Could you explain why $\frac{d}{dx} e^x = e^x$ “intuitively”?

Question

As the title implies, It is seems that $e^x$ is the only function whoes derivative is the same as itself.

thanks.

Answer 1

Well, think of exponential growth (like e.g. bacteria grow):

We know, the more bacteria exist in a colony, the faster the colony will grow. More precisely: The growth speed of the colony $B$ is proportional to it's size ... Double size, double speed.

$$\frac{dB}{dt} \sim B$$

Furthermore we know the growth is exponential, since bacteria clone themselves in fixed amounts of time, i.e.

$$B \sim 2^{k\cdot t}$$

Putting it together, we can deduce that:

$$\frac{d}{dt}2^{kt} = c \cdot 2^{kt}$$

or with $a = 2^k$

$$\frac{d}{dt}a^t = c \cdot a^t$$

---

Now, how do we get the $e$? We just ask: What base $a$ do we have to take such that $c = 1$, i.e. $\dot{B} = B$?

We simply call that base $e$. Having such an $e$ is quite useful. We could use its special derivation traits we found above to define all exponential functions to the base $e$.

$$a^x = e^{x \cdot \ln a} $$

This shows that the factor $c$ we encountered in the above equations equals $\ln a = \log_e a$ and therefore, we can easily derive all kinds of exponential terms.

After all, $e$ turns out to be $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x \approx 2.718\ldots$

Answer 2

(Looks like Niel and "J" beat me to the punch while I was composing this, but I'll post my answer anyway since I like my diagram. :) )

You can't get too much more intuitive than the Law of Exponents ($a^x a^c = a^{x+c}$) right?

To reduce symbolic (and mental) clutter for a while, let's write $k$ for $a^c$, to get

$$k \; a^x = a^{x+c}$$

The point is that

Multiplying the value of $a^x$ by something yields the same result as adding some other thing to the exponent.

Consider what this bit of algebra tells us about the geometry of the graph of the function $y=a^x$ (shown in red in the figure below). Recall some fundamental notions:

• $y=k a^x$ is the result of vertically stretching (scaling) the original graph by a factor of $k$.

• $y = a^{x+c}$ is the result of horizontally sliding (translating) the original graph by a (signed) distance of $-c$.

The Law of Exponents tells us that (provided $k$ and $c$ are related appropriately), these transformed graphs are identical! In the figure, the blue graph represents both results.[*] Importantly, what points move where aren't the same under both actions; scaling moves the red point $P$ vertically onto the blue point $Q$; translating moves the red point $R$ horizontally onto $Q$.

Now, as suggested by the diagram, imagine a tangent vector poking out of each point of the original graph, and consider what the transformations to do those vectors:

• Vertically stretching scales (only) the "rise" of the vector by factor $k$; that is, a vector with slope $m$ is pulled to achieve a slope of $m k$.
• Horizontally translating has no effect on the vector's slope.

What can we conclude here? Why, something pretty remarkable:

For any point $P$ on the original graph, the point $R$ --located $c$ units to the right[**] of $P$-- is such that the slope of the tangent vector at $R$ is $k$-times the slope of the tangent vector at $P$ (where $k$ and $c$ are related appropriately).

Let me take this opportunity to retire $k$, since it is beginning to become clutter; I'll just make the appropriate relationship explicit. Also, I'm going to retire the point name $R$, opting to describe the point instead. I'll summarize things this way:

On the graph $y=a^x$, if the slope of tangent vector at any $P$ is $m$, then the slope of the tangent vector at the point $c$ units to the right of $P$ is $m\;a^c$.

Notice that there's nothing special about the players in this game. $P$ is any point on the graph, $c$ is any (horizontal) distance you care to choose; heck, even the exact value of $a$ is up for grabs. Let's use this to our advantage.

