MHTCET 11 Apr 2026 Shift 1 Paper

Concept: Differentiate a composite square root function using the chain rule.Explanation:Write the function as $y = \left(x + \sqrt{x^2+1}\right)^{1/2}$.Differentiate both sides with respect to $x$ using the chain rule:$$\frac{dy}{dx} = \frac{1}{2}\left(x + \sqrt{x^2+1}\right)^{-1/2} \left(1 + \frac{d}{dx}\sqrt{x^2+1}\right).$$Now find the derivative of the inner square root term:$$\frac{d}{dx}\sqrt{x^2+1} = \frac{1}{2\sqrt{x^2+1}} \cdot 2x = \frac{x}{\sqrt{x^2+1}}.$$Substitute this result back into the expression for $\frac{dy}{dx}$:$$\frac{dy}{dx} = \frac{1}{2\sqrt{x+\sqrt{x^2+1}}} \left(1 + \frac{x}{\sqrt{x^2+1}}\right).$$Since $y = \sqrt{x+\sqrt{x^2+1}}$, replace the denominator by $2y$:$$\frac{dy}{dx} = \frac{1}{2y} \left(\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}\right).$$Combine the fractions to get the final derivative:$$\frac{dy}{dx} = \frac{\sqrt{x^2+1}+x}{2y\sqrt{x^2+1}}.$$Answer:The correct option is A: $\frac{\sqrt{x^2+1}+x}{2y\sqrt{x^2+1}}$.