Fractional Powers Extend the Power Rule
A power such as \(x^2\) has an integer exponent. A power such as \(\sqrt{x}\), however, can be written as \(x^{1/2}\), and its exponent is fractional. The same power-rule pattern applies, but the domain matters: a square root is not real for every real input, and its derivative formula is not valid at every point of its domain.
Here we study powers with rational exponents. We will establish the derivative rule at nonzero inputs, then treat zero separately. That separation is essential for knowing exactly where a formula applies.
For example, \(x^{1/2}=\sqrt{x}\) has domain \(x\geq 0\), while \(x^{1/3}=\sqrt[3]{x}\) has domain all real numbers. The power \(x^{-1/2}=1/\sqrt{x}\) has domain \(x>0\): the square root requires \(x\geq 0\), and the negative exponent also excludes zero.
The Power Rule for Rational Exponents
The statement at nonzero inputs follows from the difference quotient. The proof below uses only the definition of a rational power and the factorization of a difference of integer powers. It does not assume a chain rule.
Fix a nonzero \(x\) in the domain, and let \(a=\sqrt[q]{x}\). For a nearby input \(x+h\) in the domain, let \(y=\sqrt[q]{x+h}\). Then \(y\) approaches \(a\) as \(h\) approaches \(0\), and \(h=y^q-a^q\).
When \(p>0\), the quotient for \(x^{p/q}\) becomes \((y^p-a^p)/(y^q-a^q)\). Factoring each difference gives $$ \frac{y^p-a^p}{y^q-a^q} = \frac{y-a}{y-a} \frac{y^{p-1}+y^{p-2}a+\cdots+a^{p-1}} {y^{q-1}+y^{q-2}a+\cdots+a^{q-1}}. $$
As \(y\to a\), the numerator sum tends to \(p a^{p-1}\), and the denominator sum tends to \(q a^{q-1}\). The denominator limit is nonzero because \(a\ne 0\). Thus the quotient tends to \((p/q)a^{p-q}=(p/q)x^{(p-q)/q}\), which is \(r x^{r-1}\).
If \(p=-m\) for a positive integer \(m\), then $$ y^{-m}-a^{-m} = -\frac{(y-a)(y^{m-1}+y^{m-2}a+\cdots+a^{m-1})}{a^m y^m}. $$ Dividing by \(y^q-a^q=(y-a)(y^{q-1}+\cdots+a^{q-1})\) and letting \(y\to a\) gives \(-\frac{m}{q}a^{-m-q}=r x^{r-1}\). This proves the formula for negative rational exponents as well.
When \(q\) is odd, this argument also works for negative \(x\): its real \(q\)th root \(a\) is negative but nonzero, and the same factorizations and limits apply. When \(q\) is even, negative \(x\) is outside the real domain, so there is no real derivative there to consider.
Square-Root Derivatives
For \(f(x)=\sqrt{x}=x^{1/2}\), the rule gives \(f'(x)=\frac12 x^{-1/2}\). This is a real-valued derivative formula only for \(x>0\). The function itself is defined at \(0\), but the displayed formula is not.
Worked Example: Differentiate a Square Root
Let \(f(x)=\sqrt{x}\). Its domain is \(x\geq 0\). At every \(x>0\), apply the rule with \(r=\frac12\): $$ f'(x)=\frac12 x^{1/2-1} =\frac12 x^{-1/2} =\frac{1}{2\sqrt{x}}. $$ At \(x=9\), this gives \(f'(9)=1/(2\cdot 3)=1/6\).
Check the value at \(9\) from the difference quotient. For \(h\) near \(0\), \(9+h>0\), and rationalizing gives $$ \frac{\sqrt{9+h}-\sqrt{9}}{h} = \frac{\sqrt{9+h}-3}{h} \frac{\sqrt{9+h}+3}{\sqrt{9+h}+3} = \frac{1}{\sqrt{9+h}+3}. $$ As \(h\to 0\), this tends to \(1/(3+3)=1/6\), agreeing with the derivative formula.
