Q226 / 270 · NDA 2026 (Phase 1) · Mathematics

If $\sqrt[4]{y}=x+\sqrt{x^2+4}$, then what is $\sqrt{x^2+4}\,\dfrac{dy}{dx}$ equal to ?

  • A
    $\dfrac{y}{4}$
  • B
    $y$
  • C
    $2y$
  • Correct answer
    D
    $4y$

Solution

Given $y^{1/4} = x + \sqrt{x^2+4}$

Differentiating both sides with respect to $x$:

$\dfrac{1}{4} y^{-3/4} \dfrac{dy}{dx} = 1 + \dfrac{1}{2\sqrt{x^2+4}} (2x)$

$\dfrac{1}{4} y^{-3/4} \dfrac{dy}{dx} = 1 + \dfrac{x}{\sqrt{x^2+4}}$

$\dfrac{1}{4} y^{-3/4} \dfrac{dy}{dx} = \dfrac{\sqrt{x^2+4} + x}{\sqrt{x^2+4}}$

Substituting $x + \sqrt{x^2+4} = y^{1/4}$:

$\dfrac{1}{4} y^{-3/4} \dfrac{dy}{dx} = \dfrac{y^{1/4}}{\sqrt{x^2+4}}$

$\sqrt{x^2+4} \dfrac{dy}{dx} = 4 y^{1/4} y^{3/4}$

$\sqrt{x^2+4} \dfrac{dy}{dx} = 4y$

Answer: $4y$

Hence, the correct answer is option: 'D'.

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