Chapter 3.9, Problem 57E

Use the Derivative Rule in Section 3.8, Theorem 3, to derive

$\frac{d}{dx}{\sec }^{-1}x=\frac{1}{\left|x\right|\sqrt{x^2-1}},\left|x\right|>1$

Theorem 3 – The Derivative Rule for Inverses

If $f$ has an interval $I$ as domain and $f^{\text{'}}(x)$ exists and is never zero on $I$, then $f^{-1}$ is differentiable at every point in its domain (the range of $f$). The value of ${\left(f^{-1}\right)}^{\text{'}}$ at a point $b$ in the domain of $f^{-1}$ is the reciprocal of the value of $f^{\text{'}}$ at the point $a=f^{-1}\left(b\right)$.

${\left(f^{-1}\right)}^{\text{'}}\left(b\right)=\frac{1}{f^{\text{'}}\left(\left(f^{-1}\right)\left(b\right)\right)}$

Or

${\frac{d{f}^{-1}}{dx}|}_{x=b}=\frac{1}{{\frac{df}{dx}|}_{x=f^{-1}\left(b\right)}}$