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I'm teaching a preparatory course on mathematics at a university. The content is mostly calculus, manipulating expressions and solving equations and inequalities. I show a couple of simple derivations/proofs and ask the students to occasionally prove some simple equality, so the course is by no means rigorous. Most students study chemistry, biology or programming.

When teaching derivatives I have given the definition, used it once on a linear function, and drawn pictures. I give the formulas for differentiating functions as black boxes. A student asks: Is there an easy geometric way to see why differentiating the exponential function gives the exponential function?

The best I can come up with is the differential equation $f = f'$, which is not very geometric. The students are not expected to know differential equations at this point.

Does there exist a more geometric explanation?

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  • $\begingroup$ It's the curve that passes through the point $(0,1)$ and such that the slope of the tangent at every point equals the ordinate of that point. Or such that the tangent at a point of abscissa $a$ intersect the $x$-axis at $a-1$. $\endgroup$ – Paracosmiste Nov 5 '16 at 10:18
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    $\begingroup$ (Comment because it is not geometric...) I wonder whether you could use the power series definition of $e^x$? If they are comfortable differentiating term by term in this infinite series, then it becomes clear that the derivative returns the same function; plugging in $x=1$ gives $e^1 = e$ which is well-defined i.e. the series converges (comparison test with the geometric series $4 = 2 + 1 + 1/2 + 1/4 + \cdots$ for example). And you can compute (with a calculator or by hand) $\sum_{n=0}^{10} 1/n! \approx 2.7182818$. $\endgroup$ – Benjamin Dickman Nov 5 '16 at 18:22
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    $\begingroup$ @BenjaminDickman I wonder, if we just introduced series earlier and claimed they made sense (like most of beginning topics) then could we use them as you say... certainly they give the most concrete definition in the explicit sense for the exponential, sine, cosine and a host of other functions. Lagrange would be proud. Later, when maturity warranted, we could confess our sin of omission and explain to them the convergence and divergence beast. If limits don't need epsilons and deltas then why do power series need convergence theory? $\endgroup$ – James S. Cook Nov 6 '16 at 3:21
  • $\begingroup$ @JamesS.Cook Yes, this is an interesting idea. For a somewhat relevant example, if you assume [the fact] that every entire function can be represented by power series, then you can quickly prove the Fundamental Theorem of Algebra. This is done by DJ Velleman (1997) here. $\endgroup$ – Benjamin Dickman Nov 6 '16 at 5:51
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    $\begingroup$ Couldn't resist telling a joke I always use in the class in this context. Once upon a day, there was a party full of different functions. Suddenly, the derivative operator entered the party. All of the functions started shivering and crying that we are going to change now, but one function, $e^x$. It said, ha ha, I am not going to change. The derivative operator calmly replied, I differentiate with respect to $y$! $\endgroup$ – Amir Asghari Nov 19 '16 at 18:16

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Here is a method I use in all my calculus classes to show that it may be true that the derivative of $e^x$ is itself.

Sketch a graph of $y=e^x$ high up on the board and sketch coordinate axes below that on the board. Say that the lower graph will be the derivative of the upper graph.

Start at a point on the far left on the $x$-axis. Ask the class what the slope of $e^x$ is there. They will quickly say that the slope is positive but small, since the graph is increasing slowly there. Plot a point on the lower graph to correspond to this. Progress to the right, and students will see that the slope gets more and more positive. Keep graphing points.

At the end, the graph of the derivative will look very much like the original graph. Explain that mathematicians have shown that the two graphs are in fact identical.

This works well--no student has ever disputed this, and this is in classes where I give extra credit to students who successfully dispute something I say.

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    $\begingroup$ This is the canonical away to motivate this in "reform calculus" textbooks, and I agree that the intuition works very well in class. The only caveat I would add is that it would be helpful to have already introduced graphing derivatives this way. $\endgroup$ – kcrisman Nov 5 '16 at 11:45
  • $\begingroup$ Do you also know a way to distinguish the derivative of x^2 from (the derivative of) e^x? Both are increasing... $\endgroup$ – Jasper Nov 6 '16 at 10:11
  • $\begingroup$ @Jasper: The most obvious difference between $x^2$ and $e^x$ is for non-positive $x$. When I start at the left, $x^2$ slopes down so the derivative is negative there, and at zero the slope is clearly zero. The derivative of $e^x$ is clearly always positive. Since I start graphing from the left, the differences are immediately obvious. By the way, I always sketch the derivative of $x^2$ and claim it is a straight line before I sketch the derivative of $e^x$. $\endgroup$ – Rory Daulton Nov 6 '16 at 10:16
  • $\begingroup$ This is very time consuming and it gives only an approximation. $\endgroup$ – Paracosmiste Nov 6 '16 at 10:28
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    $\begingroup$ @whatever: This takes only a couple of minutes, since I have already covered how to sketch the graph of a derivative. And the issue is installing an intuition in students, so an approximation suffices. Real-world experience shows me that this method works well. I usually follow this demonstration by having students graph both $e^x$ and its numeric derivative on a graphing calculator, which shows the graphs identical to the eye. $\endgroup$ – Rory Daulton Nov 6 '16 at 10:31
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It may help to compare several exponential curves, e.g., $y=5^x$, $y=e^x$, and $y=2^x$,


            ThreeExps
and calculate slopes at a few points, as in this nice drawing from wyzant.com:
graph_slopes
It becomes plausible that there is some curve between $2^x$ and $5^x$ such that the slope at each point is exactly the $y$-coordinate of that point.

