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

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(B) Given $f(x) = x 	an^{-1} x$. Applying the product rule for differentiation,$(uv)' = u'v + uv'$,where $u = x$ and $v = 	an^{-1} x$. $f'(x) = \fra

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$\frac{1}{2} + \frac{\pi}{4}$

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(B) Given $f(x) = x 	an^{-1} x$. Applying the product rule for differentiation,$(uv)' = u'v + uv'$,where $u = x$ and $v = 	an^{-1} x$. $f'(x) = \frac{d}{dx}(x) \cdot 	an^{-1} x + x \cdot \frac{d}{dx}(	an^{-1} x)$. $f'(x) = 1 \cdot 	an^{-1} x + x \cdot \frac{1}{1 + x^2}$. $f'(x) = 	an^{-1} x + \frac{x}{1 + x^2}$. Now,substitute $x = 1$ into the derivative: $f'(1) = 	an^{-1}(1) + \frac{1}{1 + 1^2}$. Since $	an^{-1}(1) = \frac{\pi}{4}$,we have: $f'(1) = \frac{\pi}{4} + \frac{1}{2}$.

If $f(x) = x \tan^{-1} x$,then $f'(1) =$

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