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 text(.)` 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 text[(] b text[)]` 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 text(.)` 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 text(,)` then `x` must be `e`, and if `x` is `e`, then `y` must be `1`. That is, `text(ln) e = 1 text(.)`