From Radicals to Fractional Powers
A radical can often be written as a power with a fractional exponent. For example, \(\sqrt{x}=x^{1/2}\), and \(x\sqrt{x}=x^{3/2}\) whenever \(x\ge 0\). This notation makes the power rule useful for many functions that are not polynomials.
There is an important qualification: a fractional power may have a restricted real domain, and zero can behave differently from other points. The power-rule formula applies at nonzero points in the interior of the real domain. At zero, check the difference quotient separately—even if zero is an interior point of the domain.
The denominator matters because an even root of a negative real number is not real. For example, \(x^{1/2}\) has domain \([0,\infty)\), while \(x^{1/3}\) has domain \(\mathbb R\). For \(x^{2/3}\), the cube root is real for every real \(x\), so its domain is also \(\mathbb R\). In contrast, \(x^{-1/2}=1/\sqrt{x}\) has domain \((0,\infty)\): the square root requires \(x\ge 0\), and the denominator cannot be zero.
The Fractional Power Rule
For a rational exponent \(r\), the fractional power rule has the same form as the familiar power rule. Its domain conditions are essential: the rule gives the derivative at nonzero interior points of the real domain. A value at zero must be examined separately.
The formula follows the same exponent pattern as the positive-integer power rule, but it does not remove the need to respect the domain. In particular, when \(r-1\) is negative, the displayed expression may be undefined at zero even if \(x^r\) itself is defined there.
Why the Rule Works: The Square-Root Case
A difference-quotient calculation shows why the rule gives the correct derivative for \(x^{1/2}\) at positive \(x\). Fix \(x>0\). For nonzero \(h\) sufficiently close to zero, \(x+h>0\), and we can rationalize the numerator:
As \(h\) tends to zero, the denominator tends to \(2\sqrt{x}\), which is positive. Therefore,
Since \(1/(2\sqrt{x})=(1/2)x^{-1/2}\), this agrees with the fractional power rule. The restriction \(x>0\) is not incidental: the calculation requires a two-sided neighborhood of \(x\) inside the domain. It does not establish a derivative at the endpoint \(x=0\).
Worked Example: Differentiating a Square Root
Worked Example: Differentiating a Square Root
Find the derivative of \(f(x)=\sqrt{x}\), state where the formula applies, and find the derivative value at \(x=9\).
First write the radical as a fractional power:
For \(x>0\), apply the power rule:
At \(x=9\), the function value is \(f(9)=\sqrt{9}=3\), and the derivative is
The formula applies for \(x>0\), not at zero. At zero the right-hand difference quotient is \(\sqrt{h}/h=1/\sqrt{h}\) for \(h>0\), which grows without bound as \(h\to 0^+\). Thus there is no finite right-hand derivative at zero. The result is consistent with the derivative formula on its stated domain.
Worked Example: A Power Greater Than One
Worked Example: Differentiating \(x^{3/2}\)
Differentiate \(f(x)=x^{3/2}\), state the domain of the formula, and examine the endpoint \(x=0\).
Since \(x^{3/2}=x\sqrt{x}\), the real domain is \(x\ge 0\). For \(x>0\), the power rule gives
For example, at \(x=4\),
At zero, the function is defined, but the power-rule statement above only covers positive \(x\). The right-hand difference quotient is
Its limit as \(h\to 0^+\) is \(0\). Thus the right-hand derivative at the domain endpoint is \(0\). Under the usual two-sided definition, \(f\) has no ordinary derivative at \(0\), because its real domain does not contain an open interval around zero. The derivative formula for positive \(x\) approaches \(0\), in agreement with the one-sided calculation.
Zero Requires Its Own Check
The behavior at zero depends on the exponent. If \(r>0\), then \(x^r\) is defined at zero. For positive \(h\), its right-hand difference quotient at zero is
If \(r>1\), this tends to \(0\) from the right. If \(r=1\), it equals \(1\). If \(0<r<1\), it grows without bound as \(h\to 0^+\), so there is no finite right-hand derivative. These conclusions describe the right-hand behavior; for an ordinary derivative at zero, the function must also have a real domain on both sides of zero and the two-sided limit must exist.
