Let $f(x)=x+\frac{x^2}{2} + \frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}$ and let $g(x)=f^{-1} (x)$. Find $g’’’(0)$ My method is extremely inefficient, but here it is
$$g(f(x))=x$$
$$g’(f(x)).f’(x)=1$$Differentiating wrt x multiple times
$$g’’’(f(x))(f’(x))^3 + (g’’(f(x)))(2f’(x))(f’’(x)) + g’’(f(x)).f’(x).f’’(x) + g’(f(x)).f’’’(x)=0$$
I may have made some computation errors in this, but I can’t seem to find any (I apologize if there are)
$f(x)=0$ at $x=0$
$$g’’’(0) +2 g’’(0) + g’’(0) +2g’(0)=0$$
How do I proceed from here? Is there a better way to approach this ?
 A: Take $$g(f(x))=x$$
$$\Rightarrow g'(f(x)).f'(x)=1$$
$$\Rightarrow g'(f(x))=\frac{1}{f'(x)}$$
$$g''(f(x)).f'(x)=-\frac{f''(x)}{(f'(x))^2}$$
$$\Rightarrow g''(f(x))=-\frac{f''(x)}{(f'(x))^3}$$
I'll let you differentiate once again.
Note that I have not combined any other function with $g'(f(x))$ It is done so that it restricts $g'(f(x))$ from popping up in the final $g'''(x)$ expression. Same goes with $g''(x)$.
Spoilers

$$g'''(f(x)).f'(x)=-\frac{f'''(x).(f'(x))^3-3(f''(x))^2.(f'(x))^2}{(f'(x))^6}$$ $$g'''(f(x))=-\frac{f'''(x).(f'(x))^3-3(f''(x))^2.(f'(x))^2}{(f'(x))^7}$$


 $$f(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}$$ $$f'(x)=1+x^2+x^3+x^4+x^5$$ $$f''(x)=1+2x+3x^2+4x^3+5x^4$$ $$f'''(x)=2+6x+...$$


 To conclude with, substitute $0$ in the above four(as$ f(0)=0$) $$f(0)=0$$ $$f'(0)=1$$ $$f''(0)=1$$ $$f'''(0)=2$$ $$\therefore g'''(0)=-\frac{f'''(0).(f'(0))^3-3(f''(0))^2.(f'(0))^2}{(f'(0))^7}$$ $$g'''(0)=-\frac{2.1-3.1.1}{1}$$ $$\therefore g'''(0)=1$$ :)

A: If $f(x)$ is an infinite series, then
$$y=f(x)=x+x^2/2+x^3/3+x^4/4+.....=-\ln(1-x) \implies y=-\ln(1-x) $$ $$ \implies x=1-e^{-y} \implies x=f^{-1}(y)=1-e^{-y} \implies f^{-1}(x)= 1-e^{-x}=g(x).$$
Then $$g'(x)=e^{-x}, ~g''(x)=-e^{-x} \implies g'''(x)=e^{-x} \implies g'''(0)=1$$
A: If $f(x)$ is a polynomial, OP's approach is correct. The complete solution is here:
$$f(x)=x+x^2/2+x^3/3+x^4/4+x^5/5 \implies f(0)=0, f'(0)=1,f''(0)=1,=f'''(0)=2,$$
$$y=f(x) \implies x=f^{-1}{y} \implies x=g(y), gf(x)=1 \implies  g'(f(x))f'(x)=1~~~~(1)$$
$$\implies g'(0)=1/f'(0)=1$$
D. (1) w.r.t. $x$, we get
$$(2)~~~ g''(f(x)(f'(x))^2+g'(f(x)) f''(x)=0 \implies g''(0)+g'(0)f''(0)=0 \implies g''(0)=-1.$$
D. (2) w.r.t. $x$, we get
$$g'''(f(x))(f'(x))^3+3g''(f(x)) f'(x) f''(x)+g'(f(x)) f'(x) f'''(x)=0 ```(3)$$
Using values of : $g(0),g'(0),g''(0),f'(0),f''(0), f'''(0)$ in above, we get
$$g'''(0) +3g''(0)+2g'(0)=0 \implies g'''(0)=3-2=1$$
A: Hint : $g(0)=0$ and
$$g' = \frac{1}{f' \circ g}$$
You deduce $g'(0)$ and $g''$, then $g''(0)$ and $g'''$, then finally $g'''(0)$.
