If y = operatorname{Tan}^{-1}left(frac{2x}{1- | vedclass

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(D) Given $y = \operatorname{Tan}^{-1}\left(\frac{2x}{1-x^2}\right)$.We know the trigonometric identity $2\operatorname{Tan}^{-1}(x) = \operatorname{T

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$\frac{8}{5}$

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(D) Given $y = \operatorname{Tan}^{-1}\left(\frac{2x}{1-x^2}ight)$.We know the trigonometric identity $2\operatorname{Tan}^{-1}(x) = \operatorname{Tan}^{-1}\left(\frac{2x}{1-x^2}ight)$ for $|x| Substituting this into the equation,we get $y = 2\operatorname{Tan}^{-1}(x)$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = 2 	imes \frac{1}{1+x^2} = \frac{2}{1+x^2}$.Substitute $x = \frac{1}{2}$ into the derivative:$\left(\frac{dy}{dx}ight)_{x=\frac{1}{2}} = \frac{2}{1 + (\frac{1}{2})^2} = \frac{2}{1 + \frac{1}{4}} = \frac{2}{\frac{5}{4}} = \frac{8}{5}$.Thus,the correct option is $D$.

If $y = \operatorname{Tan}^{-1}\left(\frac{2x}{1-x^2}\right)$ where $|x| < 1$,then find the value of $\left(\frac{dy}{dx}\right)$ at $x = \frac{1}{2}$.

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