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So recently I was teaching high school calculus to a high school class and I was wondering about the pedagogically best way to make students actually understand why the derivatives of the exponential & logarithm functions are what they are (instead of just being told what they are). The students already knew some basic calculus including derivatives of power functions and rules of differentiation. The following is the approach that I came up with and I am curious to learn how it compares to alternative approaches and whether people think that those alternative approaches are of more pedagogical value, and why.

  1. Start by tabulating values and making precise drawings of $2^x$ and $3^x$, then draw tangents at $x=0$ and work out their slope with a ruler and calculator. You get approx. $0.7$ and $1.1$, so arguably there must be a value $2<a<3$ for which at $x=0$ the function $a^x$ has a tangent with slope exactly $1$. Call this value Euler's number $e$ (thereby establishing a working definition of $e$) and consider the corresponding exponential function $f(x)=e^x$ with $e$ as a base.

  2. From that definition of $e$ we know that $f'(0)=1$, which translates to $\lim\limits_{h \to 0}\frac{e^h-1}{h}=1$ by first principles. Rearranging this for $e$ gives $(1+h)^{\frac{1}{h}}\to e$ as $h\to 0$, from which students can work out the value of $e=2,718...$ with a calculator.

  3. Now work out the derivative of $e^x$ from first principles (differential quotient). This is easy enough, because $e^{x+h}=e^x e^h$ and the limit $\lim\limits_{h\to 0}\frac{e^h-1}{h}=1$ is known from the chosen definition of $e$. One finds that $f'(x)=e^x$, which is a cool result.

  4. Define the natural logarithm $\ln x$ as the inverse function of $e^x$ (logarithms and inverse functions are known). In order to work out the derivative of $\ln x$ start with the ansatz $e^{\ln x} = x$, which is clear from the chosen definition of $\ln x$. Differentiate both sides by applying the chain rule and using the now known fact that $e^x$ is its own derivative. This way one immediately obtains the result that $\ln x$ has derivative $\frac{1}{x}$.

  5. Given this, it is now easy to work out the derivative of general $a^x$ and $\log_a x$ by chain-ruling the expression $a^x = (e^{\ln a})^x = e^{\ln a\times x}$ and differentiating $\log_a x=\frac{\ln x}{\ln a}$. From this it becomes clear that the original exact slope values were $\ln 2$ and $\ln 3$ and could have been worked out by pressing a calculator button, which students found amusing.

In fact, this definition of $e$ can be made rigorous (assuming familiarity with uniform convergence) and thereby may be of some pedagogical value at undergraduate university level, too.

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    $\begingroup$ One thing I've done is to show how consecutive differences of $n$'th powers grow like $(n-1)$'th powers (arithmetic sequences have constant consecutive differences, quadratic functions of the positive integers have arithmetic sequences for their consecutive differences, etc. -- don't prove in general, just show for squares and cubes) whereas consecutive differences of $2^n,$ $3^n,$ etc. grow exponentially with the same base. I've written a lengthy post about this, but I can't find it, so it was probably in Math Forum's ap-calculus discussion group, whose archives appear to no longer exist. $\endgroup$ Sep 22, 2021 at 16:11
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    $\begingroup$ Use several approaches! Each approach reviews different skills and connects different concepts together. Be sure to talk about exponential growth in one of them: if a population is constantly doubling, the rate it grows depends on how large it is. $\endgroup$
    – TomKern
    Sep 22, 2021 at 20:11
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    $\begingroup$ This is essentially the approach I take in university calculus. My definition of $e$ is that it is the number for which $\lim_{h \rightarrow 0} \frac{e^h-1}{h} = 1$. Of course this can be motivated by exactly the graphs you mention. Admittedly, we use the inverse function theorem in these arguments. Nobody is proving that in Calculus I. So, Calculus 1 always has holes logically. Of course, there is also the logarithms via integral definition, but I find those calculations awkward and I really prefer to introduce exponentials and logs much earlier in the course. In short, you're good. $\endgroup$ Sep 23, 2021 at 3:25
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    $\begingroup$ I don't see a question here, so voting to close. $\endgroup$ Sep 25, 2021 at 23:45

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I normally go other way around.

