text{The domain of the derivative of the func | vedclass

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(C) $	ext{Given } f(x) = \begin{cases} \frac{1}{2}(-x-1), & x  1 \end{cases}$$	ext{T

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$R-\{-1, 1\}$

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(C) $	ext{Given } f(x) = \begin{cases} \frac{1}{2}(-x-1), & x  1 \end{cases}$$	ext{The derivative } f'(x) 	ext{ is defined as:}$$f'(x) = \begin{cases} -\frac{1}{2}, & x  1 \end{cases}$$	ext{At } x = -1: 	ext{ Left-hand derivative } = -\frac{1}{2}, 	ext{ Right-hand derivative } = \frac{1}{1+(-1)^2} = \frac{1}{2}. 	ext{ Since } -\frac{1}{2} 
eq \frac{1}{2}, f(x) 	ext{ is not differentiable at } x = -1.$$	ext{At } x = 1: 	ext{ Left-hand derivative } = \frac{1}{1+1^2} = \frac{1}{2}, 	ext{ Right-hand derivative } = \frac{1}{2}. 	ext{ However, check continuity: } f(1) = 	an^{-1}(1) = \frac{\pi}{4}, 	ext{ but } \lim_{x 	o 1^+} f(x) = \frac{1}{2}(1-1) = 0. 	ext{ Since } f(x) 	ext{ is discontinuous at } x = 1, 	ext{ it is not differentiable at } x = 1.$$	ext{Thus, the domain of } f'(x) 	ext{ is } R - \{-1, 1\}.$

$\text{The domain of the derivative of the function } f(x) = \begin{cases} \tan^{-1} x, & \text{if } |x| \le 1 \\ \frac{1}{2}(|x|-1), & \text{if } |x| > 1 \end{cases} \text{ is given by:}$

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