Question

For the curve \[\sqrt{x} + \sqrt{y} = 1, \frac{dy}{dx}\text {  at } \left( 1/4, 1/4 \right)\text {  is }\] _____________ .

विकल्प

• 1/2

• 1

• -1

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Answer 1

−1 

\[\text { We have,} \sqrt{x} + \sqrt{y} = 1\]
\[\text { Differentiating with respect to x, we get }, \]
\[\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{1}{2\sqrt{y}}\frac{dy}{dx} = - \frac{1}{2\sqrt{x}}\]
\[ \Rightarrow \frac{dy}{dx} = - \frac{1}{2\sqrt{x}} \times \frac{2\sqrt{y}}{1}\]
\[ \Rightarrow \frac{dy}{dx} = - \frac{\sqrt{y}}{\sqrt{x}}\]
\[\text { Now,} \left[ \frac{dy}{dx} \right]_\left( \frac{1}{4}, \frac{1}{4} \right) = - \frac{\sqrt{\frac{1}{4}}}{\sqrt{\frac{1}{4}}} = - 1\]