Mht CetDifferentiationMHT CET 2026 16th April Morning ShiftMCQ+2 / -02026If $y = \tan^{-1}\left[\dfrac{x - \sqrt{1 - x^2}}{x + \sqrt{1 - x^2}}\right]$ , then $\dfrac{dy}{dx} =$A$\dfrac{-1}{\sqrt{1 - x^2}}$B$\dfrac{1}{\sqrt{1 - x^2}}$C$1$D$-1$Check AnswerClear SelectionReveal AnswerShow Explanation
