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It is well known that the exponential function $\exp (x)$ has the derivative $$\frac{d}{dx} \exp (x) = \exp (x).$$ However, its inverse, the (natural) logarithm, $\log (x)$, in fact changes after derivation: $$\frac{d}{dx} \log (x) = \frac{1}{x}. $$ I understand why this is correct, but this is not intuitive to me. Why does $\exp (x) $ not change after derivation, while $\log (x) $ does and they are just mirrored at $y = x $?

EDIT: Thank you all for your answers, they really helped me understanding this!

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  • $\begingroup$ Log and experience are different functions. Why would their derivatives behave the same? Okay experience an log are inverses and inverses are reflexive over the line y=x. So the derivative of $f(x)$ at x=k will be 1 over the derivative of f inverse at x=f(k). Let j=exp k and k =log j. The derivative of exp(x) at x=k is exp (k). So the derivative of log at x=exp (k)=j is 1/exp(k) = 1/j. $\endgroup$ – fleablood Mar 20 '17 at 2:10
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If $f(a) = b$ and $f^{-1}(b) = a$ all continuous and differentiable, then $$\frac{df^{-1}}{dx}]_{x = b} = \frac{1}{\frac{df}{dx}]_{x=a}}$$

Think about that in terms of slope of the tangent. When you reflect the graph of $f(x)$ which here is $e^x$, the reflection across the $45^{\circ}$ or $y=x$ line is $f^{-1}(x)$ which here is $\ln(x)$.

The point $(a,b) = (a, e^a)$ reflects to $(b, a) = (e^a , a)$

The slope of the tangent to $f = e^x$ at $(a,b)$ is $e^a = b$

The slope of the tangent to $f^{-1}$ at $(b,a)$ is, by geometry of the reflection and definition of slope, the reciprocal of the above slope = $1/b = 1/e^a$

Being careful that $(x, y)$ have different meanings on the two graphs (which is why I specified a point $(a,b)$ rather than getting tangled in $x$'s and $y$'s), the slope of the tangent to $f^{-1} = \ln(x)$ at $(b,a)$ is $1/b = 1/x$ for the "new" x on the inverse function.

You are expecting the equations in written form to be symmetrical, which they aren't, because of difficulties of notation and our definitions of functions. But the graphs are meaningfully symmetrical.

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Heuristically, $exp(x)$ has a derivative that is equal to the image of the function, thus it quite naturally follows that its inverse will have a derivative that is the inverse of the pre-image.

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  • $\begingroup$ That's very nice way of putting it! $\endgroup$ – fleablood Mar 20 '17 at 2:38
  • $\begingroup$ I really like your answer, simple and short! $\endgroup$ – Staki42 Mar 20 '17 at 12:49
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We can look at it geometrically.

Let $f(x)=e^x$, so $f^{-1}(x)=g(x)=\log(x)$. Any function's inverse should look like its reflection over $y=x$. Let's consider a point $(x,f^{-1}(x))=(x,y)$ on the graph of log.

By definition, $f(y)=x$. As you know, the tangent line to $f$ has slope $e^x$, so the tangent line to $f$ has slope $s_f=e^y$. We want the slope of $g$.

Well, since the graphs are reflections over $y=x$, so too should be the tangent lines. So, its slope should be the reciprocal of the slope of the other line: $$ s_g = \frac{1}{s_f} = \frac{1}{e^y} = e^{-g(x)} = e^{-\log(x)} = \frac{1}{e^{\log(x)}} = \frac{1}{x} $$ Thus, based on this geometric argument, the derivative of $g$ should be $1/x$ (since that's what the slope of the tangent line should be).

See here for some pictures.

Indeed, in general: $$ \frac{d}{dx}f^{-1}(x) = [f'(f^{-1}(x))]^{-1} $$


Also, I like to look at the Taylor (or MacLaurin) series: $$ e^x=\sum_n \frac{x^n}{n!} $$ $$ \log(1-x) = -\sum_n \frac{x^n}{n} $$ $$ \frac{1}{1-x} = \sum_n x^n $$ Notice that term-by-term differentiation causes $e^x$ to become "itself", and that the same action on $\log$ causes it to become the same as $(1-x)^{-1}.

Not sure if it helps with intuition though.

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If you know the chain rule, you can start from $$ x = e^{\ln x} $$ and take the derivative on both sides: $$ 1 = e^{\ln x}(\ln'x) = x\ln'x, $$ such that $\ln'x = 1/x$. Alternatively, $$ y = \ln(e^{y}), $$ so that $$ 1 = e^{y}\ln'(e^{y}). $$ Rename $x=e^y$, and you get the same result.

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Let $x$ and $y$ be related by

$$ y = \exp(x) \qquad \qquad x = \ln(y) $$

Then, this derivative becomes

$$ \mathrm{d}y = y \, \mathrm{d} x $$

or, if you don't like differentials,

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = y $$

Either way, it's clear this isn't anything resembling being symmetric in $x$ and $y$. There's no reason that 'mirroring' them should give a result of the same form.

Instead, we have $$ \frac{\mathrm{d}x}{\mathrm{d}y} = \frac{1}{y} $$


That said, if we invert differentiation rather than inverting $\exp$, we do get something similar; e.g.

$$ \int_{-\infty}^x \exp(t) \, \mathrm{d}t = \exp(x) $$

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If $f $ and $f^{-1}$ are reflexive over the line $y=x$.

So at the point $(k,j) $ the derivative of $f$ is $f'(k) $ and so the derivative of $f^{-1} $ at $(j,k) $ will be ${f^{-1}} '(j)=\frac 1 {f'(k)} $.

So if $f=\exp $ and $f^{-1}=\log $ we have that at the point $(k,j=\exp (k))$ we know $\exp'(k)=\exp(k)$. Therefore at the point $(j=\exp (k),k)$, we know $\log'(j)=\frac 1 {\exp (k)}=\frac 1j $.

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Interesting to note that by the chain rule:

$f\circ f^{-1}(x)=x $

So $(f\circ f^{-1})'(x)=f'(f^{-1}(x)){f^{-1}}'(x)=1 $

and ${f^{-1}}'(x)=\frac 1 {f'(f^{-1}(x))} $

which is interesting.

And it means $\log'(x)=\frac 1 {\exp' (\log (x))}= \frac 1 {\exp(\log (x))}= \frac 1x $

And if we want to be perverse:

$\exp'(x)=\frac 1 {\log'(\exp (x))}=\frac 1 {1/\exp (x)}=\exp (x) $.

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