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
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. 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.
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.
Okay, I cheated. The number is, of course, e. Thus . 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. Then a = eln(a) and ax = (eln(a))x = e(ln(a)x) . Then using the Chain Rule the derivative is
Finally, going back to the first table above where a = 5, we find that the limit we found there 1.609438 = ln(5).
For a video on this topic click here.
For more on this see Logarithms