The domain of the derivative of the function | vedclass

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(C) The function $f(x)$ can be rewritten by expanding the absolute value as: $f(x) = \begin{cases} \frac{1}{2}(-x - 1), & x < -1 \ 	an^{-1}x, & -1 \

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$R - \{-1\}$

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(C) The function $f(x)$ can be rewritten by expanding the absolute value as: $f(x) = \begin{cases} \frac{1}{2}(-x - 1), & x  1 \end{cases}$ To find the domain of the derivative $f'(x)$,we check the differentiability at the transition points $x = -1$ and $x = 1$. For $x For $-1 For $x > 1$,$f'(x) = \frac{1}{2}$. At $x = -1$: Left-hand derivative $f'(-1^-) = -\frac{1}{2}$. Right-hand derivative $f'(-1^+) = \frac{1}{1+(-1)^2} = \frac{1}{2}$. Since $f'(-1^-) 
eq f'(-1^+)$,the function is not differentiable at $x = -1$. At $x = 1$: Left-hand derivative $f'(1^-) = \frac{1}{1+(1)^2} = \frac{1}{2}$. Right-hand derivative $f'(1^+) = \frac{1}{2}$. Since $f'(1^-) = f'(1^+)$,the function is differentiable at $x = 1$. Thus,the derivative $f'(x)$ exists for all $x \in R$ except $x = -1$. Therefore,the domain of $f'(x)$ is $R - \{-1\}$.

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

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