Differentiation Mastery I

Fractional Powers as a Test of Differentiation Skills

The basic differentiation rules become more powerful when they can be combined without losing track of their hypotheses. Roots are a useful test case: a root can be viewed as an inverse function, and a rational power can then be built from a root and an integer power. This approach connects the derivative rules already established in the course rather than requiring a new rule for every expression.

The domain matters. The formula for the derivative of a positive $n$th root will be stated for positive inputs. That avoids ambiguity about even roots and ensures that the function being inverted has an open interval as its domain and range. Rational powers with negative exponents also require a nonzero base. We will first establish the root formula, then use it to derive a rule for rational powers and examine separately what can happen at zero.

Definition: Let $n$ be a positive integer. For $x>0$, $x^{1/n}$ denotes the unique positive real number $y$ such that $y^n=x$. If $m$ is an integer, define $x^{m/n}=(x^{1/n})^m$ for $x>0$. Equivalent fractional representations give the same value by the laws of integer powers and uniqueness of positive roots.

The Derivative of a Positive Root

For $n\geq 1$, the function $u\mapsto u^n$ is a bijection from $(0,\infty)$ onto $(0,\infty)$; its inverse is $x\mapsto x^{1/n}$. At every $u>0$, the Power Rule gives derivative $nu^{n-1}$, which is nonzero. Thus the hypotheses of the Derivative of an Inverse Function theorem apply. The inverse theorem supplies the root derivative by taking the reciprocal of the derivative of the original function at the corresponding root.

Theorem (Derivative of a Positive $n$th Root): Let $n$ be a positive integer and define $R_n(x)=x^{1/n}$ for $x>0$. Then $R_n$ is differentiable at every $x>0$, and $$ R_n'(x)=\frac{1}{n}x^{1/n-1}. $$

Proof. Let $G(u)=u^n$ for $u>0$. The Power Rule gives $G'(u)=nu^{n-1}$, which is positive and therefore nonzero for every $u>0$. The function $G$ is a bijection from $(0,\infty)$ onto $(0,\infty)$, with inverse $R_n$. By the Derivative of an Inverse Function theorem, $R_n$ is differentiable at $x=G(u)$ and

R_n'(x)=\frac{1}{G'(u)} =\frac{1}{nu^{n-1}}, \qquad u=R_n(x)=x^{1/n}.

Since $u^{n-1}=(x^{1/n})^{n-1}=x^{(n-1)/n}$, this becomes

R_n'(x)=\frac{1}{n x^{(n-1)/n}} =\frac{1}{n}x^{1/n-1}.

This holds for every $x>0$, proving the theorem. $\square$

Worked Example: Differentiating a Seventh Root

Let $g(x)=x^{1/7}$ for $x>0$. Applying the root formula gives

g'(x)=\frac{1}{7}x^{1/7-1} =\frac{1}{7}x^{-6/7}.

For instance, $128=2^7$, so $128^{-6/7}=2^{-6}=1/64$. Therefore

g'(128)=\frac{1}{7}\cdot\frac{1}{64} =\frac{1}{448}.

The input $128$ is positive, so it lies in the domain covered by the theorem. The negative exponent in the derivative is not an error: it expresses the reciprocal $1/x^{6/7}$, which is defined because $x>0$.

From Roots to Rational Powers

A rational power $x^{m/n}$ is an integer power of the $n$th root of $x$. The Chain Rule and the Power Rule therefore give its derivative. The resulting formula resembles the familiar integer Power Rule, but it has a specific domain condition: throughout this result the base $x$ is positive.

Theorem (Power Rule for Rational Exponents on Positive Inputs): Let $m$ be an integer and $n$ a positive integer. The function $f(x)=x^{m/n}$, defined for $x>0$, is differentiable there and satisfies $$ f'(x)=\frac{m}{n}x^{m/n-1}. $$

Proof. Put $r(x)=x^{1/n}$. The root theorem gives $r'(x)=\frac{1}{n}x^{1/n-1}$ for $x>0$, and $f(x)=r(x)^m$. If $m$ is positive, the Power Rule and Chain Rule give

f'(x)=m r(x)^{m-1}r'(x) =\frac{m}{n}x^{(m-1)/n}x^{1/n-1} =\frac{m}{n}x^{m/n-1}.

