The function $f(x) = x|x|, x \in \mathbb{R}$ is differentiable:

in $\mathbb{R}$

The correct answer is Option (3) → in $\mathbb{R}$

For $x \ge 0: f(x) = x^{2}$

For $x < 0: f(x) = -x^{2}$

Now,

For $x > 0: \text{derivative} = 2x$

For $x < 0: \text{derivative} = -2x$

At $x = 0$:

• $\text{Left-hand derivative} = \lim\limits_{h \to 0^{-}} \frac{f(h) - f(0)}{h} = \lim\limits_{h \to 0^{-}} \frac{-h^{2}}{h} = -h \to 0$
• $\text{Right-hand derivative} = \lim\limits_{h \to 0^{+}} \frac{h^{2}}{h} = h \to 0$

Since both limits are equal, the derivative exists at $x = 0$.

Hence, the function is differentiable everywhere in $\mathbb{R}$.