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UPSC NDA Math Model Paper 9

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Question : 51 of 120

Marks: +1, -0

If

\sqrt{x}+\sqrt{y}=2

\frac{dy}{dx}

y=1

equal to?

5
2
4
-1

Solution:

Given,

\sqrt{x}+\sqrt{y}=2

\sqrt{x}=2-\sqrt{y}

.... (1)

or

\sqrt{y}=2-\sqrt{x}

Taking derivative on both sides, we get

\Rightarrow \frac{1}{2\sqrt{y}} \frac{dy}{dx}=0-\frac{1}{2\sqrt{x}}

\Rightarrow \frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}

From equation (1), we have

\Rightarrow \frac{dy}{dx}=-\frac{\sqrt{y}}{2-\sqrt{y}}

To find

\frac{dy}{dx}

at

y=1

, put

y=1

in above equation.

\Rightarrow \frac{dy}{dx}=-\frac{\sqrt{1}}{2-\sqrt{1}}

\Rightarrow \frac{dy}{dx}=-1

Hence, If

\sqrt{x}+\sqrt{y}=2

, then

\frac{dy}{dx}

at

y=1

equal to

-1

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