Finding The Derivative Of an Inverse

Question

I know the general formula is $$\frac{1}{f\prime(f^{-1}(x))}$$ would anyone know what the formula would be for the second derivative? is it f double prime in the denominator or do you take the entire equation and apply the quotient rule?

Answer 1

Hint: Differentiate the relation $f(f^{-1}(x)) = x$ twice using the Chain Rule, and you can solve for $(f^{-1})''(x)$.

Answer 2

Hint: let $f^{-1}(x)=g(x)$ then $g$ satisfies $g(f(x))=x$. Now differentiate this twice \begin{eqnarray*} f'(x)g'(f(x))=1 \\ f''(x)g'(f(x))+(f'(x))^2g''(f(x))=0. \end{eqnarray*} To finish calculate $f'$ and $f''$ and substitute $x=-1$.

Answer 3

"I know the general formula is "

$\frac{1}{f\prime(f^{-1}(x))}$

That is correct. (Because $f(f^{-1}(x))=x$ so by chain rule $[f(f^{-1}(x))]' = f'(f^{-1}(x))*{f^{-1}}'(x) = 1$.

"is it f double prime in the denominator or do you take the entire equation and apply the quotient rule?"

Yes.

$h(x) = \frac{1}{f\prime(f^{-1}(x))}$ so $h'(x) = -\frac {[f\prime(f^{-1}(x))]'}{[f\prime(f^{-1}(x))]^2}=-\frac {f''(f^{-1}(x))*(f^{-1}(x))}{[f\prime(f^{-1}(x))]^2}$

$=-\frac {\frac{f''(f^{-1}(x))}{f'(f^{-1}(x))}}{[f\prime(f^{-1}(x))]^2}= -\frac{f''(f^{-1}(x))}{[f\prime(f^{-1}(x))]^3}$

But the real question is, how can you evaluate $f'(f^{-1}(x))$ without determining what $f^{-1}(x)$ is. And if you can determine what $f^{-1}(x)$ is in the first place, why do all this indirect oblique derivation when we could have just expressed $f^{-1}(x)$ and derived the derivative directly?