea now to see what we can learn about the derivative of an exponential function.

We write f ( t ) = b t in order to have a name for the exponential function with base \(b\), and we write \(g(t)\) for the difference quotient of \(f\) with a suitably small Δ ⁢ t, say, Δ ⁢ t\(=0.001\):

g ( t ) = Δ ⁢ f Δ ⁢ t = f ( t + 0.001 ) - f ( t ) 0.001 .

This new function \(g\)(\(t\)) is not the same function as f ′ ( t ), but its value at each \(t\) should be very close to f ′ ( t ). Thus a graph of \(g(t)\) should closely approximate a graph of f ′ ( t ).

Recall that our objective is to show that the derivative of f ( t ) = b t is proportional to f ( t ) itself. We can write this symbolically as f ′ ( t ) = c ⁢ f ( t ) for some constant \(c\), where the value of \(c\) depends on the value of \(b\).

Activity 2   We can determine whether f ′ ( t ) = c ⁢ f ( t ) for some constant \(c\) by asking whether the ratio f ′ ( t ) / f ( t ) is a constant function. And since \(g(t)\) approximates f ′ ( t ), we can examine that question graphically by determining whether the graph of  g ( t ) / f ( t )   looks approximately constant. Use our Graph tool or your graphing calculator to graph this ratio for \(b = 2, 3,\) and \(4\). What do your graphs tell you?

Comment on Activity 2

The constant \(c\) in the formula \(f\,'(t)= cf(t)\) is not the same for every choice of \(b\). For \(b\) \(= 2, 3,\)or \(4\), \(c\) is approximately \(0.7, 1.1,\)or \(1.4\), respectively, which you can check with your graphing tool by observing where the horizontal lines cross the \(y\)-axis.

Checkpoint 2

There appears to be some (nonconstant) functional relationship between b and c. To indicate the dependence of \(c\) on \(b\), we write \(c = L(b)\), and we rewrite the formula f ′ ( t ) = c ⁢ f ( t ) in the form

d d t b t = L ( b ) b t .

Let's be clear about what this formula says. For each fixed choice of base \(b\), there is a constant, which we have just named \(L(b)\), that is the proportionality constant relating the derivative to the exponential function.

Observation   In each of the six cases that you graphed in Activity 2 and Checkpoint 2, the horizontal line \(y = L(b)\) has the same \(y\)-intercept as the graph of the (approximate) derivative. This can't be an accident!

Activity 3

1. Use a graphing tool to find the value of d d t b t at t = 0 for b = 2, 3, and 4. Compare these numbers with L ( 2 ), L ( 3 ), and L ( 4 ).
2. Use the equation d d t b t = L ( b ) b t to explain carefully why L ( b ) must be the value of d d t b t at t = 0, no matter what the value of b is.

Comment on Activity 3

The observation in Activity 3 gives us a way to calculate L ( b ) for each value of b: Find the slope of the graph (or the instantaneous rate of change) of the function f ( t ) = b t at t = 0. With a little help from your graphing tool, you can do that in Checkpoint 3 for the particular case of b = 5.

Checkpoint 3