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. 

Maybe this isn't a great reason, but why should you expect the inverse of the derivative to be the derivative of the inverse?  Take some simple examples to see why it shouldn't be a universal truth.
The function $f(x) = x$ has derivative $f'(x)=1$, which isn't even invertible.  And there are hardly nicer functions than $f(x)=x$.
For your last question, note that $f(f^{-1}(x))=x$ by definition (when $f$ is invertible). Then, assuming enough differentiability, the chain rule gives
$$
\frac{d}{dx} f(f^{-1}(x)) = f'(f^{-1}(x))\cdot \frac{d}{dx}f^{-1}(x) = 1.
$$ 
Hence 
$$
\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}.
$$
This is what you need. 
Other answerers have given nice geometric reasons why this should be the case.