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But, a geometric definition for the exponential function is the following:

The function $f$ for which $f(0)=1$ and is such that its value is equal to its slope at each point is the exponential function. Or, in the language of differential equations, it is the unique solution of $\frac{dy}{dx}=y$ for which $y(0)=1$.

As usual, the question really is... how are we defining the exponential? If we are to assume the students just know the exponential by its graph then I suppose Rory's explaination is a good approach.

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As mentioned in other answers, the starting point is the definition of the exponential you are using.

One possible definition is as the solution of $f'=f$ taking value $1$ at $0$ (or other equivalent phrasings if you don't want to involve differential equation explicitly) and then there is nothing to explain.

One more natural definition is that $\exp(x)=e^x$ where $e$ is some number (then you have to explain powers by a real number, which maybe tricky, but can be done with various level of details). This is the definition I recommend for people that will have to compute with exponential and logarithms, because it embeds the morphism property $e^{x+y}=e^x e^y$.

The number $e$ is somewhat difficult to justify a priori, so what you can aim for is explaining why an arbitrary exponential function $x\mapsto a^x$ is always proportional to its derivative (and then it is easily accepted that for some choice of $a$ the proportionality coefficient is $1$, and that makes an excellent definition of $e$).

If you aimed at a computational approach, you would write $$ \frac{a^{x+h}-a^x}{h} = a^x\frac{a^h-1}{h}$$ and then one sees immediately that if $(a^h-1)/h$ has a limit when $h\to 0$ (a small leap of faith), then $x\mapsto a^x$ is indeed proportional to its derivative.

But you ask for a geometric approach, so let us make this geometric. Draw a plausible graph for an exponential, and translate geometrically the definition of an exponential: if we move on the graph by adding any value $h$ to the $x$-axis, we obtain a fixed relative increase, i.e. the $y$-coordinate is multiplied by an amount that only depends on $h$. This exactly tells you that the slope of a chord from $(x,a^x)$ to $(x+h,a^{x+h})$ depends on the point, but the ratio of the slope to the $y$-coordinate of the leftmost point only depends on $h$. Passing to the limit, you get the desired property.

Note that if you where to define the exponential with the differential equation, this explanation could help understand why we end up with a function that takes a fixed value to the power the argument.

In any case, I think the chosen explanation should be an opportunity to manipulate the morphism property of the exponential, because it is tremendously important and often mistreated by students.

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Bunny rabbit generations: look at the discrete case $P(n)=2^n$ with $P(0)=1$. It is easy to see that $\frac{\Delta P}{\Delta n}=P(n)$ where $\Delta P=P(n+1)-P(n)$ and $\Delta n=1$. Or more generally for $P(n)=b^n$ where $b>1$ we have $\frac{\Delta P}{\Delta n}=cP(n)$ where $c=b-1$.

This discrete case for exponential growth makes it reasonable to expect that "differentiating an exponential function gives the exponential function back," at least up to a constant multiplier $c$. As you know, the fact that $e$ corresponds to the case $c=1$ in the continuous variable case is simply the definition of the number $e$.

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Here's an approach that can be illustrated with a simple picture, and also gives an actual bound on the derivative of $e^x$ as being, at the very least, fairly close to $e^x$ (which makes it much more plausible that they're in fact equal).

Compare the slope of secant line through $(x - 1, e^{x-1})$ and $(x, e^x)$ with the slope of the secant line through $(x, e^x)$ and $(x + 1, e^{x+1})$: $$\frac{e^x - e^{x - 1}}{x - (x - 1)} = e^x - e^{-1} \cdot e^x = (1 - e^{-1})e^x \approx 0.63e^x$$ and $$\frac{e^{x + 1} - e^{x}}{(x + 1) - x} = e \cdot e^x - e^x = (e - 1)e^x \approx 1.72e^x.$$ The derivative is a limit of slopes of secant lines, and it's not hard to see (from the definition or, informally, by visual inspection of the graph) that the derivative of $e^x$ is strictly increasing. Given these facts, the above slopes give us bounds on the derivative: the derivative of $e^x$ at $x$ is somewhere between $(1 - e^{-1})e^x \approx 0.63e^x$ and $(e - 1)e^x \approx 1.72e^x$.

So, even without knowing exactly what the derivative of $e^x$ is, we can see it's fairly close to $e^x$, and taking closer secant lines will give tighter bounds, making it even more plausible. You can draw a picture of the tangent line and the two secant lines at a point to show how this bounds the slope of the tangent line. (The correspondence between the above algebra and the geometric picture with secant and tangent lines also nicely illustrates the close relation between algebra and geometry.)