Suppose we take $P$ to be the point where the (original, red) graph crosses the $y$-axis; that is, we take $P$ to have $x$-coordinate $0$ (and $y$-coordinate $1$, but this doesn't really matter). Then, "the point $c$ units to the right of $P$" can be described more simply as "the point with $x$-coordinate $c$", and we have

On the graph $y = a^x$, if the slope of the tangent vector at the point with $x$-coordinate $0$ is $m$, then the slope of the tangent vector at the point with $x$-coordinate $c$ is $m\;a^c$.

In Calculusian prose:

If $f(x) = a^x$, then $f^{\prime}(c) = a^c f^{\prime}(0)$.

Observe that we don't even need "$c$" in the above formula, since it's just taking the place of some $x$-coordinate. We can simply write:

If $f(x) = a^x$, then $f^{\prime}(x) = a^x f^{\prime}(0)$.

From here, it's pretty much a matter of definition to get to the final answer to your specific question. After all, the above holds for any (non-negative) value of $a$. Clearly, some values of $a$ correspond to graphs that cross the $y$-axis very steeply; some values correspond to graphs that cross the $y$-axis very shallowly; it's not un-reasonable to believe that there's a convenient value would cause the graph to cross the $y$-axis juuuuuuuuust right ... with a slope of $1$. Of course (as I mix my folklore), to name a demon is to control him, so we'll simply assign a symbol to this "just right" value of $a$.

Let "e" be the number such that the tangent vector to $y=e^x$ at $x=0$ has slope $1$.

Assuming that such a number really does exist, you don't even have to know its exact value to conclude

If $f(x) = e^x$, then $f^{\prime}(x) = e^x$.

You can then turn your attention to figuring out why $e$ happens to have the value $2.718...$ . Other answers in this thread provide insights on how those arguments proceed.

Also left as an exercise is to determine why, in the formula for the derivative of $a^x$, the "$f^{\prime}(0)$" factor is in fact "$\log a$" (the natural logarithm of $a$).

[*] As Niel mentions, this illustrates that the graph of an exponential function is "self-similar". I think it's important to add "uni-directionally" to the description, in order to distinguish it from (conventional) similarity transformations where scaling occurs "omni-directionally".

[**] "to the left" works, too, with appropriate changes to the discussion.

Answer 3

Doesn't quite explain $e^x$, but looking at the series of $2^n$, i.e. $\{1,2,4,8,16,32,...,2^k,2^{k+1},...\}$ might help.

You see there if you take the first order difference you get back the same series, as $2^{k+1}-2^k=2^k$.

Answer 4

As J.M. says,

$$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} + \cdots $$

If you accept (it's true, but it could take a while to explain why) that this series, like polynomials, can be derived term by term

$$ p(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n \ \Longrightarrow \ p'(x) = a_1 + 2a_2x + \cdots + na_n x^{n-1} \ , $$

then

$$ \frac{d}{dx} e^x = \frac{1}{1!} + \frac{2x}{2!} + \cdots + \frac{nx^{n-1}}{n!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x \ . $$

Answer 5

One conceptual way to understand the standard definition $e^x = \lim_{n\to\infty}(1+x/n)^n$ is to view it as arising from applying Euler's approximation method to $y' = y$. The same method also works arbitrary higher-order constant coefficiant ODEs by converting them to linear system form and employing matrix exponentials. This is described quite nicely in Arnold's beautiful textbook "Ordinary differential equations". Here's an excerpt:

Answer 6

Suppose $\frac{d}{dx}f(x)=f(x)$.

Then for small $h$, $f(x+h)=f(x)+hf(x)=f(x)(1+h)$. If we do this for a lot of small intervals of length $h$, we see $f(x+a)=(1+h)^{a/h}f(x)$. (Does this ring a bell already?)

Setting $x=0$ in the above, and fixing $f(0)=1$, we then have $f(1)=(1+h)^{1/h}$, which in limit as $h\rightarrow 0$ goes to $e$. And continuing $f(x)=(1+h)^{x/h}$, which goes to $e^x$.

Answer 7

I'll act on Casebash's proposal later, but for now... care for a movie?

