If y = sin left( frac{1 + x^2}{1 - x^2} right | vedclass

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(D) Given $y = \sin \left( \frac{1 + x^2}{1 - x^2} \right)$.Using the chain rule,$\frac{dy}{dx} = \cos \left( \frac{1 + x^2}{1 - x^2} \right) \cdot \f

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$\frac{4x}{(1 - x^2)^2} \cos \left( \frac{1 + x^2}{1 - x^2} \right)$

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(D) Given $y = \sin \left( \frac{1 + x^2}{1 - x^2} \right)$.Using the chain rule,$\frac{dy}{dx} = \cos \left( \frac{1 + x^2}{1 - x^2} \right) \cdot \frac{d}{dx} \left( \frac{1 + x^2}{1 - x^2} \right)$.Applying the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$:$\frac{d}{dx} \left( \frac{1 + x^2}{1 - x^2} \right) = \frac{(1 - x^2)(2x) - (1 + x^2)(-2x)}{(1 - x^2)^2}$$= \frac{2x - 2x^3 + 2x + 2x^3}{(1 - x^2)^2} = \frac{4x}{(1 - x^2)^2}$.Therefore,$\frac{dy}{dx} = \frac{4x}{(1 - x^2)^2} \cos \left( \frac{1 + x^2}{1 - x^2} \right)$.

If $y = \sin \left( \frac{1 + x^2}{1 - x^2} \right)$,then $\frac{dy}{dx} = $

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