Why is the derivative of $e^x$ equal to $e^x$?

Question

I know that for the exponential function $e^x$ that the derivative will equal $e^x$ itself. But why? And also what is the significance of that?

Is that what gives $e$ its power? The rate of change of $e$ as it grows to the power of $x$, is $e^x$ itself. I get that the function doesn't produce $e^x$, that merely the rate at which its changes between $e^x$ and $h$ as $h \to 0$. But the intuition as to why and what is the significance to math eludes me. Mind you, I understand the math behind it, just not the intuition. For example, I understand this entire post https://mathinsight.org/exploring_derivative_exponential_function but why

I understand the math behind this post Could you explain why $\frac{d}{dx} e^x = e^x$ "intuitively"?. But I'm looking for an analogy.

Answer 1

Note $\ln e^x =x$. Use the chain rule: $\ln'(e^x)\cdot (e^x)'=1\implies \frac1{e^x}\cdot (e^x)'=1\implies (e^x)'=e^x$.

Answer 2

Without knowing anything about $e$, one could pose the question - is there a function $f(x)$ such that the derivative of the function evaluated at any point $x$ is equal to $f(x)$. The function that satisfies this property happens to be $\alpha e^x$.

$${d \over dx}f(x) = f(x)$$

$$\implies {df(x) \over f(x)} = dx$$

$$\implies ln(f(x)) = x + c$$

$$ \implies f(x) = \alpha e^x$$

Another way to look at it could be: (i) if the rate of change of a function is constant, the function is linear. (ii) if the rate of change is polynomial, the function is a polynomial (of a higher order than the derivative), and (iii) if the rate of rate change is exponential, the function is exponential.

$$ {d \over dx}f(x) = ae^x$$

$$ \implies df(x) = ae^x dx$$

$$ \implies f(x) = ae^x + c$$

Answer 3

I find cosine and sine very natural and significant in mathematics, and they can be represented by Euler's formula:

$$ e^{i\theta} = \cos(\theta) + i\sin(\theta)$$

And from this we know the derivatives of these natural pair of functions easily thanks to derivative of $e^z$ being itself:

$$\frac{d}{d\theta}e^{i\theta} = \frac{d}{d\theta}\cos(\theta) + i\frac{d}{d\theta}\sin(\theta)$$ $$ ie^{i\theta} = -\sin{(\theta)} + i\cos(\theta) = \frac{d}{d\theta}\cos(\theta) + i\frac{d}{d\theta}\sin(\theta)$$

so

$$ \frac{d}{d\theta}\cos(\theta) = -\sin{(\theta)} \tag 1$$ $$ \frac{d}{d\theta}\sin(\theta) = \cos(\theta) \tag 2$$

Perhaps it was the necessity of (1) and (2) that fueled nature to mandate

$$ \frac{d}{dz}e^z = e^z$$