If y = (1 + x^2) tan^{-1} x - x,then frac{dy} | vedclass

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(A) Given,$y = (1 + x^2) 	an^{-1} x - x$. Applying the product rule for the first term and the power rule for the remaining terms: $\frac{dy}{dx} = \

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$2x 	an^{-1} x$

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(A) Given,$y = (1 + x^2) 	an^{-1} x - x$. Applying the product rule for the first term and the power rule for the remaining terms: $\frac{dy}{dx} = \frac{d}{dx} [(1 + x^2) 	an^{-1} x] - \frac{d}{dx} (x)$ $\frac{dy}{dx} = [(1 + x^2) \cdot \frac{d}{dx}(	an^{-1} x) + 	an^{-1} x \cdot \frac{d}{dx}(1 + x^2)] - 1$ $\frac{dy}{dx} = [(1 + x^2) \cdot \frac{1}{1 + x^2} + 	an^{-1} x \cdot (2x)] - 1$ $\frac{dy}{dx} = [1 + 2x 	an^{-1} x] - 1$ $\frac{dy}{dx} = 2x 	an^{-1} x$.

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

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