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UPSC NDA I 2026 Mathematics Paper

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Question : 76 of 120

Marks: +1, -0

If

\sqrt[4]{y}=x+\sqrt{x^{2}+4}

\sqrt{x^{2}+4} \frac{d y}{d x}

equal to ?

$\frac{y}{4}$
$y$
$2 y$
$4 y$

Solution:

Concept:

Let

u = x + \sqrt{x^2+4}

. Then

y^{1/4} = u

, so

y = u^4

. Differentiate using the chain rule.

Explanation:

Let

u = x+\sqrt{x^2+4}

, so

y = u^4

.

Differentiate

u

:

\frac{du}{dx}=1+\frac{x}{\sqrt{x^2+4}}=\frac{x+\sqrt{x^2+4}}{\sqrt{x^2+4}}=\frac{u}{\sqrt{x^2+4}}.

Now,

\frac{dy}{dx}=4u^3\frac{du}{dx}=4u^3\cdot\frac{u}{\sqrt{x^2+4}}=\frac{4u^4}{\sqrt{x^2+4}}=\frac{4y}{\sqrt{x^2+4}}.

Therefore,

\sqrt{x^2+4}\,\frac{dy}{dx}=4y.

Answer:

4y

, which is option D.

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