The derivative of f(x) = x|x| is | vedclass

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(D) Given the function $f(x) = x|x|$.We can define the absolute value function as:$f(x) = \begin{cases} x(-x) = -x^2, & x < 0 \ x(x) = x^2, & x \ge 0

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$2|x|$

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(D) Given the function $f(x) = x|x|$.We can define the absolute value function as:$f(x) = \begin{cases} x(-x) = -x^2, & x Now,we find the derivative $f'(x)$ for both cases:For $x For $x > 0$,$f'(x) = \frac{d}{dx}(x^2) = 2x$.At $x = 0$,we check the left-hand derivative and right-hand derivative:$LHD = \lim_{h 	o 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^-} \frac{-h^2 - 0}{h} = \lim_{h 	o 0^-} (-h) = 0$.$RHD = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{h^2 - 0}{h} = \lim_{h 	o 0^+} (h) = 0$.Since $LHD = RHD = 0$,the derivative at $x = 0$ is $0$.Combining these results,$f'(x) = \begin{cases} -2x, & x < 0 \ 2x, & x \ge 0 \end{cases} = 2|x|$.

The derivative of $f(x) = x|x|$ is

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