• If the real function $f$ is defined in the point $x_0$ and on some interval left from this and if the left-hand one-sided limit $\lim_{h\to 0-}\frac{f(x_0+h)-f(x_0)}{h}$ exists, then this limit is defined to be the left-sided derivative of $f$ in $x_0$ .
• If the real function $f$ is defined in the point $x_0$ and on some interval right from this and if the right-hand one-sided limit $\lim_{h\to 0+} \frac{f(x_0+h)-f(x_0)}{h}$ exists, then this limit is defined to be the right-sided derivative of $f$ in $x_0$ .

It's apparent that if $f$ has both the left-sided and the right-sided derivative in the point $x_0$ and these are equal, then $f$ is differentiable in $x_0$ and $f'(x_0)$ equals to these one-sided derivatives. Also inversely.

Example. The real function $x \mapsto x\sqrt{x}$ is defined for $x \geqq 0$ and differentiable for $x > 0$ with $f'(x) \equiv \frac{3}{2}\sqrt{x}$ . The function also has the right derivative in $0$ : $$\lim_{h \to 0+}\frac{h\sqrt{h}- 0\sqrt{0}}{h} = \lim_{h \to 0+}\sqrt{h} = 0$$

Remark. For a function $f\!: [a,\,b] \to \mathbb{R}$ , to have a right-sided derivative at $x = a$ with value $d$ , is equivalent to saying that there is an extension $g$ of $f$ to some open interval containing $[a,\,b]$ and satisfying $g'(a) = d$ . Similarly for left-sided derivatives.