-1
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

marked as duplicate by Ramiro, gt6989b, Xander Henderson, Lord Shark the Unknown, A. Pongrácz Jan 21 at 7:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

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.

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

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

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

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ "Exponential functions describe population growth, radioactive decay and interest" Why does e do that. What's wrong with just using 2^x or 3^x or 4^x $\endgroup$ – Seph Jan 21 at 4:38
  • $\begingroup$ There certainly appears to be something interesting about $e$. You are right. I , like i said, think of it as a phenomenon. $\endgroup$ – Chris Custer Jan 21 at 4:42
  • 1
    $\begingroup$ @Seph You could use $2^x$ or $3^x$, etc, to describe population growth and radioactive decay and interest. But if we use $e^x$ then the calculations are a little simpler and more clean, because the derivative of $e^x$ is just $e^x$. $\endgroup$ – littleO Jan 21 at 4:49
  • 2
    $\begingroup$ The important thing is that $e$ is the unique constant with $\lim_{x \to 0} \frac{e^x-1}{x} =1$. THIS is why it is equal to its own derivative. $\endgroup$ – Randall Jan 21 at 4:50
  • $\begingroup$ There are many interesting patterns in nature, including mathematics. The Bell curve is interesting. It seems kind of natural for things to be distributed that way. More around the average, fewer at the extremes. I remember a joke to the effect that God must like average people: he made an awful lot of them. But normal is just a setting on your hair dryer. But I digress. $\endgroup$ – Chris Custer Jan 21 at 4:52

Not the answer you're looking for? Browse other questions tagged or ask your own question.