If $\sqrt{x + y} + \sqrt{y - x} = \sqrt{2}a$, then $\dfrac{d^2 y}{d x^2}$ is equal to …

Differentiate both sides: $\dfrac{1}{2\sqrt{x + y}}(1 + \dfrac{dy}{dx}) + \dfrac{1}{2\sqrt{y - x}}(\dfrac{dy}{dx} - 1) = 0$. Simplify to get $\dfrac{dy}{dx} = \dfrac{\sqrt{y - x} - \sqrt{x + y}}{\sqrt{y - x} + \sqrt{x + y}}$. Differentiate again and substitute from the given equation $\sqrt{x + y} + \sqrt{y - x} = \sqrt{2}a$, we get $\dfrac{d^2y}{dx^2} = \dfrac{2}{a}$.