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!

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.
