# How do you find the derivative of the function: arcsec(x/2)?

May 2, 2016

$\frac{d}{\mathrm{dx}} \text{arcsec} \left(\frac{x}{2}\right) = \frac{2}{x \sqrt{{x}^{2} - 4}}$

#### Explanation:

Using implicit differentiation, we start by letting $y = \text{arcsec} \left(\frac{x}{2}\right)$

$\implies \sec \left(y\right) = \frac{x}{2}$

$\implies \frac{d}{\mathrm{dx}} \sec \left(y\right) = \frac{d}{\mathrm{dx}} \frac{x}{2}$

$\implies \sec \left(y\right) \tan \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sec \left(y\right) \tan \left(y\right)}$

We already know $\sec \left(y\right) = \frac{x}{2}$, and if we construct a right triangle with an angle $y$ such that $\sec \left(y\right) = \frac{x}{2}$ we find that $\tan \left(y\right) = \frac{\sqrt{{x}^{2} - 4}}{2}$. Substituting these in, we have

$\frac{d}{\mathrm{dx}} \text{arcsec} \left(\frac{x}{2}\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$

$= \frac{1}{2 \left(\frac{x}{2}\right) \left(\frac{\sqrt{{x}^{2} - 4}}{2}\right)}$

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