Start with a logarithm, which by definition maps multiplication to addition: $\ln{ab} = \ln{a} + \ln{b}$.

A detour into a history could be also useful: multiplication is hard, addition is easy; this mapping, they say, prolonged the lives of astronomers.

Prove, by a sheer geometrical argument, that the $\displaystyle \int_1^x\dfrac {1}{t}dt$ satisfy the above definition. You may shamelessly use the idea of an area of a curvilinear trapezoid. They will understand.

Use a fundamental theorem to demonstrate that $(\ln{x})' = \dfrac{1}{x}$

Introduce the inverse function, $f$. Show that it maps addition to multiplication. Show how it naturally extends the idea of power. Use the the inverse function derivative theorem to demonstrate that $f' = f$

Make sure to spend time to discuss how $e$ emerges, how other bases come to play, and what makes $e$ special.

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Here is the approach which seems most natural to me, and it wasn't a disaster the two times I used it in a classroom. The basic idea is to hold off on $e$ until it is motivated.

Start with numerical data: $(2^{0.1}, 2^{0.01}, 2^{0.001}) = (1.071773, 1.006956, 1.000693)$. It looks like $2^h \approx 1 + 0.693 h$ for $h$ small. Ask class: What do we call that $0.693$? Hopefully, they know this is the derivative.

Now shift to another point on $2^x$: $(2^{1.1}, 2^{1.01}, 2^{1.001}) = (2.143547, 2.013911, 2.001387)$. What does it look like we can say about $2^{1+h}$? If you are lucky, you get two answers: $2^{1+h} \approx 2+1.387 h$ and $2^{1+h} = 2 \times 2^h \approx 2(1+0.693 h)$. If you are not so lucky (I was one time and not the other), point out that $1.387$ is twice $0.693$ and ask if this is a coincidence.

Keep playing around until students are happy that $2^{x+h} \approx 2^x +0.693 \times 2^x h$. So $\tfrac{d}{dx} 2^x = 0.693 \times 2^x$.

One could play the same game with powers of $3$, or jump to the general computation: $$\frac{d}{dx} a^x = \lim_{h \to 0} \frac{a^{x+h} - a^x}{h} = a^x \lim_{h \to 0} \frac{a^{h} - 1}{h}.$$ So $\tfrac{d}{dx} a^x = L(a) a^x$ where $L(a) = \lim_{h \to 0} \frac{a^{h} - 1}{h}$. (This wasn't the sort of course where we proved limits exist.)

What is $L(a)$? If $a = b^k$, then $L(a) = \lim_{h \to 0} \tfrac{a^{h} - 1}{h} =\lim_{h \to 0} \frac{b^{kh} - 1}{h}= k \lim_{h \to 0} \frac{b^{kh} - 1}{kh} = k L(b)$. In other words, $L(a) = \log_b(a) L(b)$. So, if we know $L(b)$ for any $b$, we can compute all the other $L$'s.

We could just compute one value once and for all, say $L(10) = 2.30259$. But, instead, we introduce $e$ as the number where $L(e) = 1$ ...


The warning I would give is that, while all of our students had seen exponentials, it soon became clear that many of them didn't really remember how they work. I spend a lot of time writing "$(a^3)^2 \neq a^9$!" and similar comments.

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The approach I tend to use is similar to that presented by user58697, and agrees with that presented in several calculus texts (e.g. Thomas' Calculus, but not any of the versions with the "Early Transcendentals" sub-titles).

