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

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(B) Given the function $y = (1 + x^2)	an^{-1}x - x.$ To find $\frac{dy}{dx},$ we apply the product rule to the first term $(1 + x^2)	an^{-1}x$ and t

Vedclass · Accepted Answer

$2x	an^{-1}x$

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(B) Given the function $y = (1 + x^2)	an^{-1}x - x.$ To find $\frac{dy}{dx},$ we apply the product rule to the first term $(1 + x^2)	an^{-1}x$ and the power rule to the second term $-x.$ Using the product rule $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx},$ where $u = (1 + x^2)$ and $v = 	an^{-1}x$: $\frac{dy}{dx} = \frac{d}{dx}[(1 + x^2)	an^{-1}x] - \frac{d}{dx}(x)$ $\frac{dy}{dx} = (1 + x^2) \cdot \frac{d}{dx}(	an^{-1}x) + 	an^{-1}x \cdot \frac{d}{dx}(1 + x^2) - 1$ Since $\frac{d}{dx}(	an^{-1}x) = \frac{1}{1 + x^2}$ and $\frac{d}{dx}(1 + x^2) = 2x$: $\frac{dy}{dx} = (1 + x^2) \cdot \frac{1}{1 + x^2} + 	an^{-1}x \cdot (2x) - 1$ $\frac{dy}{dx} = 1 + 2x	an^{-1}x - 1$ $\frac{dy}{dx} = 2x	an^{-1}x.$ Thus,the correct option is $B$.

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

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