The derivative of y = 1 - |x| at x = 0 is | vedclass

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

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Does not exist

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(D) The function is defined as $y = f(x) = 1 - |x|$.We can express this as a piecewise function:$f(x) = \begin{cases} 1 - x, & x \ge 0 \ 1 + x, & x To check for differentiability at $x = 0$,we find 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{(1 + h) - 1}{h} = \lim_{h 	o 0^-} \frac{h}{h} = 1$$RHD = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{(1 - h) - 1}{h} = \lim_{h 	o 0^+} \frac{-h}{h} = -1$Since $LHD 
eq RHD$,the derivative of the function at $x = 0$ does not exist.

The derivative of $y = 1 - |x|$ at $x = 0$ is

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