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NDA 2026
Exam
NDA
Year
2026
Subject
Mathematics
Chapter
Differentiation

NDA 2026 · Mathematics · Differentiation

If [4] y =x+ x^2+4 , then what is x^2+4 \,(dy)/(dx) equal to ?

If y4=x+x2+4\sqrt[4]{y}=x+\sqrt{x^2+4}, then what is x2+4 dydx\sqrt{x^2+4}\,\dfrac{dy}{dx} equal to ?

Options

  1. Ay4\dfrac{y}{4}
  2. Byy
  3. C2y2y
  4. D4y4y

Correct answer

D. 4y4y

Step-by-step solution

Given y1/4=x+x2+4y^{1/4} = x + \sqrt{x^2+4}

Differentiating both sides with respect to xx:

14y−3/4dydx=1+12x2+4(2x)\dfrac{1}{4} y^{-3/4} \dfrac{dy}{dx} = 1 + \dfrac{1}{2\sqrt{x^2+4}} (2x)

14y−3/4dydx=1+xx2+4\dfrac{1}{4} y^{-3/4} \dfrac{dy}{dx} = 1 + \dfrac{x}{\sqrt{x^2+4}}

14y−3/4dydx=x2+4+xx2+4\dfrac{1}{4} y^{-3/4} \dfrac{dy}{dx} = \dfrac{\sqrt{x^2+4} + x}{\sqrt{x^2+4}}

Substituting x+x2+4=y1/4x + \sqrt{x^2+4} = y^{1/4}:

14y−3/4dydx=y1/4x2+4\dfrac{1}{4} y^{-3/4} \dfrac{dy}{dx} = \dfrac{y^{1/4}}{\sqrt{x^2+4}}

x2+4dydx=4y1/4y3/4\sqrt{x^2+4} \dfrac{dy}{dx} = 4 y^{1/4} y^{3/4}

x2+4dydx=4y\sqrt{x^2+4} \dfrac{dy}{dx} = 4y

Answer: 4y4y
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