Differentiate the following with respect of x:
$y = \tan^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}})$

Solution

$Let x = \cos 2 θ \Rightarrow θ = \frac{1}{2} \cos^{- 1} x . . . . . . . . . . . . . . . . . (1) ∴ \sqrt{1 + x} = \sqrt{1 + \cos 2 θ} = \sqrt{1 + 2 \cos^{2} θ - 1} = \sqrt{2} cosθ \sqrt{1 - x} = \sqrt{1 - \cos 2 θ} = \sqrt{1 - 1 - 2 \sin^{2} θ} = \sqrt{2} sinθ Let y = \tan^{- 1} [\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}] = \tan^{- 1} [\frac{\sqrt{2} cosθ - \sqrt{2} sinθ}{\sqrt{2} cosθ + \sqrt{2} sinθ}] = \tan^{- 1} [\frac{1 - tanθ}{1 + tanθ}] = \tan^{- 1} \{\tan (\frac{π}{4} - θ)\} = \frac{π}{4} - θ = \frac{π}{4} - \frac{1}{2} \cos^{- 1} x . . . . . . . . . From (1) ∴ \frac{dy}{dx} = - \frac{1}{2} (\frac{- 1}{\sqrt{1 - x^{2}}}) = \frac{1}{2 \sqrt{1 - x^{2}}}$

For Daily Free Study Material Join wiredfaculty

Differential Equations

Differentiate the following with respect of x:
$y = \tan^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}})$

Some More Questions From Differential Equations Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Differential Equations

Differentiate the following with respect of x:y = tan-1 1 + x - 1 - x1 + x + 1 - x

Some More Questions From Differential Equations Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Differentiate the following with respect of x:
$y = \tan^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}})$