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

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(B) Given the function $f(x) = \frac{x}{1+|x|}$.To find $f^{\prime}(0)$,we check the left-hand derivative and right-hand derivative at $x = 0$.For $x

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(B) Given the function $f(x) = \frac{x}{1+|x|}$.To find $f^{\prime}(0)$,we check the left-hand derivative and right-hand derivative at $x = 0$.For $x > 0$,$|x| = x$,so $f(x) = \frac{x}{1+x}$. Then $f^{\prime}(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} = \frac{1}{(1+x)^2}$. Thus,$f^{\prime}(0^+) = \lim_{x 	o 0^+} \frac{1}{(1+x)^2} = 1$.For $x Since $f^{\prime}(0^+) = f^{\prime}(0^-) = 1$,the derivative $f^{\prime}(0)$ exists and is equal to $1$.

If $f(x) = \frac{x}{1+|x|}$,for $x \in \mathbb{R}$,then $f^{\prime}(0)$ is equal to :

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