If $m=0$, then $f(x)=1$ and $f'(x)=0$, which agrees with the formula. If $m<0$, the Negative Integer Power Rule applied to $r(x)^m$, followed by the Chain Rule, gives the same expression $m r(x)^{m-1}r'(x)$. The displayed simplification still holds because $r(x)>0$; its negative integer powers are defined. Thus the formula holds for every integer $m$ and every positive integer $n$. $\square$

Worked Example: Simplifying Before Differentiating

For $x>0$, consider

f(x)=\frac{x^{5/3}-2x^{2/3}}{x^{1/3}}.

The denominator is nonzero on this domain. Using the laws of rational powers, simplify first:

f(x)=x^{5/3-1/3}-2x^{2/3-1/3} =x^{4/3}-2x^{1/3}.

The rational-power rule now gives

f'(x)=\frac{4}{3}x^{1/3} -\frac{2}{3}x^{-2/3}.

At $x=8$, $8^{1/3}=2$ and $8^{-2/3}=1/4$, so

f'(8)=\frac{4}{3}\cdot 2-\frac{2}{3}\cdot\frac{1}{4} =\frac{8}{3}-\frac{1}{6} =\frac{16}{6}-\frac{1}{6} =\frac{5}{2}.

Simplifying first avoids applying the Quotient Rule to a quotient whose terms are all powers of the same positive variable. It also makes the domain restriction clear: the original denominator requires $x>0$.

Using the Chain Rule with Rational Powers

The rational-power formula can be applied to an inner function, provided its values stay positive wherever the derivative is being calculated. If $u$ is differentiable at $a$ and $u(a)>0$, continuity of $u$, which follows from differentiability, ensures that $u(x)$ remains positive for all $x$ sufficiently close to $a$. The Chain Rule then gives differentiability at $a$ for $u(x)^{m/n}$.

Corollary (Rational Power of a Positive Differentiable Function): Let $u$ be differentiable at an interior point $a$, with $u(a)>0$, and let $m$ be an integer and $n$ a positive integer. Near $a$, the function $u(x)^{m/n}$ is defined, and it is differentiable at $a$, with $$ \left.\frac{d}{dx}\bigl(u(x)^{m/n}\bigr)\right|_{x=a} =\frac{m}{n}u(a)^{m/n-1}u'(a). $$

Indeed, differentiability of $u$ implies continuity at $a$, so there is a neighborhood of $a$ on which $u(x)>0$. At $a$, apply the rational-power theorem to the outer function and the Chain Rule. The positivity hypothesis is what makes this direct composition valid, including when $m$ is negative.

Worked Example: A Rational Power of a Polynomial

Let $F(x)=(2x^2+3)^{5/3}$. The inner function $u(x)=2x^2+3$ is positive for every real $x$, since $2x^2\geq0$ and hence $2x^2+3\geq3>0$. Its derivative is $u'(x)=4x$. The Chain Rule and rational-power formula give

F'(x)=\frac{5}{3}(2x^2+3)^{2/3}(4x) =\frac{20x}{3}(2x^2+3)^{2/3}.

At $x=-1$, the inner value is $2(-1)^2+3=5$, and substitution yields

F'(-1)=\frac{20(-1)}{3}\,5^{2/3} =-\frac{20}{3}5^{2/3}.

The negative input $x=-1$ causes no difficulty: the base of the fractional power is not $x$, but $2x^2+3$, which remains positive.

What Changes at Zero?

