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 $g'''(0)$ 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 attempt:As $g(x)=f^{-1}(x)\Rightarrow fog(x)=x\Rightarrow f'(g(x)).g'(x)=1$
$g'(x)=\frac{1}{f'(g(x))}$ .Similarly i found,$g''(x)=-\frac{f''(g(x))}{(f'(g(x)))^3}$ and Similarly i found $g'''(x)=-\frac{f'(g(x))f'''(g(x)).g'(x)-3(f''(g(x)))^2g'(x)}{(f'(g(x)))^4}$
So $g'''(0)=-\frac{f'(g(0))f'''(g(0)).g'(0)-3(f''(g(0)))^2g'(0)}{(f'(g(0)))^4}$
Now i am stuck here,i cannot find values of $g'(0),f'(g(0)),f''(g(0)),f'''(g(0))$.Please help me.
 A: Say $f(y) = y+y^2/2+y^3/3+y^4/4+y^5/5 = x$, then $f^{-1}(x) = y = g(x)$.
So $g(x)+g(x)^2/2+g(x)^3/3+g(x)^4/4+g(x)^5/5 = x$
First note that $g(0) = 0$. 
Now doing implicit differentiation with chain rule, you would be able to obtain 
$g'(x)(1+4g(x)) = 1.$
Plug in $0$ and you can get $g'(0) = \frac{1}{1+4g(0)} = 1 $.
Continue with implicit differential and you would be able to find $g''(0)$ then $g'''(0)$
$g'(x)(4g'(x)) + g''(x) (1+4g(x)) = 0$
$4(g'(x))^2 + g''(x) + 4g(x)g''(x) = 0$
so $4+g''(0)+4g''(0)=0$ and $g''(0) = -4/5$ 
Differentiate once more, 
$8(g'(x))(g''(x))+ g'''(x) + 4g(x)g'''(x)+ 4g'(x)g'''(x) = 0$
$8(-4/5) + g'''(0) + 4g'''(0) + 4(-4/5)g'''(0) = 0$
so $g'''(0) = 32/9$
A: If $f(x)$ and $g(x)$ are Inverse of each other, Then $$\displaystyle f(g(x)) = g(f(x)) = x$$
Now $$\displaystyle g'(f(x))\cdot f'(x) = 1.......................................(1)$$
Now Given $$\displaystyle f(x) = x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}\;,$$ So $\displaystyle f'(x) = 1+x+x^2+x^3+x^4.$
So $f'(0) = 1$ and $f(0) =0\;,$ Now Put into $(1)\;,$ We get $g'(0) = 1$
Now Again Differentiate $(1)\;,$ with respect to $x\;,$ We get
$$g'(f(x))\cdot f''(x)+f'(x)\cdot g''(f(x))\cdot f'(x)=0.......(2)$$
Now $$f''(x) = 1+2x+3x^2+4x^3\;,$$ So we get $f''(0) = 1$
Now Put $x=0$ in $(2)\;,$ We get $$g'(f(0))\cdot f''(0)+g''(f(0))\cdot f''(0)=0$$
So we get $1+g''(0)\cdot 1 = 0\Rightarrow g''(0) = -1$
Now Again Differentiate $(2)$ with respect to $x\;,$ We get
$$g'(f(x))\cdot f'''(x)+f''(x)\cdot g''(f(x))\cdot f'(x)+g''(f(x))\cdot f"(x)+f'(x)\cdot g'''(f(x))\cdot f'(x)=0$$
Now put $x=0\;,$ We get $$g'(f(0))\cdot f'''(0)+f''(0)\cdot g''(f(0))\cdot f'(0)+g''(f(0))\cdot f"(0)+f'(0)\cdot g'''(f(0))\cdot f'(0)=0$$