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I would say to first sketch the curve $4^x$ and it's derivative $4^x \ln 4$, and the same for $3^x$ and $2^x$, then it will be easy to see that the case $e^x$ is a particular case. For the first two functions the derivative graph is a little higher than the function, but with $2^x$ the derivative is lower, so there must be a number between 2 and 3 such that the derivative graph is equal to the function itself. After this it will be evident that the responsible for this difference is the logarithm that appears in the derivative, the final stroke is the fact that $\ln e=1$ which makes the two functions identical.

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Take the graph of an exponential function such as $2^x$. If, for example, you shift it to the right by 3 units and then stretch it vertically by a factor of 8, you get the same graph back again. But the vertical stretch also increases the slope of the tangent line by a factor of 8. This holds for any shift, and therefore the derivative of this function is proportional to the function itself.

The same argument holds for an exponential function with any base. We define $e$ as the base for which the constant of proportionality is 1.

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here is a way i teach about the natural exponential function. we look at the slope of the graphs $y = 2^x$ and $y = 3^x$ at $x = 0, y = 1$ by evaluating the average rate of change(slope of the secant line) of these functions on $[0, h]$ for small values of $h.$ conclude that there is a number $e$ with the property that the slope of the graph $y = e^x$ is one. we make a small characteristic triangle with one corner at $(0,1)$ and the others at $(1+h, 1), (h, e^h).$

now we use addition property of the exponential function specialized to moving one unit to the right and what it does to the $y$-value embodied in $e^{1+x} = e e^x.$ that is, the $y$ value is multiplied by the base $e.$ they can, and usually do, see that the characteristic triangle at $(1, e)$ is the vertically stretched version of characteristic triangle at $(0,1).$ that is the side parallel to $y$ axis is multiplied by $e$ while the side parallel to the $x$ stay the same.

they conclude that the slope of the graph of $y = e^x$ at $(1, e)$ is $e.$ you can now generalize this to idea that the slope of $y = e^x$ at any point is $y$ coordinate of the point.

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This is not so geometric, but if you have students who can actually do some difference quotients (numerically only, I mean) then just having them do a table where they estimate the derivative (say, $(f(3.001)-f(3))/(3.001-3)$) for various values, and then compare with the actual value of $e^x$, they might get surprised. No, this is not an original idea at all.

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Geometrically the relation $f^{\prime} = f$ means that the slope at $(x, f(x))$ of the graph of $f$ is equal to $f(x)$. We look for a function $f$ that has this property. Consider $g(x) = 1 + x$. Its graph is $(x, 1 + x)$ and the slope is $1$ at every point. Hence for $x$ very small this function almost has the desired property; it does have the desired property for $x = 0$. So we zoom in at the point $(0, 1)$ by replacing $g(x)$ by $h(x) = 1 + x/n$ for a big $n$. This doesn't work because the slope is $1/n$ not $1$, as we would like for $x$ near $0$. However, since the derivative of $h(x)^{n}$ is $nh(x)^{n-1}h^{\prime}(x)$, and $nh(0)^{n-1}h^{\prime}(0) = 1$, the function \begin{equation} g_{n} = \left(1 + \frac{x}{n}\right)^{n} \end{equation} should approximately have the desired property. Indeed, \begin{equation} g_{n}^{\prime} = \left(1 + \frac{x}{n}\right)^{n-1} = g_{n}^{1-\tfrac{1}{n}}, \end{equation} so the slope of the graph of $g_{n}$ at $(x, g_{n}(x))$ is almost $g_{n}(x)$. This suggests that when $n \to \infty$ there results a function that has the desired property. This in fact works, and yields the exponential function.

There is nothing special about the initial function $g(x) = 1 + x$. Any $g$ differentiable at the origin and satisfying $g(0) = g^{\prime}(0) = 1$ would work. So the idea can be summarized as follows: we start at a point on a graph that has the desired property, and we zoom in at this point in an appropriate way. The remarkable thing is that we always get the same result.

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    $\begingroup$ Interesting point of view (but certainly more suitable for a real analysis course than a calculus lecture for non-math majors). $\endgroup$ – Benoît Kloeckner Nov 8 '16 at 11:46
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    $\begingroup$ @BenoîtKloeckner: Certainly it is not appropriate for such students, at least as sketched here. However, one can view it as an exercise in understanding how internal rescaling and external exponentiation affect the graph of a function. From this point of view it might be possible to adapt it to something that could be presented to such students. $\endgroup$ – Dan Fox Nov 8 '16 at 17:50
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Geometricaly, ln(x) is the area from 1 to x under the graf of y=1/x. Geometricaly, derivation of the area under f(x) is f(x). Hence, d(ln(x))/dx = 1/x. exp(x) is the inverse of ln(x) i.e. if y=ln(x) then x=exp(y). Hence, if dy/dx=1/x then dx/dy=x i.e. the derivative of exp is exp.

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