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प्रश्न
Differentiate \[\sqrt{\frac{1 - x^2}{1 + x^2}}\] ?
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उत्तर
\[\text{ Let } y = \sqrt{\frac{1 - x^2}{1 + x^2}}\]
\[ \Rightarrow y = \left( \frac{1 - x^2}{1 + x^2} \right)^\frac{1}{2} \]
\[\text{ Differentiate it with respect to x we get }, \]
\[\frac{d y}{d x} = \frac{d}{dx} \left( \frac{1 - x^2}{1 + x^2} \right)^\frac{1}{2} \]
\[ = \frac{1}{2} \left( \frac{1 - x^2}{1 + x^2} \right)^{\frac{1}{2} - 1} \times \frac{d}{dx}\left( \frac{1 - x^2}{1 + x^2} \right) \left[ \text{ Using chain rule } \right]\]
\[ = \frac{1}{2} \left( \frac{1 - x^2}{1 + x^2} \right)^\frac{- 1}{2} \times \left\{ \frac{\left( 1 + x^2 \right)\frac{d}{dx}\left( 1 - x^2 \right) - \left( 1 - x^2 \right)\frac{d}{dx}\left( 1 + x^2 \right)}{\left( 1 + x^2 \right)^2} \right\} \]
\[ = \frac{1}{2} \left( \frac{1 + x^2}{1 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 2x\left( 1 + x^2 \right) - 2x\left( 1 - x^2 \right)}{\left( 1 + x^2 \right)^2} \right\}\]
\[ = \frac{1}{2} \left( \frac{1 + x^2}{1 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 2x - 2 x^3 - 2x + 2 x^3}{\left( 1 + x^2 \right)^2} \right\}\]
\[ = \frac{1}{2} \left( \frac{1 + x^2}{1 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 4x}{\left( 1 + x^2 \right)^2} \right\}\]
\[ = \frac{- 2x}{\sqrt{1 - x^2} \left( 1 + x^2 \right)^\frac{3}{2}}\]
