If y={x/a}+{a/x}, then 2 x y d y/d x is equal to

Correct Answer is : xa−axx/a-a/xax−xa

Correct Answer is : xa−axx/a-a/xax−xa

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COMEDK 2025 Morning Solved Paper

Question : 145 of 180

Marks: +1, -0

y=\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}, \quad

2 x y \frac{d y}{d x}

is equal to

Solution:

Concept:

Rewrite

y

as sum of two terms and differentiate using standard power rule.

Explanation:

Given

y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}} = \frac{\sqrt{x}}{\sqrt{a}} + \frac{\sqrt{a}}{\sqrt{x}}

Differentiate term by term:

\frac{d}{dx}\left(\frac{\sqrt{x}}{\sqrt{a}}\right) = \frac{1}{2\sqrt{a}}x^{-1/2} = \frac{1}{2\sqrt{ax}}

\frac{d}{dx}\left(\frac{\sqrt{a}}{\sqrt{x}}\right) = \sqrt{a}\cdot \left(-\frac12 x^{-3/2}\right) = -\frac{\sqrt{a}}{2x^{3/2}}

Thus

\frac{dy}{dx} = \frac{1}{2\sqrt{ax}} - \frac{\sqrt{a}}{2x^{3/2}}

Now compute

2xy\frac{dy}{dx}

2xy\frac{dy}{dx} = 2x\left(\frac{\sqrt{x}}{\sqrt{a}}+\frac{\sqrt{a}}{\sqrt{x}}\right)\left(\frac{1}{2\sqrt{ax}}-\frac{\sqrt{a}}{2x^{3/2}}\right)

The factor

2

cancels with

\frac12

, and distributing

x

gives:

= \left(\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}\right)\left(\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}}\right)

This is of the form

(u+v)(u-v)=u^2-v^2

. Hence

= \frac{x}{a} - \frac{a}{x}

Answer:

2xy\frac{dy}{dx} = \frac{x}{a} - \frac{a}{x}

, which is option C.

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