frac{d}{d x} left{ (1+x^2) tan^{-1}(x) right} | 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 \cdot \frac{dv}{dx} + v \cdot \

Vedclass · Accepted Answer

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

Vedclass · Answer

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

$\frac{d}{d x} \left\{ (1+x^2) \tan^{-1}(x) \right\} =$

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