# The Derivatives of Exponential Functions

Our problem for today is to differentiate ax with the (usual) restrictions that a is a positive number and not equal to 1. The reasoning here is very different from that for finding other derivatives and therefore I hope you and your students find it interesting.

The definition of derivative followed by a little algebra gives tells us that

$\displaystyle \frac{d}{dx}{{a}^{x}}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{a}^{x+h}}-{{a}^{x}}}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{a}^{x}}{{a}^{h}}-{{a}^{x}}}{h}={{a}^{x}}\underset{h\to 0}{\mathop{\lim }}\,\frac{{{a}^{h}}-1}{h}$.

Since the limit in the expression above is a number, we observe that the derivative of ax is proportional to ax. And also, each value of a gives a different constant. For example if a = 5 then the limit is approximately 1.609438, and so $\displaystyle \frac{d}{{dx}}{{5}^{x}}\approx \left( {1.609438} \right){{5}^{x}}$.

I determined this by producing a table of values for the expression in the limit near x = 0. You can do the same using a good calculator, computer, or a spreadsheet.

h                            $\frac{{{5}^{h}}-1}{h}$

-0.00000030            1.60943752

-0.00000020            1.60943765

-0.00000010             1.60943778

0.00000000             undefined

0.00000010             1.60943804

0.00000020             1.60943817

0.00000030             1.60943830

That’s kind of messy and would require us to find this limit for whatever value of a we were using. It turns out that by finding the value of a for which the limit is 1 we can fix this problem. Your students can do this for themselves by changing the value of a in their table until they get the number that gives a limit of 1.

Okay, that’s going to take a while, but may be challenging. The answer turns out to be close to 2.718281828459045…. Below is the table for this number.

h                            $\frac{{{a}^{h}}-1}{h}$

-0.00000030            0.99999985

-0.00000020            0.99999990

-0.00000010            0.99999995

0.00000000            undefined

0.00000010             1.00000005

0.00000020             1.00000010

0.00000030             1.00000015

Okay, I cheated. The number is, of course, e. Thus,

$\displaystyle \frac{d}{{dx}}{{e}^{x}}={{e}^{x}}\left( {\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{{{e}^{h}}-1}}{h}} \right)={{e}^{x}}(1)={{e}^{x}}$.

The function ex is its own derivative!

And from this we can find the derivatives of all the other exponential functions. First, we define a new function (well maybe not so new) which is the inverse of the function ex called ln(x), the natural logarithm of x. (For more on this see Logarithms.) Then a = eln(a) and ax = (eln(a))x = e(ln(a)x) . Then using the Chain Rule the derivative is

$\frac{d}{dx}{{a}^{x}}={{e}^{(\ln (a))x}}\ln (a)={{\left( {{e}^{\ln (a)}} \right)}^{x}}\ln (a)$

$\frac{d}{dx}{{a}^{x}}={{a}^{x}}\ln \left( a \right)$

Finally, going back to the first table above where a = 5, we find that the limit we found there 1.609438 = ln(5).

Revised 8-28-2018, 6-2-2019

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