For a natural $n \ge 1$, the derivative of $x \mapsto x^{1/n}$ on $(0,\infty)$ is $\frac{1}{\iota(n)}x^{1/n - 1}$, obtained from the inverse rule applied to $x \mapsto x^{n}$; in particular $(\sqrt{x})' = 1/(\iota(2)\sqrt{x})$

Let $n \in \mathbb{N}$ with $n \ge 1$, let $\iota$ be the canonical natural of The canonical natural $\iota(n) = n \cdot 1_F$ of a field, and let rational powers be those of Rational powers $a^r$ of a positive base, so that $u^{1/n}$ is the unique nonnegative $n$-th root of $u$ (Existence and uniqueness of $n$-th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$).

Claim. The function

$g : (0,\infty) \to \mathbb{R}, \qquad g(u) := u^{1/n},$

is differentiable at every $b \in (0,\infty)$ (The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$, and differentiability on a set), and

$g'(b) \;=\; \frac{1}{\iota(n)}\; b^{\,1/n - 1} .$

In particular at $n = 2$, writing $\sqrt{u} = u^{1/2}$,

$g'(b) \;=\; \frac{1}{\iota(2)}\,b^{-1/2} \;=\; \frac{1}{\iota(2)\sqrt{b}} .$

The domain is $(0,\infty)$ and not $[0,\infty)$, and the reason depends on $n$. For $n \ge 2$ the exponent $1/n - 1$ is a negative rational, and Rational powers $a^r$ of a positive base leaves $0^{r}$ undefined for rational $r < 0$, so at $b = 0$ the displayed formula is not a statement at all; and the root really is not differentiable there, by claim 2 of Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f'(c) \ne 0$, then the inverse $g$ is differentiable at $f(c)$ with $g'(f(c)) = 1/f'(c)$; and if $f'(c) = 0$ then $g$ is not differentiable at $f(c)$ applied on $[0,\infty)$, since $x \mapsto x^{n}$ has derivative $\iota(n)\,0^{\,n-1} = 0$ at $0$ for $n \ge 2$ (For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$; for $n = 0$ it is the constant $1$, with derivative $0$; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Integer powers $a^m$). At $n = 1$ neither obstruction arises: the exponent $1/n - 1$ is $0$, not negative; $u^{1/1} = u$ is the identity (Existence and uniqueness of $n$-th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$); and the formula reads $g'(b) = b^{0} = 1$, which is correct at every real. So for $n = 1$ the restriction to $(0,\infty)$ is a convenience of the uniform statement rather than a necessity. Nothing below asserts anything about the root at $0$ in either case.

• Why the inverse rule and not a direct estimate. A direct computation of $\lim_{u \to b}(u^{1/n} - b^{1/n})/(u - b)$ has to rationalise the numerator using the factorisation of a difference of $n$-th powers, and then has to know that $u^{1/n} \to b^{1/n}$ to evaluate the limit, which is the continuity of the root. Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f'(c) \ne 0$, then the inverse $g$ is differentiable at $f(c)$ with $g'(f(c)) = 1/f'(c)$; and if $f'(c) = 0$ then $g$ is not differentiable at $f(c)$ packages both, and its own proof gets the continuity from Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f[I]$, and the inverse $g : f[I] \to I$ is continuous and strictly monotone in the same sense as $f$ rather than proving it again.

• Where the hypothesis $f'(c) \ne 0$ is spent. At $c > 0$ the derivative $\iota(n)c^{\,n-1}$ is positive, so the hypothesis costs nothing on $(0,\infty)$. It is exactly the hypothesis that fails at $c = 0$ when the domain is enlarged to $[0,\infty)$ and $n \ge 2$, and there the inverse rule says the root is not differentiable at $0$, which is claim 2 of that theorem.

• The exponent arithmetic is rational arithmetic, not real arithmetic. The identity $(n-1)/n = 1 - 1/n$ is an identity of rationals, and $\bigl(b^{1/n}\bigr)^{\,n-1} = b^{\,(n-1)/n}$ is claim 5 of Laws of rational exponents. Real exponents are not available at this page's position in the reading order, so every step above stays inside $\mathbb{Q}$ as Rational powers $a^r$ of a positive base requires. The later Real powers for positive bases, with the zero-base positive-exponent convention ↗ does not alter this proof boundary.