If f(x) = frac{x}{1 + |x|} for x in R,then f' | vedclass

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(B) The function is defined as $f(x) = \frac{x}{1 + |x|}$.We examine the left-hand derivative and right-hand derivative at $x = 0$.For $x < 0$,$|x| =

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$1$

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(B) The function is defined as $f(x) = \frac{x}{1 + |x|}$.We examine the left-hand derivative and right-hand derivative at $x = 0$.For $x The derivative is $f'(x) = \frac{(1-x)(1) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2}$.Thus,the left-hand derivative $f'(0^-) = \lim_{x 	o 0^-} \frac{1}{(1-x)^2} = 1$.For $x > 0$,$|x| = x$,so $f(x) = \frac{x}{1 + x}$.The derivative is $f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} = \frac{1}{(1+x)^2}$.Thus,the right-hand derivative $f'(0^+) = \lim_{x 	o 0^+} \frac{1}{(1+x)^2} = 1$.Since $f'(0^-) = f'(0^+) = 1$,the derivative $f'(0) = 1$.

If $f(x) = \frac{x}{1 + |x|}$ for $x \in R$,then $f'(0) = $

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