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

**dgmath** · November 18th 2009, 07:03 AM

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

Thanks in advance.

**Drexel28** · November 18th 2009, 12:34 PM

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 $\left\{f^{-1}(x)\right\}''$ .

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

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

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

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