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प्रश्न
If \[y = \sqrt{x + 1} + \sqrt{x - 1}\] , prove that \[\sqrt{x^2 - 1}\frac{dy}{dx} = \frac{1}{2}y\] ?
उत्तर
\[\text{We have }, y = \sqrt{x + 1} + \sqrt{x - 1}\]
Differentiating with respect to x,
\[\frac{d y}{d x} = \frac{d}{dx}\left( \sqrt{x + 1} \right) + \frac{d}{dx}\left( \sqrt{x - 1} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{2} \left( x + 1 \right)^\frac{- 1}{2} + \frac{1}{2} \left( x - 1 \right)^\frac{- 1}{2} \]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{2}\left( \frac{1}{\sqrt{x + 1}} + \frac{1}{\sqrt{x - 1}} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{2}\left( \frac{\sqrt{x - 1} + \sqrt{x + 1}}{\left( \sqrt{x + 1} \right)\left( \sqrt{x - 1} \right)} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{2}\left( \frac{y}{\left( \sqrt{x^2 - 1} \right)} \right)\]
\[ \Rightarrow \left( \sqrt{x^2 - 1} \right)\frac{d y}{d x} = \frac{1}{2}y\]
\[\text{Hence proved }\]
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