  1. Define the natural logarithm by the integral formula: if $x > 0$, then $$ \log(x) := \int_{1}^{x} \frac{1}{t} \,\mathrm{d}t. $$ In terms of putting this definition into the "story" or "narrative" that I try to develop in class, the natural logarithm fills a hole in the power rule. For any $n \ne -1$, $$ \int x^n \,\mathrm{d}x = \frac{1}{n+1} x^{n+1}. $$ This rule fails for $n=-1$, but the function $x \mapsto x^{-1}$ is continuous (and therefore differentiable) on $(0,\infty)$, so the function $$ x \mapsto \int_{1}^{x} t^{-1} \,\mathrm{d}t $$ is a perfectly well-defined function. For mysterious reasons that I usually just kind of skip on by, we call this function the natural logarithm.

  2. It is not too difficult to show that the natural logarithm has a lot of nice properties, such as

    • $\log$ is a continuous function,
    • $\log(ab) = \log(a) + \log(b)$,
    • $\log(1/a) = -\log(a)$, and
    • $\log(a^r) = r \log(a)$,

    where $a,b \in (0,\infty)$ and $r \in \mathbb{Q}$ (it is harder to show that last identity holds when $r \in \mathbb{R}$, as it requires working carefully with some limits, but this, too, can be done if you feel like it).

  3. At this point in the course, students have been introduced to the inverse function theorem (IFT), so it is not unreasonable to ask about the inverse of the logarithm. From the, $$ (\log^{-1})'(x) = \frac{1}{\log' (\log^{-1} x)} = \frac{1}{1/\log^{-1}(x)} = \log^{-1}(x). $$ In other words, the inverse of the natural logarithm is a function which has the property that it is its own derivative. For more mysterious reasons (again, sometimes I talk about this), define the function $\exp$ to be the inverse of the natural logarithm (this is the natural exponential function).

  4. Running through properties of the logarithm (only backwards), it is possible to show that

    • $\exp(a+b) = \exp(a)\exp(b)$,
    • $\exp(-a) = 1 / \exp(a)$, and
    • $\exp(a^r) = \exp(a)^r$.

    These properties are very reminiscent of how rational exponents work over the real numbers. That is, if $a,b \in\mathbb{R}$ and $c \in (0,\infty)$, then

    • $c^{a+b} = c^a c^b$,
    • $c^{-a} = 1/c^a$, and
    • $c^{ar} = (c^a)^r$.

    This suggests that it might be reasonable to define a class of functions: for each $c \in (0,\infty)$, define $$ \exp_c : \mathbb{Q} \to \mathbb{R} : x \mapsto c^x. $$ This is the exponential function with base $c$.

    I am a little careful about this, because the tools available in the class up to this point make it harder to extend this function to something continuous in a tractable manner (for intro students). Rational exponentiation can be defined inductively, but the continuous extension is a bit of a leap. However...

  5. This discussion suggests that there may be a special number (which I am going to call $\mathrm{e}$) which has the property that $$ \exp(x) = \exp_{\mathrm{e}}(x). $$ Actually, this step is kind of not that hard to believe. Really, all I am doing is asserting that there is a number $\mathrm{e}$ which has the property that $$ \exp(x) = \mathrm{e}^x$$ for any rational number $x$. But $\exp$ is continuous on $\mathbb{R}$, which gives us the continuous extension we wanted above.

    Note that it is possible to approximate $\mathrm{e}$ in a number of ways. One possibility is to note that $$\exp(1) = \mathrm{e} \iff 1 = \log(\mathrm{e}),$$ and then approximate its value using Riemann sums over partitions with increasingly small mesh. A better approach is outlined in step (2) of the question (I usually do both in order to demonstrate what a pain in the ass Riemann sums are if you have to work with them directly).

    At this point, it is entirely reasonable to assert that $$\exp_{c}(x) = c^x$$ is perfectly well-defined for all $x \in \mathbb{R}$, and that the function $\exp_c$ is continuous, and has the desired properties (most importantly, that $$\exp_{c}(a+b) = \exp_{c}(a)\exp_{c}(b) \leftrightarrow c^{a+b} = c^ac^b. $$ for any real $a$ and $b$).