(granted, I cheated and spaced the two plots of $\exp(x)$ to clearly show the moving tangent line... but you guys should be able to understand this).

(thanks to Stan Wagon!)

Answer 8

Let us define a function $\exp$ as the solution to $D \exp(x) = \exp(x)$ with initial value $\exp(0) = 1$

Clearly $D \exp \circ f = D f \cdot (\exp \circ f)$ but also notice that $$ \begin{align} D (\exp(a(x)) \exp(b(x))) &= D a(x) \exp(a(x)) \exp(b(x)) + D b(x) \exp(a(x)) \exp(b(x)) \\ &= (D (a + b) (x)) \cdot \exp(a(x)) \exp(b(x)) \end{align} $$ so we see that $\exp(a)\exp(b)=\exp(a+b)$ which lets us write $\exp(x) = e^x$ for some as yet undetermined constant $e$.

Geometrically we can see that the gradient of the tangent line of the curve $y = e^x$ must be between the finite differences, i.e. $\frac{e^x - e^{x-h}}{h} < e^x < \frac{e^{x+h} - e^x}{h}$. With some numerical exploration we see that $e$ is between $2$ and $3$. Specifically, the finite differences for $e=2$ and $h=0.9$ are too small and the finite differences for $e=3$ and $h=0.1$ are too big. To narrow it down more let $e=2.7$ and $h = 0.01$ to see that $2.7 < e < 3$.

Answer 9

There are two components to the original question:

• Why do exponential functions generally (the class $x \mapsto ab^x$ for $b>0$) have the property of being eigenfunctions of the derivative operator? That is: why do these functions f have the property $\frac{\mathrm d}{\mathrm dx} f(x) = \lambda f(x)$?

• Why are the functions $x \mapsto a \mathrm e^x$ in particular +1 eigenfunctions --- i.e. why is the scalar $\lambda$ equal to 1 for these functions in particular? (To put it another way: why is this the case when the base is equal to the particular constant e ≈ 2.71828?)

I can give an intuitive answer to the first of these questions, by combining two observations.

1. Consider the slope at a paticular point, e.g. at x = 0. Multiplying the function f(x) by some scalar now only amplifies the function by that factor, but also amplifies the slope by that same factor. (This is just a restatement of the product rule, for constant scalar factors.)

2. Exponential functions f(x) are self-similar: multiplying them by a positive scalar is equivalent to translating them to the left, or to the right, depending on the scalar by which you multiply: more precisely, if you multiply $\mathrm e^x$ by c, why you get is $\mathrm e^{[x + \ln(c)]}$, or the same function translated ln(c) to the left. More generally, for $f(x) = ab^x$, we have cf(x) = f(x + log_b(c)); that is, multiplying f(x) by c is equivalent to a shift to the left by log_b(c).

Now, combine these two facts. The slope of f(x) at x = log_b(c) is c times the slope of f(x) at x = 0, because of the self-similarity of the exponential function, and because the slope of cf(x) at x = 0 is c times the slope of f(x) at x = 0. Thus the slope of f(x) is a function which is proportional to f(x), QED.

Answer 10

$d/dx (e^{-x} \cdot y)=-e^{-x} \cdot y+e^{-x} \cdot dy/dx=e^{-x}(dy/dx-y)$

So if $dy/dx=y$, then $d/dx (e^{-x} \cdot y)=0$, ie $e^{-x} \cdot y=c$ or $y=c \cdot e^x$

Answer 11

A particle on a trigonometric hyperbola has acceleration described by the same hyperbola.

That is, differentiation is idempotent on the hyperbolic cosine. Since $e^{x} = (\cosh x)^{\prime} + \cosh x$, the claim follows immediately.

Answer 12

Here's my 2 cents, just look at the def of derivative

df/dx = lim f(x+d)-f(x)/d

now the defining property of e^x is that for very small x, e^x ~ 1 + x

and since e^{x + d} = e^xe^d ~ e^x(1+d), it's clear that by plugging everything into the definition of derivative you will get the required result.