# Thread: how would I find a general formula for f^(-1)''(x) ?

1. ## how would I find a general formula for f^(-1)''(x) ?

how would I find a general formula for f^(-1)''(x) ?

2. Originally Posted by dgmath
The title explains it all. Thanks in advance.
I assume the following is what you are asking for. I will give you the outline without all the neccessary filler

Problem: Find a general formula for $\displaystyle \left\{f^{-1}(x)\right\}''$.

Lemma: $\displaystyle \left\{f^{-1}(x)\right\}'=\frac{1}{f'\left(f^{-1}(x)\right)}$

Proof: Notice that $\displaystyle f\left(f^{-1}(x)\right)=x$ so that $\displaystyle f'\left(f^{-1}(x)\right)\cdot \left\{f^{-1}(x)\right\}'=1$. The conclusion follows. $\displaystyle \blacksquare$

Proof:So then $\displaystyle \left\{f^{-1}(x)\right\}''=\left\{\frac{1}{f'\left(f^{-1}(x)\right)}\right\}'$$\displaystyle =\frac{-1}{\left(f'\left(f^{-1}(x)\right)\right)^2}\cdot\left\{f'\left(f^{-1}(x)\right)\right\}'=\frac{-1}{\left(f'\left(f^{-1}(x)\right)\right)^2}\cdot f''\left(f^{-1}(x)\right)\cdot\frac{1}{f'\left(f^{-1}(x)\right)}$$\displaystyle =\frac{-f''\left(f^{-1}(x)\right)}{\left(f'\left(f^{-1}(x)\right)\right)^3}$