taking derivative of inverse function

Asked Jul 09 '17 at 22:29

Active Jul 09 '17 at 22:43

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Let $f(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}$, and $g(x)=f^{-1}(x)$. Compute $g^{(3)}(0)$ (the 3rd derivative of g).

I have a solution but its not giving the correct answer. My question is not how to solve the problem but what I did wrong in my solution.

First $g(0)=f^{-1}(0)=0$

$g^{'}(x)= (f^{'}(f^{-1}(x)))^{-1}$, so $g^{'}(0)=1$

$g^{(2)}(x)=-(f^{'}(f^{-1}(x)))^{-2} \cdot f^{(2)}(f^{-1}(x)) \cdot g^{'}(x)= -(g^{'}(x))^{-3} \cdot f^{(2)}(f^{-1}(x))$, so $g^{(2)}(0)=-1$

Finally $g^{(3)}(x)=3(g^{'}(x))^{-4} \cdot g^{(2)}(x) \cdot f^{(2)}(f^{-1}(x))+ f^{(3)}(f^{-1}(x)) \cdot g^{'}(x) \cdot -(g^{'}(x))^{-3}$, so $g^{(3)}(0)=-3-2=-5$. The correct answer is 1, however. I've checked my work so many times but still can't figure out what I did incorrect.

edited Jul 09 '17 at 22:43

asked Jul 09 '17 at 22:29

chrismc

1

possible dublicate: https://math.stackexchange.com/questions/1380972/third-derivative-of-inverse-function – serg_1 Jul 09 '17 at 22:34
But my question is where did I go wrong in my solution – chrismc Jul 09 '17 at 22:35
By $f^{-1}(x)$ do you mean the inverse function or the reciprocal ? – Donald Splutterwit Jul 09 '17 at 22:36
Inverse function – chrismc Jul 09 '17 at 22:36
Note that $f(x)$ is the order-5 Taylor polynomial for $-\log(1-x)$, which has inverse $1-e^{-x}$. So the first few terms of $f^{-1}(x)$ will match $1-e^{-x}$. – Jair Taylor Jul 09 '17 at 22:53

taking derivative of inverse function

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