Intuition why the derivative of $e^x$ is itself

Question

Is there an intuitive reason why the constant $e$ to the power of $x$ has a derivative that equals the value of the function? I know that this is the result of differentiating, and I've seen several proofs of how you work out the derivative, I was just wondering why this is and why it is the case for a number estimated from constant growth?

Answer 1

Maybe a nice way to see it is the power series expansion: $$\forall x\in \mathbb{R} \colon e^x =1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \cdots $$

Answer 2

By definition,$$\frac{d}{dx}a^x=\lim_{h\to0}\frac{a^{x+h}-a^x}{h}.$$Each exponential function has the nice property $a^{x+h}=a^xa^h$, so we can take out an $h$-independent $a^x$ factor, making the above limit the original function $a^x$ times just a number, namely $\lim_{h\to0}\frac{a^h-1}{h}$. This limit is in turn, by definition, the derivative of $a^x$ at $x=0$. Now if we gradually increase $a$ from just above $0$ to not quite $\infty$, $a^x$ will get steeper and steeper at $x=0$. And $e$ is just the choice of $a$ for which the slope is $1$, so that $e^x$ is its own derivative.

Answer 3

Something that might lend some intuition, is that $$ e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n=\lim_{n\to\infty}\left(1+\frac xn\right)^{n-1} $$ and $$ \frac{\mathrm{d}}{\mathrm{d}x}\left(1+\frac xn\right)^n=\left(1+\frac xn\right)^{n-1} $$ It takes a little bit to make this rigorous, but intuition and rigor often follow different paths.

Answer 4

If $x$ changes by $1$, then the function changes from $e^x$ to $e^{x+1} = e\cdot e^x.$

If $x$ changes by $2$, then the function changes from $e^x$ to $e^{x+2} = e^2 e^x.$

If $x$ changes by $h$, then the function changes from $e^x$ to $e^{x+2} = e^h e^x.$ In each case the change in the function is $e^x(e^h-1)$, which is $e^x$ times some number. This is true for any base, but for $e$, the "some number" over $h$ happens to tend to $1$.

Answer 5

The slope of the tangent line at point $x$ is always equal to the value of $f(x) = e^x$. This may be the intuition you're looking for. And this is true for every point $x$. And it is true for every point $x$ only for functions of the form $e^x + const$. That makes $f(x) = e^x$ quite "special" if you think about it from geometrical perspective.

How to see this? Draw in one plane: 1) the graph of $e^x$, 2) the unit circle, 3) the line S with the equation $x=1$ (on which values of $\tan$ are measured).

Take any point $x_0$ and draw the tangent line $L_0$ to the graph (of $e^x$) at the point $(x_0, e^{x_0})$. Now draw a line $L_1$ passing through $(0,0)$ and parallel to the tangent line $L_0$. Note the height at which $L_1$ intersects the line $S$. The special ting here is that this height is always equal to $e^{x_0}$ i.e. to the value (at point $x_0$) of the function itself. In a way that property is quite remarkable especially since it holds for every value of $x$ ($x_0$ was taken arbitrary here).

Why remarkable? Because as we know no other function (apart from the functions $g(x) = e^x + const$) has this property (no other function has it for every $x$, I mean).

Answer 6

It depends entirely on how you define the function $x \mapsto \mathrm{e}^x$. For example, looking at other answers to this question:

• robjohn defines the exponential function in terms of the limit $\lim_{n\to\infty} \left( 1+\frac{x}{n} \right)^n$—from this definition, the intuition is that the derivative can pass through the limit, and then the power rule gives the desired result;
• S. Maths defines the exponential function via a power series—from this definition, it is again possible to pass the derivative through the limit (i.e. the limit can be brought into the sum) and then term-by-term differentiation gives the result;
• J.G. doesn't quite give a definition of the exponential function, but looks at the difference quotient directly and uses the "fact" that $a^0 = 1$.

Personally, I like to define the exponential function by the property that it is its own derivative. More precisely, the exponential function $x \mapsto \exp(x)$ is defined to be the unique solution to the initial value problem $$ \begin{cases} u' = u \\ u(0) = 1, \end{cases}$$ where $u$ is defined on the real numbers. Note that this initial value problem exactly encapsulates the idea of "constant growth"—the growth rate of a population is equal to the size of the population; each individual reproduces at a constant rate.

The "intuition" I have about this is that there should be some function which behaves this way with respect to differentiation—in a fair and just world, there should be a function which is its own derivative.

But then the question is, perhaps, why does this function look so much like exponential functions that are taught in high school? I was taught, at one point or another in my education, that if $a$ is a real number and $n$ is a natural number, then $$ a^n = \underbrace{a\cdot a\cdot \dotsb \cdot a}_{\text{$n$ times}}; $$ that is, exponentiation is repeated multiplication. What does $\mathrm{e}^x$ have to do with repeated multiplication!?

Typically, some of the first things that are done when the exponential-is-repeated-multiplication function is introduced is to show that it satisfies certain properties:

• $a^{m+n} = a^m \cdot a^n$,
• $(a^m)^n = a^{mn}$,
• $a^{-1} = 1/a$,
• $a^0 = 1$,
• for fixed $a > 0$, the function $n \mapsto a^n$ is increasing,

and so on. Note that some of these require some adjustments to the possible values of $a$, so the definition of the exponential-as-repeated-multiplication function is typically extended a little to make this all make sense. Then, after a bit more work, it can be shown that $\exp(x)$ has all of the same properties—this has already been done on this site, so I won't repeat the arguments here. This implies that there is some real number (call it $\mathrm{e}$) such that $$ \exp(x) = \mathrm{e}^x,$$ where $\exp$ (on the left) is the solution to the initial value problem, and $\mathrm{e}^x$ (on the right) is an appropriate extension of the exponentiation-as-repeated-multiplication function.

Note that all of the above is a bit hand-wavy (the question is about intuition, not rigor, after all), but it can all be made rigorous with a bit of work.

TL;DR: The exponential function is the solution to an initial value problem, and from this definition, the exponential function is its own derivative. A function which is its own derivative has all of the same properties as a particular exponential function (as defined in high school via repeated multiplication), and is therefore the same thing.

Answer 7

Differentiation is a linear operation. If you double the function, the derivative doubles. So if there is a function such that a shift of the variable is equivalent to a doubling, $f(x+1)=2f(x)$, the derivative will follow that trend.

$$f(1)=2f(0)\implies f'(1)=2f'(0).$$

More generally, by induction

$$f(k)=2^kf(0)\to f'(k)=2^kf'(0).$$

The iterative doubling makes the exponential appear.