frac{d}{dx}[(1 + x^2)tan^{-1}x] = | vedclass

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Question

(C) To find the derivative of the product $(1 + x^2)	an^{-1}x$,we use the product rule: $\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}$

Vedclass · Accepted Answer

$2x 	an^{-1}x + 1$

Vedclass · Answer

(C) To find the derivative of the product $(1 + x^2)	an^{-1}x$,we use the product rule: $\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}$.Let $u = (1 + x^2)$ and $v = 	an^{-1}x$.Then $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = \frac{1}{1 + x^2}$.Applying the product rule:$\frac{d}{dx}[(1 + x^2)	an^{-1}x] = (1 + x^2) \cdot \frac{d}{dx}(	an^{-1}x) + 	an^{-1}x \cdot \frac{d}{dx}(1 + x^2)$$= (1 + x^2) \cdot \frac{1}{1 + x^2} + 	an^{-1}x \cdot (2x)$$= 1 + 2x 	an^{-1}x$.Thus,the correct option is $C$.

$\frac{d}{dx}[(1 + x^2)\tan^{-1}x] = $

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