How find this value $(f^{-1}(y))^{(4)}|_{y=1}$ let $$f(x)=\sin{x}+x^2+1$$.
find  the value $(f^{-1}(y))^{(4)}|_{y=1}$
My try: since
$$f(x)=\sin{x}+x^2+1$$
 then we can't find $x=g(y)$,so I can't find it.Thank you very much!
 A: If $g(f(x)) = x$, then:
$$ g^{(1)}(f(x)) f^{(1)}(x) = 1 $$
$$g^{(2)}(f(x)) (f^{(1)}(x))^2 + g^{(1)}(f(x))f^{(2)}(x) = 0 $$
etc.  (Two more derivatives for you to calculate)
Then plug in $x=0$.  You have to calculate $f(0)$, $f^{(1)}(0)$, $f^{(2)}(0)$, $f^{(3)}(0)$ and $f^{(4)}(0)$ manually.  The $g^{(k)}(f(x))$ terms can be calculated one after another using the equations above in order.
If we wish to make the working out a little neater, assume argument to all $g$ is $f(x)$ and argument to all $f$ is $x$.  Then:
$$ g = x $$
$$ g' f' = 1 $$
$$ g'' (f')^2 + g' f'' = 0 $$
$$ g''' (f')^3 + 3g'' f' f'' + g' f''' = 0 $$
$$ g'''' (f')^4 + 6g''' (f')^2 f'' + 3g'' (f'')^2 + 4g'' f' f''' + g' f'''' = 0$$
Hope I got that all right.. verify for yourself
A: The equation $f(x)=1$ has two real solutions, namely $x=0$ and $x=\xi\doteq-0.8767262$. Therefore we have two local inverses defined in the neighborhood of $y=1$, one given by
$$x=\phi(y)=a_1(y-1)+a_2(y-1)^2+a_3(y-1)^3+a_4(y-1)^4+?(y-1)^5\ ,$$
and the other by
$$x=\psi(y)=\xi+b_1(y-1)+b_2(y-1)^2+b_3(y-1)^3+b_4(y-1)^4+?(y-1)^5\ .$$
I shall deal with $\phi$. We are asked to determine $\phi^{(4)}(1)=24a_4$.
To this end we write $y-1=:z$ and then have
$$z=\sin x +x^2=x+x^2-{x^3\over6}+?x^5\ .\tag{1}$$
Inverting this series by hand, or using a known formula, or a computer, gives
$$x=z-z^2+{13z^3\over6}-{35z^4\over6}+?z^5\ ,\tag{2}$$
so that we now can read off $a_4=-{35\over6}$, which leads to $\phi^{(4)}(1)=-140$.
There is no "easy way" leading from $(1)$ to $(2)$.
A: Hint: $f^{-1}(f(x))=x \to (f^{-1}(f(x)))' = f'(x)(f^{-1}(f(x)))^{(1)}=1 \to (f^{-1}(f(x)))^{(1)} = \frac{1}{f'(x)}$
$f(0)=1$
