Let f be defined on D = mathbb{R} - {-1, 1} b | vedclass

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(B) We have $f(x) = \frac{|x|}{1-|x|}$. For $x \in D$,we can write this as:$f(x) = \begin{cases} \frac{-x}{1+x}, & x < 0 \ \frac{x}{1-x}, & x \geq 0

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$f$ is differentiable on $D$ except at $x = 0$

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(B) We have $f(x) = \frac{|x|}{1-|x|}$. For $x \in D$,we can write this as:$f(x) = \begin{cases} \frac{-x}{1+x}, & x To check differentiability at $x = 0$:$	ext{LHD (at } x=0) = \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{h}{1-h} \cdot \frac{1}{-h} = \lim_{h 	o 0^+} \frac{-1}{1-h} = -1$$	ext{RHD (at } x=0) = \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 $	ext{LHD} 
eq 	ext{RHD}$,$f(x)$ is not differentiable at $x = 0$.For all other points in $D$,$f(x)$ is a rational function with non-zero denominators,hence it is differentiable.Thus,$f$ is differentiable on $D$ except at $x = 0$.

Let $f$ be defined on $D = \mathbb{R} - \{-1, 1\}$ by $f(x) = \frac{|x|}{1-|x|}$,then

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