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Question
Differentiate \[\sin \left( \frac{1 + x^2}{1 - x^2} \right)\] ?
Solution
\[\text{Let } y = \sin\left( \frac{1 + x^2}{1 - x^2} \right)\]
\[\text{Differentiate it with respect to x we get }, \]
\[\frac{d y}{d x} = \frac{d}{dx}\left[ \sin\left( \frac{1 + x^2}{1 - x^2} \right) \right]\]
\[ = \cos x\left( \frac{1 + x^2}{1 - x^2} \right)\frac{d}{dx}\left( \frac{1 + x^2}{1 - x^2} \right) \left[ \text{Using chain rule } \right]\]
\[ = \cos x\left( \frac{1 + x^2}{1 - x^2} \right)\left[ \frac{\left( 1 - x^2 \right)\frac{d}{dx}\left( 1 + x^2 \right) - \left( 1 + x^2 \right)\frac{d}{dx}\left( 1 - x^2 \right)}{\left( 1 - x^2 \right)^2} \right] \left[ \text{ Using quotient rule } \right]\]
\[ = \cos x\left( \frac{1 + x^2}{1 - x^2} \right)\left[ \frac{\left( 1 - x^2 \right)\left( 2x \right) - \left( 1 + x^2 \right)\left( - 2x \right)}{\left( 1 - x^2 \right)^2} \right]\]
\[ = \cos x\left( \frac{1 + x^2}{1 - x^2} \right)\left[ \frac{2x - 2 x^3 + 2x + 2 x^3}{\left( 1 - x^2 \right)^2} \right]\]
\[ = \frac{4x}{\left( 1 - x^2 \right)^2}\cos x\left( \frac{1 + x^2}{1 - x^2} \right)\]
\[So, \frac{d}{dx}\left\{ \sin\left( \frac{1 + x^2}{1 - x^2} \right) \right\} = \frac{4x}{\left( 1 - x^2 \right)^2}\cos x\left( \frac{1 + x^2}{1 - x^2} \right)\]
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