Explanation of why $\frac{d}{dx} e^x=e^x$

Question

I'm taught that all the way back when I'm in high school that $\dfrac{de^x}{dx}=e^x$ and $\int e^x dx=e^x$.

Can someone explain why this is the case?

Answer 1

If you differentiate any exponential function, you can write something like this:

$$ \frac{d}{dx} 2^x = \lim_{h\to0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to0}\frac{2^{x+h} - 2^x}{h} = \lim_{h\to0}2^x\frac{2^h-1}{h}.\tag{1} $$

Now an odd thing: The last expression in ($1$) is equal to $$ 2^x\lim_{h\to0} \frac{2^h-1}{h}\tag{2} $$ because $2^x$ is "constant", but "constant" in this case means not depending on $h$, i.e. $2^x$ does not change as $h$ goes to $0$. But next we will say that the expression ($2$) is $$ (2^x\cdot\text{constant}) $$ where this time "constant" means not depending on $x$, i.e. that last quantity does not change as $x$ changes.

In that way we see that $\dfrac{d}{dx}2^x = (2^x\cdot\text{constant})$.

But what number is this constant?

Letting $f(x)=2^x$ and $f'(x)=(2^x\cdot\text{constant})$, we see that $f'(0)=(2^0\cdot\text{constant})$. Since $2^0=1$, the "constant" is $f'(0)$; it's how fast $f$ is changing at that point.

As $x$ goes from $-1$ to $0$ to $1$, $f(x)$ goes from $1/2$ to $1$ to $2$, and so its average rate of change between $-1$ and $0$ is $1/2$ and its average rate of change between $0$ and $1$ is $1$, and so its exact rate of change at $0$ is somewhere between $0$ and $1$.

If we had considered $4^x$ instead of $2^x$, we would have had a bigger constant. By considering $f(x)$ when $x=-1/2$ and when $x=0$, we can see that $f'(0)>1$.

When the base is $2$, the "constant" is somewhere between $1/2$ and $1$.

When the base is $4$, the "constant" is bigger than $1$.

When the base is $e$ the constant is $1$. That's what's "natural" about $e$.

The number $e$ must therefore be somewhere between $2$ and $4$. With more work we can narrow it down to $e=2.71828\ldots$, but this particular method of narrowing it down is not efficient.

Answer 2

Apart from the fact that the notation has been reversed and, as has been pointed out, the $dx$ is missing from the integral expression - which is considered extremely bad Maths, the question is back to front. The problem is to find a function that is it's own derivative, ie find $f(x)$ such that $f'(x) = f(x)$. Looking at various graphs and their gradient functions soon gives you the idea that the function in question must start small and then rise - the gradient must increase as the function does. Thus you get the idea of it having to be an exponential function. Then, using Newtonian notation, $\lim([a^{x+h}-a^x]/h)$ will give you the derivative of $a^x$. This turns out to be $k[a^x)$ From here you can investigate various values of "$a$" or using a binomial expansion to find the value for which $k = 1$. In fact there are only two functions that are identical to their derivatives. The second one is $f(x) =0$. It is the uniqueness of $e^x$ in this respect that makes it special.

Answer 3

Let me continue the discussion Julien mentioned in the comment.

The discovery for $(e^x)' = e^x$ is to look for a function $f$:

$f$ is differentiable, and $$\text{the rate of change of $f(x)$ is itself.}$$ i.e., $f' = f$, also it satisfies $f(0)=1$.

For $f' = f$, hence $f'$ is differentiable again, and $f''= (f')' = f' = f$, and we have $$ f^{(n)}(x) = \cdots = f'(x) = f(x),\tag{1} $$ for any $n$. Now writing this $f(x)$ in Taylor expansion: $$ f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!}x^2 + \cdots + \frac{f''(0)}{n!}x^n + \cdots. $$ Now by (1): for any $n$ $$ f^{(n)}(0) = \cdots = f'(0) = f(0), $$ hence: $$ f(x) = f(0) + f(0) x + \frac{f(0)}{2!}x^2 + \cdots + \frac{f(0)}{n!}x^n + \cdots \\ = 1 + x + \frac{1}{2!}x^2 + \cdots\frac{1}{n!}x^n + \cdots. \tag{2} $$

(2) implies:

For such a function whose rate of change is itself, it must increases faster than any degree $n$ polynomial!

Therefore a reasonable guess for such a function is that it is exponential function: $$ f(x) = a^x, $$ with a magical yet mysterious number $a$, and the discussion goes on with Michael Hardy's answer.

Answer 4

Let $f$ and $g$ differentiable functions such that $$f(g(x))=x=g(f(x))$$ Then applying chain's rule $$f'(g(x)) g'(x)=1$$ replacing $x=f(y)$ we have: $$f'(g(f(y))) g'(f(y))=1$$ Therefore $$f'(y) g'(f(y))=1 $$ Note that $\log e^x=x=e^{\log x}$ , so if we take $e^x=f(x)$ and since $\frac{d}{dx}log x=\frac{1}{x}$ we get $$(e^x)'(\frac{1}{e^x})=1$$ therefore $$(e^x)'=e^x$$ The other equation $\int e^xdx = e^x + c$ is a simple application of the fundamental theorem of calculus.