The positive-input rule does not by itself determine whether a rational power has a derivative at zero. Zero is not in the open domain $(0,\infty)$ used in the theorem, and the difference quotient at zero must be checked directly. For a positive exponent, the behavior depends on whether that exponent is greater than, equal to, or less than $1$.

Proposition (Right Derivative at Zero for a Positive Rational Power): Let $\alpha$ be a positive rational number, and define $f:[0,\infty)\to\mathbb{R}$ by $f(0)=0$ and $f(x)=x^\alpha$ for $x>0$. The right difference quotient at zero is $x^{\alpha-1}$. It tends to $0$ if $\alpha>1$, equals $1$ if $\alpha=1$, and is unbounded as $x\to0^+$ if $0<\alpha<1$. Thus a finite right derivative exists exactly when $\alpha\geq1$.

Proof. For $h>0$, the right difference quotient is

\frac{f(0+h)-f(0)}{h} =\frac{h^\alpha}{h} =h^{\alpha-1}.

If $\alpha>1$, then $\alpha-1>0$, so $h^{\alpha-1}\to0$ as $h\to0^+$. If $\alpha=1$, the quotient is $h^0=1$ for every $h>0$. If $0<\alpha<1$, then $1-\alpha>0$ and $h^{\alpha-1}=1/h^{1-\alpha}$; the denominator tends to zero through positive values, so the quotient grows without bound and cannot have a finite limit. These cases exhaust all positive $\alpha$, proving the proposition. $\square$

Worked Example: Two Different Behaviors at the Endpoint

On $[0,\infty)$, define $p(x)=x^{4/3}$ for $x>0$ and $p(0)=0$. Since $4/3>1$, the proposition gives a right derivative at zero equal to $0$. Directly,

\lim_{h\to0^+}\frac{p(h)-p(0)}{h} =\lim_{h\to0^+}h^{1/3} =0.

In contrast, define $q(x)=x^{2/3}$ for $x>0$ and $q(0)=0$. Here $0<2/3<1$, and

\frac{q(h)-q(0)}{h}=h^{-1/3} =\frac{1}{h^{1/3}}\longrightarrow+\infty \qquad\text{as }h\to0^+.

Therefore $q$ has no finite right derivative at zero, even though it is differentiable at every positive input. A derivative formula valid for $x>0$ must not be substituted at $x=0$ without a separate difference-quotient check.

A Practical Checklist

When differentiating an expression involving roots or rational powers, first identify the base and its domain. Then choose the rule that matches the structure: the root theorem for a positive root, the rational-power rule for a single positive variable, or the Chain Rule when the base is a differentiable function. Finally, treat boundary points separately. In particular, the formula on positive inputs says nothing automatically about a point where the base is zero.

Check the base: even roots require a nonnegative input for real values, while the formulas proved here use strictly positive inputs.
Check denominators: negative rational powers and quotients require a nonzero base.
Apply the Chain Rule only where defined: for a rational power of $u(x)$, verify that $u(x)>0$ near the point in question.
Check boundary derivatives from the definition: at zero, the positive-input derivative formula cannot replace the difference quotient.

These checks are not extra formalities. They explain why the same-looking expression can be differentiable on one interval but not another, and why a correct algebraic derivative formula can still be used outside its domain if the hypotheses are ignored. The root and rational-power rules are most reliable when the domain is established before differentiation, not after it.

Check Your Understanding

Use the root and rational-power results to answer the following questions.

Why can the Derivative of an Inverse Function theorem be applied to $u\mapsto u^n$ on $(0,\infty)$?
For $x>0$, what is the derivative of $x^{-3/4}$, and which hypotheses allow the negative exponent?
If $u(a)>0$, why is $u(x)^{m/n}$ defined on some neighborhood of $a$ when $u$ is differentiable at $a$?
Does $x^{5/4}$, defined to be zero at zero on $[0,\infty)$, have a finite right derivative there? Explain using the difference quotient.
Why does the formula for the derivative of $x^{2/3}$ on $x>0$ not establish a finite derivative at zero?

What You'll Learn