Intiutive argument that $\exp' = \exp$

Question

Is there any intuitive argument or visual "proof" that $\exp' = \exp$? Suppose you have defined the Euler number $\mathrm{e}$ as limit of the sequence $(a_n)$ where $a_n = \left (1 + \frac{1}{n} \right)^n \quad \forall n > 0$, and that the $\exp(x)$ is introduced as $\mathrm{e}^x$. The use of the power series of $\exp$ should be avoided.

Answer 1

Since $$ \frac{e^{x+h}-e^x}{h}=\frac{e^xe^h-e^x}{h}=e^x\frac{e^h-1}{h} $$ you only need to prove that $\exp'(0)=1$.

Perhaps this helps.

Answer 2

$$ \exp(x) = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n $$

Answer 3

Once you have defined $exp(x)=e^x$ we can work our way trhough the definition of derivative (which is quite intuitive) for demostrating that $exp'(x)=exp(x)$. The formal definition of derivative is:

$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

When $h$ tends to $0$.

So, substituting $f(x)$ by $e^x$ we get that:

$$f'(x)=\lim_{h\to0}\frac{exp(x+h)-exp(x)}{h}$$

Working with the expression a bit we can rearrange it to:

$$f'(x)=\lim_{h\to0}\frac{e^x(e^h-1)}{h}$$

Now notice that as $h$ approaches $0$ the denominator of the fraction $h$ and the term $e^h-1$ also approach $0$. Furthermore, when being really close to $0$ they approach it at 'the same speed' and we say they are 'equivalent infinitesimals' which means that we can approximate one to the other. We can substitute then $e^h-1$ with $h$.

With this in mind we end up with:

$$f'(x)=\lim_{h\to0}\frac{e^x·h}{h}$$

We can cancel the $h$ and get to the final result:

$$exp'(x)=exp(x)$$

Answer 4

Provided that you have defined general exponentiation in the first place, you can write

$$\frac{d}{dx} a^x = \lim_{h \to 0} \frac{a^{x+h} - a^x}{h} = a^x \lim_{h \to 0} \frac{a^h - 1}{h}.$$

You now need to intuitively argue that this last limit, as a function of $a$, ranges from $0$ to $\infty$ as $a$ ranges from $1$ to $\infty$. It is also clearly continuous. Therefore by the intermediate value theorem there is some $a$ such that the limit is $1$. We call this number $e$.

Defining general exponentiation in the first place is a huge mess. Indeed, I think that from the perspective of real analysis, it is easier to define $\exp$ by the ODE it satisfies, then prove it has an inverse $\ln$, then prove that $\exp(q \ln x) = x^q$ whenever $q$ is rational. (Provided you can independently prove the power rule for rational powers $q$, this last argument is actually quite simple.)

Answer 5

There is an elegant way to define e which is to start from

$$\lim_{x\to\infty} (1+1/x)^x = e$$

This definition allows the normal intuition associated do e (euler number), that makes the concept easy to explain

It starts from a hypothetical situation where has 100% interest in one year over some money. In this simple case, one double his money. 100 turns in 200. After, if one consider 2 periods ($x=2$), the factor that multiply the money is $(1 + 1/2 )^2 \approx 2.25$, because the interest sum to the principal at half period. In this case, $\$100$ turns $\$225$.

Now supposes that one uses $x=4$ periods. The total is $(1+1/4)^4 \approx 2.44$, i.e., 100 turns 244. If one uses $N=10$, the value is $(1+1/10)^10 \approx 2.59$ ($\$100$ converts in $\$259$)

Now there are 10 periods where the interest join with the money and serves as a basis for calculating subsequent interest, increasing the gain. For $n=100$, one reachs $2,70$ ($\$100$ changes to $\$270$). The money growth rate is lowering...

So one does the number of periods to increase towards infinity. In that situation, it's easy to notice that the mysterious number approaches $2.71828...$. $\$100$ transforms in almost $\$271.83$.

So

$$\lim_{x\to\infty} (1/x)^x = e$$

Now, replace $x$ for $1/h$. If $x\to\infty$, now one has $h\to 0$

$$\lim_{h\to 0} (1+h)^{1/h} = e$$

Is the starting point for the next steps:

$$ \lim_{h\to 0} (1+h) = \lim_{h\to 0}e^h $$ $$ \lim_{h\to 0} (h) = \lim_{h\to 0}(e^h - 1)$$

How $h\to 0$ but it's not 0, one can divide both sides for $h$ and invert the sides:

$$\lim_{h\to 0}(e^h-1) / h=1$$

Now we can calculate (e^x)' using the derivative definition:

$$\lim_{h\to 0}\left(\frac{e^{x+h}-e^x}{h}\right)=\lim_{h\to 0}\left(\frac{e^xe^h-e^x}{h}\right)= e^x\lim_{h\to 0}\left(\frac{e^h-1}{h}\right)$$

Well, this limit was calculated above and is equal to 1, so

$$(e^x)' = e^x $$

There is logic in this result since we are in a exponential function: the $e^x$ slope is increasing forever in increasing rate, that increases in increasing rate etc. It never stops.

If one has $a^x$ compared with $x^b$, where $a$ is a constant $> 1$ and $b$ is another constant, at some point the exponential exceeds the power, i.e., $a^x > x^b$ (for large value of $x$)

If one uses $log_e$ in both sides, this feature becomes clearly visible:

$$x\; \ln a = b \; \ln x$$

If $a>1$, the distance between $b$ and $\ln(a)$ can be huge but they are constants. However at some point $x$ will prevail over $\ln(x)$ in such a proportion that the exponential will leave the power function well behind.

In formal terms, it is said that:

$$\lim_{x\to \infty} a^x/x^b = \infty $$

PS: It can be easily proved by L'Hôpital's rule after apply $log_e$ in the numerator and denominator, but it's very intuitive.

Speed is variation of distance, acceleration is variation of speed etc. So a second degree function stops varying after second derivative, e.g, the acceleration is constant.

If one supposes $b=100$ exponent in a power function, if one differentiate 100 times gets a constant, i.e, it stops the rate of variation of variation of variation... The exponential function never stops. For that reason the derivative needs to be a exponential function, because the reasoning is the same. It's a recursive argument.

Any constant in exponent, can be converted to other constant in the base.

So

$$(a^x)' = (e^{x \ln a})' = \ln a \;e^{x \ln a} = \ln a\; a^x$$

it keeps completely the logic of exponential function.

One can imagine hyper exponential function, with an even more radical profile of increase, but that's it for now