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

Example

Let $n \in N$ with $n \geq 1$ , let $ι$ be the canonical natural of The canonical natural $ι (n) = n \cdot 1_{F}$ 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 $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ).

Claim. The function

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

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

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

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

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

The domain is $(0, \infty)$ and not $[0, \infty)$ , and the reason depends on $n$ . For $n \geq 2$ 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 $f^{'} (c) \neq 0$ , 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 $ι (n) 0^{n - 1} = 0$ at $0$ for $n \geq 2$ (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; 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 $a^{1 / n} \geq 0$ 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.

Facts & Assumptions

Given: A natural $n \geq 1$ , the set $I : = (0, \infty)$ , the function $f : I \to R$ , $f (x) : = x^{n}$ , and the function $g : I \to R$ , $g (u) : = u^{1 / n}$ .

[L1]

Roots (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ): for every real $a \geq 0$ and every natural $n \geq 1$ there is a unique real $s \geq 0$ with $s^{n} = a$ , written $a^{1 / n}$ ; and $a^{1 / n} > 0$ when $a > 0$ . By Rational powers $a^{r}$ of a positive base the rational power $a^{1 / n}$ is that same number.

[L2]

Rational power laws (Laws of rational exponents): for $a > 0$ and rationals $r, s$ one has $a^{r} > 0$ , $(a^{r})^{s} = a^{r s}$ , $a^{r + s} = a^{r} a^{s}$ and $a^{- r} = 1 / a^{r}$ ; and rational powers extend integer powers on positive bases (Rational powers $a^{r}$ of a positive base, Integer powers $a^{m}$ ).

[L3]

Monotonicity of integer powers (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ): for a natural $n \geq 1$ the map $x \mapsto x^{n}$ is strictly increasing on ${x \geq 0}$ , hence injective there (claim 2); and $x > 0$ implies $x^{n} > 0$ (claim 1).

[L4]

Continuity (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 5, and Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point): $x \mapsto x^{n}$ is continuous at every point of its domain, and continuity passes to a subset of the domain.

[L5]

Power rule and restriction (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, claim 2, with the restriction clause of The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set): $x \mapsto x^{n}$ on $R$ is differentiable at every real $c$ with derivative $ι (n) c^{n - 1}$ , and if $c$ lies in a subset of $R$ having $c$ as a limit point then the restriction is differentiable there with the same derivative.

[L6]

Derivative of an inverse (Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , 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 $f (c)$ , claim 1): for $I$ order-convex with at least two elements and $f : I \to R$ continuous and injective with inverse $g : f [I] \to I$ , if $f$ is differentiable at $c \in I$ with $f^{'} (c) \neq 0$ then $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ .

[L7]

The continuous inverse theorem (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 $f$ ): $f : I \to f [I]$ is a bijection with a unique two-sided inverse; and a right inverse of a bijection is that unique inverse (Injection, surjection, bijection).

[L8]

$I = (0, \infty)$ is order-convex with at least two elements (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), and every point of it is a limit point of it (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ); and $ι (n) > 0$ , hence $ι (n) \neq 0$ , for $n \geq 1$ (Canonical naturals are positive and strictly increasing).

Verification

technique · direct

1.1

$I = (0, \infty)$ is order-convex with at least two elements, and every point of $I$ is a limit point of $I$ .

1.2

$f$ is injective on $I$ by [L3], continuous on $I$ by [L4], and takes only positive values by [L3].

L3L4

1.3

$f [I] = I$ . For $x \in I$ one has $f (x) = x^{n} > 0$ by [L3], so $f [I] \subseteq I$ ; and for $u \in I$ the number $u^{1 / n}$ is positive by [L1], hence lies in $I$ , and $f (u^{1 / n}) = (u^{1 / n})^{n} = u$ by [L1], so $u \in f [I]$ .

L1L3

2.1

The map $g : I \to I$ , $u \mapsto u^{1 / n}$ , is the inverse of $f : I \to f [I] = I$ . By [L7] and step 1.2 that bijection has a unique two-sided inverse; by step 1.3 the map $g$ takes values in $I$ and satisfies $f (g (u)) = u$ for every $u \in I$ , so it is a right inverse of the bijection and therefore is that unique inverse.

step 1.2step 1.3L1L7

2.2

$f$ is differentiable at every $c \in I$ with $f^{'} (c) = ι (n) c^{n - 1}$ , by [L5] together with step 1.1; and $f^{'} (c) \neq 0$ , since $ι (n) > 0$ by [L8] and $c^{n - 1} > 0$ by [L3] as $c > 0$ .

step 1.1L3L5L8

3.1

Let $b \in I$ and put $c : = b^{1 / n}$ , an element of $I$ by [L1], with $f (c) = b$ by [L1]. By step 1.2, step 2.2 and [L6], applied on $I$ at $c$ , the inverse $g$ is differentiable at $b = f (c)$ with $g^{'} (b) = 1 / f^{'} (c) = 1 / (ι (n) c^{n - 1})$ .

step 2.1step 2.2L1L6

4.1

Rewriting in terms of $b$ : since $c = b^{1 / n}$ and $n - 1$ is a natural, [L2] gives $c^{n - 1} = (b^{1 / n})^{n - 1} = b^{(n - 1) / n} = b^{1 - 1 / n}$ , a positive real. Hence $g^{'} (b) = 1 / (ι (n) b^{1 - 1 / n}) = \frac{1}{ι (n)} b^{- (1 - 1 / n)} = \frac{1}{ι (n)} b^{1 / n - 1}$ , using $a^{- r} = 1 / a^{r}$ from [L2] and $ι (n) \neq 0$ from [L8].

step 3.1L2L8

5.1

At $n = 2$ the map $g$ is $u \mapsto u^{1 / 2} = \sqrt{u}$ , and step 4.1 reads $g^{'} (b) = \frac{1}{ι (2)} b^{1 / 2 - 1} = \frac{1}{ι (2)} b^{- 1 / 2} = \frac{1}{ι (2) \sqrt{b}}$ , again by [L2].

step 4.1L2∎

Remarks

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 $f^{'} (c) \neq 0$ , 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 $f^{'} (c) \neq 0$ is spent. At $c > 0$ the derivative $ι (n) c^{n - 1}$ 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 $n \geq 2$ , 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 $(b^{1 / n})^{n - 1} = b^{(n - 1) / n}$ 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 $Q$ 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.

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73 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Nth root (Wikipedia) (standard reference, not scraped)
Inverse function rule (Wikipedia) (standard reference, not scraped)
Power rule (Wikipedia) (standard reference, not scraped)
J. Hunter, An Introduction to Real Analysis (standard reference, not scraped)
J. Lebl, Basic Analysis I, The Derivative (standard reference, not scraped)

For a natural n≥1n≥1, the derivative of x↦x1/nx↦x1/n on (0,∞)(0,∞) is 1ι(n)x1/n−1ι(n)1​x1/n−1, obtained from the inverse rule applied to x↦xnx↦xn; in particular (x)′=1/(ι(2)x)(x​)′=1/(ι(2)x​)