What is the derivative of y="arcsec"(x)?

May 19, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{x}^{2} \cdot \sqrt{1 - {\left(\frac{1}{x}\right)}^{2}}}$

Explanation:

show that

$y = a r c \sec x = \frac{1}{\arccos x} = \arccos \left(\frac{1}{x}\right)$

$\frac{d}{\mathrm{dx}} \left[\arccos u\right] = \frac{1}{\sqrt{1 - {u}^{2}}} \cdot u '$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{1 - {\left(\frac{1}{x}\right)}^{2}}} \cdot \left[- \frac{1}{x} ^ 2\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{x}^{2} \cdot \sqrt{1 - {\left(\frac{1}{x}\right)}^{2}}}$