# Question #d4611

Oct 8, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{{x}^{2} + 1}$ or $\frac{- 2 {x}^{- 1}}{x + {x}^{- 1}}$

#### Explanation:

$y = {\cos}^{- 1} \left[\frac{x - {x}^{- 1}}{x + {x}^{- 1}}\right]$

=${\cos}^{- 1} \left[\frac{{x}^{2} - 1}{{x}^{2} + 1}\right]$

Take cosine both sides,

$\cos y = \frac{{x}^{2} - 1}{{x}^{2} + 1}$

Take differentiation both sides,

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

$- \sqrt{1 - {\left(\cos y\right)}^{2}} \cdot \mathrm{dy} = \frac{4 x \cdot \mathrm{dx}}{{x}^{2} + 1} ^ 2$

$- \sqrt{1 - {\left[\frac{{x}^{2} - 1}{{x}^{2} + 1}\right]}^{2}} \cdot \mathrm{dy} = \frac{4 x \cdot \mathrm{dx}}{{x}^{2} + 1} ^ 2$

$- \sqrt{\frac{4 {x}^{2}}{{x}^{2} + 1} ^ 2} \cdot \mathrm{dy} = \frac{4 x \cdot \mathrm{dx}}{{x}^{2} + 1} ^ 2$

$- \frac{2 x \cdot \mathrm{dy}}{{x}^{2} + 1} = \frac{4 x \cdot \mathrm{dx}}{{x}^{2} + 1} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 x \cdot \mathrm{dx}}{{x}^{2} + 1} ^ 2 \cdot - \frac{{x}^{2} + 1}{2 x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{{x}^{2} + 1}$ or $\frac{- 2 {x}^{- 1}}{x + {x}^{- 1}}$