Prove the relation involving derivative of inverse of a function I want to prove the following result:
$$ {f^{-1}}'(x)=\frac{1}{f'({f^{-1}}(x))}$$ 
Is simple application of chain rule a valid proof of it?
i.e. $$f({f^{-1}}(x))=x \implies \frac{df({f^{-1}}(x))}{dx}=1$$ and hence the result. Or is this not a standard proof? Is there additional conditions necessary, expect of course that the function is bijective, or that the inverse exists.
 A: The chain rule "derivation" is invalid, because it assumes $f^{-1}$ is differentiable, which is begging the question (circular reasoning).
Here is a valid proof:
Conditions:
i) Function $f$ defined and continuous on a neighborhood of point $x$.
ii) Inverse function $f^{-1}$ defined and continuous on a neighborhood of $y = f(x)$.
iii) $f$ differentiable at point $x$, and $f'(x) \not = 0$.
By the differentiability theorem:
$$f(x + h) - f(x) = h(f'(x) + g(h))$$
where $g(h)$ goes to zero as $h$ goes to zero.
Define $k := h(f'(x) + g(h))$
By limit theorem $k$ also goes to zero as $h$ does.
Since $f'(x)$ is non-zero, you can restrict $|h|$ so that that $|g(h)|$ is less than $|f'(x)|$ (and thus $k$ is non-zero), and add a further restriction so that $f^{-1}$ is defined and continuous on the interval $[y-k, y + k]$. 
Now you can show that the derivative exists:
$$\frac{1}{f'(x)}  = \lim_{h\to 0} \frac{h}{f(x+h) - f(x)} = \lim_{h\to 0} \frac{f^{-1}(f(x + h)) - f^{-1}(f(x))}{f(x+h)-f(x)} = \lim_{k\to 0} \frac{f^{-1}(y+k) - f^{-1}(y)}{k}.$$
Note: If a function is continuous and strictly increasing or strictly decreasing on a neighborhood of $x$, the inverse function theorem says that $f^{-1}$ is then also strictly increasing/decreasing on a neighborhood of $y$. Thus the inverse function is defined and continuous on a neighborhood of $y$.
A: For differentiable functions $\mathbb{R} \to \mathbb{R}$ it is easy using geometrical interpretation. Take the graph of $f$, then the graph of $f^{-1}$ is symmetrical with respect to $\{y = x\}$ line. However, $f'(x) = \tan \alpha$ implies that $${f^{-1}}'(y) = \tan\left(\frac{\pi}{2}-\alpha\right) = \cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{f'(x)}.$$
Hope that helps ;-)
A: The question is: Why is $f^{-1}$ differentiable? In the theorem you probably learned in a lecture this isn't assumed. However if you assume that $f^{-1}$ and $f$ are differentiable at and all the expressions make sense $f^{-1}$ should be a function  and $f'\neq 0$ (both at least locally).
But nevertheless, with the chainrule the formula is easy to remember.
A: Let $y_0\in \mathbb{R}$. Then there is $x_0\in\mathbb{R}$ such that $f(x_0)=y_0$. If $x\rightarrow x_0$ and by continuity of $f$ (because $f$ is differentiable) we have $y=f(x)\rightarrow f(x_0)=y_0$.
Now, let's check the differentiability of $f^{-1}$ by the definition:
$$(f^{-1})'(y_0)=\lim_{y\rightarrow y_0}\frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=\lim_{y\rightarrow y_0}\frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\\=\lim_{x\rightarrow x_0}\frac{x-x_0}{f(x)-f(x_0)}=\frac{1}{f'(x_0)}=\frac{1}{f'(f^{-1}(y_0))}.$$
A: Here is a proof of the rule using the definitions a little more carefully.
Let $f^{-1}$ and $f$ be defined and differentiable over some domain $A \times f(A)$. It follows that $f$ is continuous at any $x_0\in A$: for $\alpha>0$ there exists some $\delta>0$ such that
$$0<|x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\alpha.$$
For $f(x_0)=y_0\in f(A)$, let $L=\left(f^{-1}(y_0)\right)'$. It follows that for any $\epsilon>0$ there exists some $\beta>0$ such that
$$0<|y-y_0|<\beta \Rightarrow \left|\frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}-L\right|<\epsilon.$$
For $\epsilon>0$ choose $\beta=\alpha$, there exists therefore some $\delta>0$ such that
$$0<|x-x_0|<\delta \Rightarrow \left|\frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}-L\right|<\epsilon.$$
This is exactly, however, that for any $\epsilon>0$ there exists some $\delta>0$ such that $\frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=\frac{x-x_0}{f(x)-f(x_0)}=\frac{1}{f'(x_0)}$ gets infinitesimally close to $L$.
Therefore, $\left(f^{-1}(y_0)\right)'=\frac{1}{f'(x_0)}$.
