Question

Find d/dx tan⁻¹ √1+x-√1-x/√1+x+√1-x, | x | < 1

Answer 1

so this is the answer

Answer 2

$y = {tan}^{ - 1} ( \frac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } ) \\ \\ put \: \: \: \: x = cos2 \theta \\ \\ 2 \theta = {cos}^{ - 1} x \\ \\ \theta = \frac{1}{2} {cos}^{ - 1}x \\ \\ y = {tan}^{ - 1} ( \frac{ \sqrt{1 + cos2 \theta} - \sqrt{1 - cos 2\theta} }{ \sqrt{1 + cos2 \theta} + \sqrt{1 - cos \theta} } ) \\ \\ y= {tan}^{ - 1} ( \frac{ \sqrt{2 {cos}^{2} \theta } - \sqrt{ 2{sin}^{2} \theta} }{ \sqrt{2 {cos}^{2} \theta} + \sqrt{2 {sin}^{2} \theta } } ) \\ \\ y= {tan}^{ - 1} ( \frac{cos \theta - sin \theta}{cos \theta + sin \theta} ) \\ \\ y = {tan}^{ - 1} ( \frac{1 - tan \theta}{1 + tan \theta} ) \\ \\y = {tan}^{ - 1} tan( \frac{\pi}{4} - \theta) \\ \\ y= \frac{\pi}{4} - \theta \\ \\ y = \frac{\pi}{4} - \frac{1}{2} {cos}^{ - 1} x \\ \\ \frac{dy}{dx} = 0 - \frac{1}{2} \frac{ - 1}{ \sqrt{1 - {x}^{2} } } \\ \\ \frac{dy}{dx} = \frac{1}{2 \sqrt{1 - {x}^{2} } }$