If f(x) = x tan^{-1} x,then lim_{x rightarrow | vedclass

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(D) The given expression is the definition of the derivative of $f(x)$ at $x = 1$,i.e.,$f'(1)$.Given $f(x) = x 	an^{-1} x$.Using the product rule for

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$\frac{\pi + 2}{4}$

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(D) The given expression is the definition of the derivative of $f(x)$ at $x = 1$,i.e.,$f'(1)$.Given $f(x) = x 	an^{-1} x$.Using the product rule for differentiation,$f'(x) = \frac{d}{dx}(x) \cdot 	an^{-1} x + x \cdot \frac{d}{dx}(	an^{-1} x)$.$f'(x) = 1 \cdot 	an^{-1} x + x \cdot \frac{1}{1 + x^2} = 	an^{-1} x + \frac{x}{1 + x^2}$.Now,evaluate at $x = 1$:$f'(1) = 	an^{-1}(1) + \frac{1}{1 + 1^2} = \frac{\pi}{4} + \frac{1}{2}$.$f'(1) = \frac{\pi}{4} + \frac{2}{4} = \frac{\pi + 2}{4}$.

If $f(x) = x \tan^{-1} x$,then $\lim_{x \rightarrow 1} \frac{f(x) - f(1)}{x - 1}$ is equal to

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