Why is area under hyperbola $\frac{1}{x}$ given by output of function $e^x$? In other words, what does exponential function have in common with area under $y = \frac{1}{x}$? I see that hyperbolic functions also have $e^x$ in their formulas but I don't understand why.

Also, is there intuitive explanation for this?

  • $\begingroup$ This answer may (or may not) help with some of the intuition. $\endgroup$
    – Blue
    Commented Jun 8, 2020 at 7:42

4 Answers 4


If I understand correctly, the OP is asking two questions:

  1. What is the anti-derivative of (1/x)?

Answer : ln(x).

  1. Is there an intuitive (informal) explanation for this?

Answer :

$e$ is chosen so that $y = e^x \Rightarrow y' = e^x.$

Thus, (informally), $x = \ln(y)$ and
$dx/dy = \frac{1}{dy/dx} = \frac{1}{e^x} = \frac{1}{y}.$

Addendum : Response to OP's followup question.
First of all, a function $y = f(x)$ will have an inverse function
$x = g(y)$ for a specific region if and only if $f(x)$ is either strictly increasing or strictly decreasing in that region.
$f(x) = e^x$ is a strictly increasing function.

This explains, for example, why arcsin $x$ is normally required to be an element in $[-\pi/2, \pi/2].$ That is, in that region, the sine function is strictly increasing.

Secondly, there is a theorem in calculus that when $y = f(x)$ has an inverse function $x = g(y)$, with $f(x)$ differentiable at a specific point $x_1$, with $f'(x_1) \neq 0$, and with $y_1 = f(x_1)$
[so that $g(y_1) = x_1$]
then $g(y)$ is differentiable at $y_1$ and $g'(y_1) = 1/f'(x_1).$

I don't think it is constructive to give a formal proof of this theorem, because it would assume too much context. I think that any standard calculus book should have this theorem. This means that to verify the proof of the theorem in the calculus book, you need to start on page 1 of the book.

  • $\begingroup$ Wow, is this trick actually road to rigorous proof? I am asking this because I saw many wrong "intuitions" on $e.$ Thanks $\endgroup$
    – 1b3b
    Commented Jun 7, 2020 at 19:30
  • 1
    $\begingroup$ @1b3b I edited my answer with an addendum to respond to your question. $\endgroup$ Commented Jun 7, 2020 at 19:57
  • $\begingroup$ Yes, I know that function must be bijection. Thanks $\endgroup$
    – 1b3b
    Commented Jun 7, 2020 at 20:23

$$A =\int_1^x \dfrac{1}{x}\cdot dx= \log x $$

The area is the logarithm of $x$ but not its exponential inverse function... $ e^x \; !$

Yes, the inverse hyperbolic functions must involve logarithms.

Like e.g., if $ y = \cosh^{-1} x, $ then that also equals after solving a quadratic starting with its definition...

$$ y = \log (x \pm \sqrt{x^2-1}) $$

This is the case with other inverse hyperbolic functions as well.


First things first, the integral (Area Under) of the hyperbola $\frac{1}{x}$ is not $e^x$, but actually its inverse ln(|x|) (Absolute as $\frac{1}{x}$ exists for x<0 but not ln(x)). Now why is this the case?

To look at this intuitively, we have to see not the area of $\frac{1}{x}$, but the derivative (Gradient) of ln(x) instead. If you have seen the $e^x$, you can visualise ln(x) as the exponential mirrored by the y-axis and rotated 90 degrees clockwise(or you can google ln(x) and it will graph it for you).

You see as x tends towards 0, ln(x) just heads to $-\infty$ exponentially, while on the opposite, it heads to positive infinity(at a decreasing rate). Now on the flipside, $\frac{1}{x}$, for x->0, it will head to positive infinity. That means the slope of ln(x) and be increasingly positive towards infinity, as we clearly see in the uphill of ln(x).

Meanwhile tending towards infinity, the hyperbola will increasingly head towards 0, the slope increasingly flat while never touching the x-axis. This is mirrored in ln(x) heading to infinity at an excruciatingly slow pace.

Now looking at this in a rigorous way, we see $y=ln(x)$, so we can write $$e^y = x$$ We can, in some other time, prove $\frac{d}{dx}(e^x) = e^x$, so we can show $$\frac{d}{dx}(e^y) = \frac{d}{dx}(x)$$ Via the chain rule, which is difficult to prove elegantly $\frac{d}{dx}(e^y) = \frac{d}{dy}(e^y)\cdot \frac{dy}{dx}$ $$(e^y)\frac{dy}{dx} = 1$$ $$\frac{dy}{dx}= \frac{1}{e^y}$$ We can now put y=ln(x) back in $$\frac{dy}{dx}= \frac{1}{e^{ln(x)}}$$ as $e^x$ is the inverse of ln(x) $$\frac{dy}{dx}= \frac{1}{x}$$ So we have proved, for x>0(as ln(x) is only defined for positive numbers) , the derivative of ln(x).

  • 2
    $\begingroup$ I don't understand how did you get $\frac{dx}{dx}$... $\endgroup$
    – 1b3b
    Commented Jun 7, 2020 at 20:24
  • $\begingroup$ basically in a function you consider both sides as equivalent, so as you differentiate both sides by x, it's basically $\frac{d}{dx}(x)$ which is the coefficient of x, that is 1 $\endgroup$ Commented Jun 8, 2020 at 4:41

Let's approach it from the direction of $f(x)=\int_1^x\frac{dt}t$

First, note that by substituting $t\to\frac uy$, we get $$f(x)=\int_y^{yx}\frac{\frac{du}y}{\frac uy}=\int_1^{yx}\frac{dx}x-\int_1^y\frac{dx}x=f(yx)-f(y)$$, hence $$f(x)+f(y)=f(xy)$$

Now consider $f\left(x^n\right)$, by induction we can show that $f\left(x^n\right)=nf(x)$. Similarly consider $f\left(x\right)=f\left(\prod_{i=1}^nx^{\frac1n}\right)$(taking product of $x^{\frac1n}$ $n$ times), we get $f(x)=n\left(x^n\right)$, hence by combining these 2, for any rational $q$, $$f\left(x^q\right)=qf(x)$$

This looks exactly like the logarithm function! If $f(y)=1$ for some $y$, then $f(x)=\log_y(x)$

So now we have seen that $\log_y(x)=\int_1^x\frac{dt}t$ for some $y$, so what is $y$? Suppose $x$ is some small number, let's look at $\log_y(1+x)$. $$\log_y(1+x)=\int_1^{1+x}\frac{dt}t\approx x\cdot\left(\frac11\right)=x$$ This gets more and more accurate as $x$ approaches $0$, so it gives us a way to find $y$: $$\lim_{x\to0}frac1x\log_y(1+x)=1$$ $$\lim_{x\to0}\log_y\left((1+x)^{\frac1x}\right)=1$$ $$\lim_{x\to0}(1+x)^{\frac1x}=y$$ And this is exactly(substituting $n=\frac1x$) $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$$ So $\log_ex=\int_1^x\frac{dy}y$


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