2.4.5 Symbolic Differentiation of Exponential Functions: Other Bases

We know how to differentiate e 2 ⁢ t, e 3 ⁢ t, and e 5 ⁢ t, but what about 2 t, 3 t, and 5 t? The formula

d d t e k ⁢ t = k ⁢ e k ⁢ t

will help us answer this question.

Example 2   Differentiate 3 t.

Solution   We begin by writing \(3\) in the form e k. In fact, 3 = e k, where k = log e ⁢ 3. (That's the definition of log e ⁢ 3.) Now for this \(k\),

3 t = ( e k ) t = e k ⁢ t.

Thus,

d d t 3 t = d d t e k ⁢ t = k ⁢ e k ⁢ t = ( log e 3 ) e k ⁢ t = ( log e ⁢ 3 ) 3 t.

 

Checkpoint 5

There is nothing special about \(b = 3\). In general,

d d t b t = ( log e ⁢ b ) b t.

Notice that we have now solved a problem we left dangling earlier: how to calculate \(L (b)\) in the formula

d d t b t = L ( ⁢ b ) b t.

Specifically, L ( b ) = log e ⁢ b.

In Chapter 1 we reviewed logarithms with an arbitrary base \(b\). In this chapter we have already determined a special interest in the natural base \(e \). The corresponding logarithm is also called “natural.” Its abbreviated name is ln, which stands for “logarithm, natural” (but which is read “natural logarithm”). Thus ln ⁢ x = log e ⁢ x. At the risk of belaboring the obvious, we repeat the definition of logarithm for this important case.

Note, in particular, that if \(y = 1 \), then \(x\) must be \(e\), and if \(x\) is \(e\), then \(y\) must be \(1\). That is, \(\ln e = 1 \).