Q79 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
If $y = \sqrt{x + \sqrt{x^2 + 1}}$, then the value of $\dfrac{dy}{dx}$ is
- Correct answerA$\dfrac{\sqrt{x^2 + 1} + x}{2y\sqrt{x^2 + 1}}$
- 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$\dfrac{\sqrt{x^2 + 1} + x}{2y\sqrt{x^2 - 1}}$
Solution
Given $y = \sqrt{x + \sqrt{x^2 + 1}}$
Squaring both sides:
$y^2 = x + \sqrt{x^2 + 1}$
Differentiating both sides with respect to $x$:
$2y \dfrac{dy}{dx} = 1 + \dfrac{1}{2\sqrt{x^2 + 1}} \cdot 2x$
$2y \dfrac{dy}{dx} = 1 + \dfrac{x}{\sqrt{x^2 + 1}}$
$2y \dfrac{dy}{dx} = \dfrac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}}$
$\dfrac{dy}{dx} = \dfrac{\sqrt{x^2 + 1} + x}{2y\sqrt{x^2 + 1}}$
Answer: $\dfrac{\sqrt{x^2 + 1} + x}{2y\sqrt{x^2 + 1}}$
Squaring both sides:
$y^2 = x + \sqrt{x^2 + 1}$
Differentiating both sides with respect to $x$:
$2y \dfrac{dy}{dx} = 1 + \dfrac{1}{2\sqrt{x^2 + 1}} \cdot 2x$
$2y \dfrac{dy}{dx} = 1 + \dfrac{x}{\sqrt{x^2 + 1}}$
$2y \dfrac{dy}{dx} = \dfrac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}}$
$\dfrac{dy}{dx} = \dfrac{\sqrt{x^2 + 1} + x}{2y\sqrt{x^2 + 1}}$
Answer: $\dfrac{\sqrt{x^2 + 1} + x}{2y\sqrt{x^2 + 1}}$
Hence, the correct answer is option: 'A'.