Intuition difference between derivatives of $\exp (x) $ and $\log (x)$ 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!

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) $.
