Is it Possible to Develop an inverse function using the function it self Is it Possible to Develop (taylor expansion) of  an inverse function by knowing  the function it self ?
If Yes ,Can you illustrate with a simple function 
 I know that we use the identity formula 
$$ f(f^-1(y))= x $$
 A: I'm going to assume that the inverse is not easy to compute or find.
Lets call the inverse function $g(x)$
an inverse function is one that
$f(x) = y \iff g(y) = x$ 
Note that there can be left and right inverses
ie:
$f(x) = y \implies g(y)=x$
or the other way around
$g(y) = x \implies f(x)=y$
$f(x) = x^2$ is an example of a function which can't always be reversed
since $f(2) = 4$ and $f(-2) = 4$
Keep left and right inverses in mind.
For a Taylor series, you just need derivatives of the function at certain points
$g(x) \equiv \sum_{n=0}^{n=\infty} \frac{g^n(0)}{n!}x^n$
As long as you can find the derivatives of $g$ at zero, you're in the clear.
If on the other hand you cant do this, then things get more interesting. 
You can find them from the derivatives of $f$
for example, if we are considering a left inverse function
$x = g(f(x))$ 
differentiate with respect to $x$
$1 = g'(f(x))f'(x)$
This gives us 
$1 = g'(y)f'(x)$
$\frac{1}{f'(x)} = g'(y)$ So we have the first derivative
Now differentiate 
$1 = g'(f(x))f'(x)$ with respect to x again
See here for working
The result is 
$g''(y) = \frac{-f''(x)}{[f'(x)]^3}$
You can continue in this fashion or investigate the generalized Faa di Bruno formula
which makes my head hurt.
