Exponential Derivatives and Statistics

This post gives a different way I developed years ago to determine the form of the derivative of exponential functions, y=b^x.  At the end, I provide a copy of the document I use for this activity in my calculus classes just in case that’s helpful.  But before showing that, I walk you through my set-up and solution of the problem of finding exponential derivatives.


I use this lesson after my students have explored the definition of the derivative and have computed the algebraic derivatives of polynomial and power functions. They also have access to TI-nSpire CAS calculators.

The definition of the derivative is pretty simple for polynomials, but unfortunately, the definition of the derivative is not so simple to resolve for exponential functions.  I do not pretend to teach an analysis class, so I see my task as providing strong evidence–but not necessarily a watertight mathematical proof–for each derivative rule.  This post definitely is not a proof, but its results have been pretty compelling for my students over the years.

Sketching Derivatives of Exponentials:

At this point, my students also have experience sketching graphs of derivatives from given graphs of functions.  They know there are two basic graphical forms of exponential functions, and conclude that there must be two forms of their derivatives as suggested below.

When they sketch their first derivative of an exponential growth function, many begin to suspect that an exponential growth function might just be its own derivative.  Likewise, the derivative of an exponential decay function might be the opposite of the parent function.  The lack of scales on the graphs obviously keep these from being definitive conclusions, but the hypotheses are great first ideas.  We clearly need to firm things up quite a bit.

Numerically Computing Exponential Derivatives:

Starting with y=10^x, the students used their CASs to find numerical derivatives at 5 different x-values.  The x-values really don’t matter, and neither does the fact that there are five of them.  The calculators quickly compute the slopes at the selected x-values.

Each point on f(x)=10^x has a unique tangent line and therefore a unique derivative.  From their sketches above, my students are soundly convinced that all ordered pairs \left( x,f'(x) \right) form an exponential function.  They’re just not sure precisely which one. To get more specific, graph the points and compute an exponential regression.

So, the derivatives of f(x)=10^x are modeled by f'(x)\approx 2.3026\cdot 10^x.  Notice that the base of the derivative function is the same as its parent exponential, but the coefficient is different.  So the common student hypothesis is partially correct.

Now, repeat the process for several other exponential functions and be sure to include at least 1 or 2 exponential decay curves.  I’ll show images from two more below, but ultimately will include data from all exponential curves mentioned in my Scribd document at the end of the post.

The following shows that g(x)=5^x has derivative g'(x)\approx 1.6094\cdot 5^x.  Notice that the base again remains the same with a different coefficient.

OK, the derivative of h(x)=\left( \frac{1}{2} \right)^x causes a bit of a hiccup.  Why should I make this too easy?  <grin>

As all of its h'(x) values are negative, the semi-log regression at the core of an exponential regression is impossible.  But, I also teach my students regularly that If you don’t like the way a problem appears, CHANGE IT!  Reflecting these data over the x-axis creates a standard exponential decay which can be regressed.

From this, they can conclude that  h'(x)\approx -0.69315\cdot \left( \frac{1}{2} \right)^x.

So, every derivative of an exponential function appears to be another exponential function whose base is the same as its parent function with a unique coefficient.  Obviously, the value of the coefficient depends on the base of the corresponding parent function.  Therefore, each derivative’s coefficient is a function of the base of its parent function.  The next two shots show the values of all of the coefficients and a plot of the (base,coefficient) ordered pairs.

OK, if you recognize the patterns of your families of functions, that data pattern ought to look familiar–a logarithmic function.  Applying a logarithmic regression gives

For y=a+b\cdot ln(x), a\approx -0.0000067\approx 0 and b=1, giving coefficient(base) \approx ln(base).

Therefore, \frac{d}{dx} \left( b^x \right) = ln(b)\cdot b^x.

Again, this is not a formal mathematical proof, but the problem-solving approach typically keeps my students engaged until the end, and asking my students to  discover the derivative rule for exponential functions typically results in very few future errors when computing exponential derivatives.

Feedback on the approach is welcome.

Classroom Handout:

Here’s a link to a Scribd document written for my students who use TI-nSpire CASs.  There are a few additional questions at the end.  Hopefully this post and the document make it easy enough for you to adapt this to the technology needs of your classroom.  Enjoy.


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