  6. Note that $$ \exp_c(x) = c^x = \exp(\log(c))^x = \exp(x \log(c)), $$ where $\log(c)$ is "just" a constant. Thus, from the chain rule, $$ \frac{\mathrm{d}}{\mathrm{d}x} \exp_c(x) = \exp(x \log(c)) \frac{\mathrm{d}}{\mathrm{d}x} [ x \log(c)] = \log(c) \exp(x \log(c)) = \log(c) c^x. $$

  7. Finally, note that the whole machine can be run "backwards": define the logarithm with base $c$ by $$ \log_c := \exp_{c}^{-1} $$ (that is, the logarithm with base $c$ is the inverse of the exponential with base $c$), and run with it. It is not too difficult to deduce properties of this logarithm, including the change of base formula (which should be mentioned): $$ \log_c(a) = \frac{\log_d(a)}{\log_d(c)}, $$ where $c, d > 0$.

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A historical perspective may be interesting to students. Here's how Napier's definition of the logarithm (which he worked out a century before calculus), translated into terms of the natural logarithm & calculus, reveals the derivative of the logarithm.

Historical background

John Napier (1550-1617) published his table of logarithms Mirifici Logarithmorum Canonis Descriptio in 1614 after some twenty years of work and described his method of construction in Mirifici Logarithmorum Canonis Constructio, published posthumously in 1619 (Edinburgh) by his son Robert, with appendices by Napier and Henry Briggs (1561-1630). In our modern terms, we can represent Napier's definition of a logarithm (Prop. 26):

The "artificial number" (logarithm) of a given sine is that which has increased Arithmetically always with the same velocity as the total sine [began] to decrease Geometrically and in the same time as the total sine decreased to that given sine.

Ignoring what is exactly meant by the given sine and total sign, what he is building on is a fundamental fact about geometric sequences that has been known from ancient times: If you take a geometric sequence starting from $1$ and number the terms with consecutive integer indices start from $0$, then the product of two given terms of the geometric sequence will be the term in the geometric sequences who index is the sum of the indices of the given terms. In algebraic notation, this is the familiar law $a^ja^k=a^{j+k}$. What Napier is saying is that if the index (the logarithm, which he first called an "artificial number") increases a at a constant rate (arithmetically), then the term (called a sine) changes at a geometric rate, that is, proportional to its own magnitude. In fact, he has it backwards from the way I am going to do it: If $x$ is the sine, then the rate is proportional to $r-x$, where $r$ is the total sine. Napier uses trigonometrical language for reasons I can only guess, which I won't go into. Remarkably from this, Napier is able to prove some inequalities about logarithms and construct quite an accurate table. A key invention for his success is that he is imagining two quantities in continuous motion under two different laws for their rates of motion.

Modern version

Let $y$ change at an "arithmetic" (constant) rate starting from $y=0$, and let $x$ change at a "geometric" rate (proportional to $x$) starting from $x=1$; and we'll let the constants both be $1$: $${dy \over dt} = 1\,,\quad {dx \over dt} = x\,;\quad y(0)=0\,,\ x(0)=1 \,.$$ Then the natural logarithm at a given $x$ is defined to be $y$.

Clearly, we have $${d\over dx}\log x = {dy \over dx} = {dy/dt \over dx/dt} = {1 \over x}\,.$$ Likewise, since $x = \exp y$, we have $${d\over dy}\exp y = {dx \over dy} = {dx/dt \over dy/dt} = x = \exp y\,.$$ Objections about the need to prove the existence of $x$ are quite technical, and I would question whether they are appropriate for the OP's purposes.

Note that one can prove properties of $\ln x$ and $e^x$ from this definition, but that would seem to go beyond the scope of the question.

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