For example, \(x^{4/3}\) has real domain \(\mathbb R\), and its zero difference quotient is \(h^{1/3}\), which tends to \(0\) from both sides. Its derivative at zero is therefore \(0\). But \(x^{2/3}\) also has domain \(\mathbb R\), and its difference quotient at zero does not have a finite limit.
Worked Example: An Interior Zero Where the Derivative Fails
Worked Example: Checking \(x^{2/3}\) at Zero
Find the derivative of \(f(x)=x^{2/3}\) for nonzero \(x\), then decide whether \(f'(0)\) exists.
Because the denominator of the reduced exponent \(2/3\) is odd, the function has real domain \(\mathbb R\). At every nonzero \(x\), the power rule gives
At zero, \(f(0)=0\). The difference quotient for \(h\ne 0\) is
For \(h>0\), this is \(h^{-1/3}\), which tends to \(+\infty\) as \(h\to0^+\). For \(h<0\), it is \(-|h|^{-1/3}\), which tends to \(-\infty\) as \(h\to0^-\). The two-sided limit is not finite, so \(f'(0)\) does not exist. Zero is an interior point of the domain, but that fact alone does not ensure differentiability.
Worked Example: A Negative Fractional Power
Worked Example: Differentiating \(x^{-1/2}\)
Differentiate \(g(x)=x^{-1/2}\), state its domain, and evaluate the derivative at \(x=4\).
Rewrite the function as a reciprocal radical:
The square root requires \(x\ge0\), and the denominator cannot equal zero. Thus the domain is \(x>0\). Applying the power rule on that domain gives
At \(x=4\),
The domain check confirms that \(x=4\) is a valid point for the formula. At zero, the function itself is undefined, so there can be no derivative there.
A Common Mistake: Substituting Zero into the Formula
A tempting but invalid step is to use the rule at zero whenever the original function is defined there. For \(f(x)=x^{2/3}\), one might write
This does not produce a derivative value: \(0^{-1/3}\) is undefined. More importantly, the difference quotient shows that the derivative does not exist, rather than merely leaving its value unknown. The correction is to state the power-rule formula only for \(x\ne0\) and test zero directly, as above.
The same caution applies to endpoint values. For \(\sqrt{x}\), substituting zero into \(1/(2\sqrt{x})\) is impossible, and the right-hand quotient grows without bound. For \(x^{3/2}\), the formula on \(x>0\) also cannot be substituted at zero, but the right-hand difference quotient has finite limit \(0\). Checking the quotient distinguishes these cases.
How to Apply the Rule Reliably
For example, \(\sqrt{x}=x^{1/2}\) and \(x\sqrt{x}=x^{3/2}\), with the original real-domain restrictions preserved.
Check whether the reduced denominator is even or odd, and exclude zero when the exponent is negative.
Multiply by the exponent and subtract one from the exponent: \(x^r\) becomes \(r x^{r-1}\).
If zero is in the domain, evaluate the difference quotient there and determine whether a finite two-sided or one-sided limit is appropriate.
Fractional powers are useful because they let the power rule handle many radical functions in a compact form. The rule is not a substitute for domain analysis: the exponent determines where the function is real, and the difference quotient determines what happens at zero. Keeping those checks separate prevents a valid formula on \(x\ne0\) from being mistaken for a statement about every point in the domain.
Check Your Understanding
For each question, state any domain restriction as well as the derivative conclusion.
- Rewrite \(\sqrt[3]{x^2}\) using a fractional exponent and differentiate it for \(x\ne0\).
- What is the real domain of \(x^{5/2}\), and where does the fractional power rule apply directly?
- Does \(x^{5/2}\) have a finite right-hand derivative at zero? Use its difference quotient to decide.
- Why is the power-rule formula for \(x^{2/3}\) not valid at zero, even though the function is defined there?
- State the domain and derivative of \(x^{-3/2}\).
