Chapter 3.5, Problem 64E

(a) One way of defining sec^–1x is to say that $y={\sec }^{-1}x\iff \sec y=x\text{and}\text{0}\le y<\pi /2\text{or}\pi \le y<3\pi /2$ Show that, with this definition,

$\frac{d}{dx}\left({\sec }^{-1}x\right)=\frac{1}{\sqrt[x]{x^{2-1}}}$

(b) Another way of defining sec^–1x that is sometimes used is to say that $y={\sec }^{-1}x\iff \sec y=x$ and 0 ≤ y ≤π, y ≠ π/2. Show that, with this definition,

$\begin{array}{l}\frac{d}{dx}\left({\sec }^{-1}x\right)=\frac{1}{\left|x\right|\sqrt{x^2-1}}\\ \end{array}$