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COMEDK · Maths · 26. Differentiation

If \(y=\sqrt{\dfrac{x}{a}}+\sqrt{\dfrac{a}{x}}, \quad\) then \(2 x y \dfrac{d y}{d x}\) is equal to

  1. A \(x+\dfrac{a}{x}\)
  2. B \(\dfrac{a}{x}-\dfrac{x}{a}\)
  3. C \(\dfrac{x^2+a^2}{a x}\)
  4. D \(\dfrac{x}{a}-\dfrac{a}{x}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\dfrac{x}{a}-\dfrac{a}{x}\)

Step-by-step Solution

Detailed explanation

Given \(y = \sqrt{\dfrac{x}{a}} + \sqrt{\dfrac{a}{x}} = \dfrac{\sqrt{x}}{\sqrt{a}} + \dfrac{\sqrt{a}}{\sqrt{x}}\). Differentiating with respect to \(x\): \(\dfrac{dy}{dx} = \dfrac{1}{\sqrt{a}} \cdot \dfrac{1}{2\sqrt{x}} + \sqrt{a} \cdot \left( -\dfrac{1}{2} x^{-3/2} \right)\)…