Why is the inverse of the derivative of f not the actual derivative of the inverse of f? So, I've explored this a little, but it is still confusing. When you calculate the inverse of a function, f, that is one-to-one, the points switch: a point (2,8) on f would be (8,2) on the inverse. So, one would assume that the derivatives of the functions would also constitute the reversal of points. However, that is not the case. For example, you have:
$f (x) = 5x^2 \phantom{=}\text{ for  $x\geq0$}$
$f '(x) = 10x$
and 
$(f^{-1}) (x) = \sqrt{\frac{x}{5}}$
Here is my question: Why is finding the inverse of the derivative of $f$, $f '(x)$, and taking its inverse not the real derivative of the inverse? I would think $(f^{-1}) '(x) = \frac{x}{10}$, but that is not the case. The real inverse would be taking the derivative of $(f^{-1}) (x)$ and finding $(f^{-1}) '(x) = (\frac{1}{10\sqrt{x/5}})$. In my mind, both of these seem like they could be the derivatives of the inverse, yet only the latter is true. Why is this? 
Also, maybe I missed out in class, but is there some sort of quick relationship between (besides the formula) $f '(x)$ and $(f^{-1}) '(x)$ similar to how points switch between $f (x)$ and $(f^{-1}) (x)$.
Thanks. 

Intuitive thoughts to reflect on: draw the graph of $f$ and mark a point on it (say $(a,f(a))$). Draw the tangent at that point. It will have slope $f'(a)$.
Now flip the entire plane around the line $y=x$. The graph of $f$ has now become the graph of $f^{-1}$, the marked point has become $$(f(a),a)= (f(a),f^{-1}(f(a)))=(b,f^{-1}(b))$$ where $b=f(a)$. The tangent line is still the tangent line, but its slope is inverted ($\Delta y$ and $\Delta x$ have swapped roles for the line, so their ratio is inverted).
Putting this together, we get $f'(a)=\frac{1}{f^{-1}(b)}$. The fact that $a$ and $b$ both appear here is what makes the expressions for $f'(x)$ and $(f^{-1})'(x)$ look less related than they are. Geometrically, the derivative of $f$ and of $f'$ at the same point in the plane (allowing for flipping the plane, of course) are very related. Algebraically, the derivative of $f$ and of $f'$ at the same input value are less so.
