Let f(x) = frac{x}{1 + |x|} be differentiable | vedclass

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(A) The function is defined as $f(x) = \frac{x}{1 + |x|}$.We can write this as a piecewise function:$f(x) = \begin{cases} \frac{x}{1 - x}, & x < 0 \

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$( - \infty, \infty )$

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(A) The function is defined as $f(x) = \frac{x}{1 + |x|}$.We can write this as a piecewise function:$f(x) = \begin{cases} \frac{x}{1 - x}, & x Now,we find the derivative $f'(x)$ for both cases:For $x For $x > 0$,$f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} = \frac{1+x-x}{(1+x)^2} = \frac{1}{(1+x)^2}$.At $x = 0$,we check the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$:$LHD = \lim_{h 	o 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^-} \frac{\frac{h}{1-h} - 0}{h} = \lim_{h 	o 0^-} \frac{1}{1-h} = 1$.$RHD = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{\frac{h}{1+h} - 0}{h} = \lim_{h 	o 0^+} \frac{1}{1+h} = 1$.Since $LHD = RHD = 1$,the function is differentiable at $x = 0$ as well.Therefore,the function is differentiable for all $x \in ( - \infty, \infty )$.

Let $f(x) = \frac{x}{1 + |x|}$ be differentiable at . . . .

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