Question

Why do some textbooks say that the derivative of arcsec(x) =

Answer 1

\[\frac{ 1 }{ \left| x \right| \sqrt{x^2-1}}\]

Answer 2

Where did the absolute value come from?

Answer 3

Let's review how that derivative is obtained.$$y = \sec^{-1} x \iff \sec y = x $$ After taking derivative we have$$ \sec y \tan y \cdot y' = 1
\\ \, \\ y' = \frac{1}{\sec y \tan y}
$$ The product of sec y and tan y is always positive.

Answer 4

Ohh, I did not know that sec(y) * tan(y) is always positive. Interesting. Thanks for the answer.

Answer 5

if you mash out the derivative you end up here eventually

\(y' = \dfrac{\cos^2 y}{\sin y} = \dfrac{(\frac{1}{x})^2}{( 1 - \frac{1}{x^2})^{1/2}}\)

when you then simplify into the form you quote, you lose the fact that the value of \(x\) in the numerator is squared, so you preserve that by using the abs value sign

Answer 6

@steve816 this is all about the domain of arcsec(x)
In order to make sec a one-to-one function and hence invertible, the standard domain of sec is usually restricted to [0,pi/2) or (pi/2, pi], with a the range [1, infinity) or (-inf, -1]. The inverse will have swapped the domain and range such that the range will always be positive. One way to ensure this is by placing an absolute value sign on the x (on could place it as a squared form inside the square root).