Answer
$$\left.\frac{df^{-1}}{dx}\right|_{x=f(5)}=\frac{1}{6}$$
Work Step by Step
$$f(x)=x^2-4x-5\hspace{1cm}x\gt2$$
To find $df^{-1}/dx$ at $x=0=f(5)$, or $\left.\frac{df^{-1}}{dx}\right|_{x=0=f(5)}$, we need to apply Theorem 3. In detail,
1) Find $\left.\frac{df}{dx}\right|_{x=5}$:
We have $$\frac{df}{dx}=2x-4$$
So, $$\left.\frac{df}{dx}\right|_{x=5}=2\times5-4=6$$
2) According to Theorem 3:
$$\left.\frac{df^{-1}}{dx}\right|_{x=f(5)}=\frac{1}{\left.\frac{df}{dx}\right|_{x=5}}=\frac{1}{6}$$
