Single Variable Calculus, Chapter 7, 7.1, Section 7.1, Problem 42

If $\displaystyle G(x) = \frac{1}{f^{-1}(x)}, \quad f(3) = 2 \quad \text{and} \quad f'(3) = \frac{1}{9}$, find $G'(2)$


$
\begin{equation}
\begin{aligned}
\text{if } G(x) &= \frac{1}{f^{-1}(x)}, \text{ then}\\
\\
G'(x) &= \frac{-\left[ \left( f^{-1} \right)\ (x)\right]}{\left[ f^{-1} (x) \right]^2}
\end{aligned}
\end{equation}
$


If $f(3) = 2$, then $f^{-1}(2) =3$
From the theorem in Inverse Function,

$
\begin{equation}
\begin{aligned}
G'(2) &= \frac{-\left[ \frac{1}{f'\left(f^{-1}(2)\right)} \right]}{\left[f^{-1}(2)\right]^2}\\
\\
G'(2) &= \frac{\frac{-1}{f'(3)}}{3^2} = \frac{\frac{-1}{\frac{1}{9}}}{9} = -1
\end{aligned}
\end{equation}
$