Mht CetDifferentiationMHT CET 2026 11th April Morning ShiftMCQ+2 / -02026If $y = \sqrt{x + \sqrt{x^2 + 1}}$, then the value of $\dfrac{dy}{dx}$ isA$\dfrac{\sqrt{x^2 + 1} + x}{2y\sqrt{x^2 + 1}}$B$\dfrac{\sqrt{x^2 + 1} + x}{y\sqrt{x^2 + 1}}$C$\dfrac{\sqrt{x^2 + 1} - x}{2y\sqrt{x^2 + 1}}$D$\dfrac{\sqrt{x^2 + 1} + x}{2y\sqrt{x^2 - 1}}$Check AnswerClear SelectionReveal AnswerShow Explanation
