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Question
Differentiate \[\sqrt{\frac{1 + x}{1 - x}}\] ?
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Solution
\[\text{ Let } y = \sqrt{\frac{1 + x}{1 - x}}\]
\[ \Rightarrow y = \left( \frac{1 + x}{1 - x} \right)^\frac{1}{2} \]
\[\text{Differentiate it with respect to x we get }, \]
\[\frac{d y}{d x} = \frac{d}{dx} \left( \frac{1 + x}{1 - x} \right)^\frac{1}{2} \]
\[ = \frac{1}{2} \left( \frac{1 + x}{1 - x} \right)^{\frac{1}{2} - 1} \times \frac{d}{dx}\left( \frac{1 + x}{1 - x} \right) \left[ \text{ Using chain rule } \right]\]
\[ = \frac{1}{2} \left( \frac{1 + x}{1 - x} \right)^\frac{- 1}{2} \times \left\{ \frac{\left( 1 - x \right)\frac{d}{dx}\left( 1 + x \right) - \left( 1 + x \right)\frac{d}{dx}\left( 1 - x \right)}{\left( 1 - x \right)^2} \right\} \left[ \text{Using quotient rule} \right]\]
\[ = \frac{1}{2} \left( \frac{1 - x}{1 + x} \right)^\frac{1}{2} \left\{ \frac{\left( 1 - x \right)\left( 1 \right) - \left( 1 + x \right)\left( - 1 \right)}{\left( 1 - x \right)^2} \right\}\]
\[ = \frac{1}{2} \left( \frac{1 - x}{1 + x} \right)^\frac{1}{2} \left\{ \frac{1 - x + 1 + x}{\left( 1 - x \right)^2} \right\}\]
\[ = \frac{1}{2}\frac{\left( 1 - x \right)^\frac{1}{2}}{\left( 1 + x \right)^\frac{1}{2}} \times \frac{2}{\left( 1 - x \right)^2}\]
\[ = \frac{1}{\sqrt{1 + x} \left( 1 - x \right)^\frac{3}{2}}\]
\[So, \frac{d}{dx}\left( \sqrt{\frac{1 + x}{1 - x}} \right) = \frac{1}{\sqrt{1 + x} \left( 1 - x \right)^\frac{3}{2}}\]
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