# What is the derivative of arcsec(x/2)?

Jul 29, 2018

$\frac{d}{\mathrm{dx}} \left(a r c \sec \left(\frac{x}{2}\right)\right) = \frac{2}{\left\mid x \right\mid \sqrt{{x}^{2} - 4}}$

#### Explanation:

Let $y = a r c \sec \left(\frac{x}{2}\right)$, then:

$2 \sec y = x$

Differentiate implicitly:

$2 \sec y \tan y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sec y} \frac{1}{\tan} y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x} \frac{1}{\tan} y$

using now the identity:

${\tan}^{2} y = {\sec}^{2} y - 1 = {x}^{2} / 4 - 1 = \frac{{x}^{2} - 4}{4}$

we have that for $x \in \left[2 , + \infty\right)$, that is for $y \in \left[0 , \frac{\pi}{2}\right)$:

$\tan y = \frac{\sqrt{{x}^{2} - 4}}{2}$, so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{x \sqrt{{x}^{2} - 4}}$

while for $x \in \left(- \infty , - 2\right]$, that is for $y \in \left(\frac{\pi}{2} , \pi\right]$:

$\tan y = - \frac{\sqrt{{x}^{2} - 4}}{2}$, so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{x \sqrt{{x}^{2} - 4}}$

We can then write the derivative for both intervals as.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{\left\mid x \right\mid \sqrt{{x}^{2} - 4}}$