Cube Roots and Negative Inputs
Unlike a square root, a cube root accepts negative real inputs. For \(f(x)=\sqrt[3]{x}=x^{1/3}\), the power rule gives \(f'(x)=\frac13x^{-2/3}\) when \(x\ne 0\). The excluded point is zero, not the negative part of the domain.
Worked Example: Differentiate a Cube Root at a Negative Input
Let \(f(x)=\sqrt[3]{x}\), and find the derivative at \(x=-8\). Since the cube root is real, \(f(-8)=-2\). The power rule gives $$ f'(-8)=\frac13(-8)^{-2/3} =\frac{1}{3(\sqrt[3]{-8})^2} =\frac{1}{3\cdot(-2)^2} =\frac{1}{12}. $$ To check from the definition, set \(y=\sqrt[3]{-8+h}\). Then \(h=y^3+8=(y+2)(y^2-2y+4)\), so $$ \frac{f(-8+h)-f(-8)}{h} = \frac{y+2}{(y+2)(y^2-2y+4)} = \frac{1}{y^2-2y+4}. $$ As \(h\to 0\), \(y\to -2\), and this tends to \(1/(4+4+4)=1/12\). The check confirms both the value and its positive sign.
What Happens at Zero?
The power-rule formula at nonzero inputs cannot simply be substituted at zero when it contains a negative power. To decide differentiability there, return to the difference quotient. For a positive exponent \(r\), the quotient at zero is $$ \frac{h^r-0^r}{h}=h^{r-1}, $$ for nonzero \(h\) allowed by the domain.
If \(r>1\), the magnitude of this quotient approaches \(0\) as \(h\to 0\), so the derivative at zero is \(0\). If \(0<r<1\), its magnitude grows without bound as \(h\to 0\), so there is no finite derivative at zero. For an even denominator, only positive \(h\) are allowed near zero, and the same conclusions follow using the right-hand difference quotient. For an odd denominator, the domain includes inputs on both sides; the magnitude conclusion still holds.
Worked Example: Differentiate \(x^{3/2}\), Including at Zero
Let \(f(x)=x^{3/2}\), with domain \(x\geq 0\). For \(x>0\), the power rule gives $$ f'(x)=\frac32x^{1/2}=\frac32\sqrt{x}. $$ At \(x=0\), the right-hand difference quotient is \(h^{3/2}/h=h^{1/2}=\sqrt{h}\) for \(h>0\), which tends to \(0\). Thus the derivative at the domain endpoint, in the right-hand sense, is \(f'_+(0)=0\).
Check the formula at \(x=4\). It gives \(f'(4)=\frac32\cdot2=3\). To verify by the difference quotient, let \(y=\sqrt{4+h}\), so \(h=y^2-4=(y-2)(y+2)\). Then $$ \frac{(4+h)^{3/2}-4^{3/2}}{h} = \frac{y^3-8}{y^2-4} = \frac{(y-2)(y^2+2y+4)}{(y-2)(y+2)}. $$ As \(h\to0\), \(y\to2\), and the quotient tends to \((4+4+4)/(2+2)=3\), as required.
Negative Fractional Powers
A negative fractional exponent requires a nonzero input, so its derivative rule applies only at points in that nonzero domain. The formula may contain another negative exponent; that does not add zero to the domain.
Worked Example: Differentiate a Negative Fractional Power
Let \(f(x)=x^{-1/2}=1/\sqrt{x}\), whose real domain is \(x>0\). The power rule gives $$ f'(x)=-\frac12x^{-3/2}. $$ At \(x=4\), $$ f'(4)=-\frac12\cdot4^{-3/2} =-\frac12\cdot\frac{1}{(\sqrt{4})^3} =-\frac{1}{16}. $$ For a check, set \(y=\sqrt{4+h}\). Then \(h=(y-2)(y+2)\), and $$ \frac{1/y-1/2}{h} = \frac{2-y}{2y(y-2)(y+2)} = -\frac{1}{2y(y+2)}. $$ As \(h\to0\), \(y\to2\), so the quotient tends to \(-1/(2\cdot2\cdot4)=-1/16\), matching the rule.
Common Pitfall: Using the Formula Where It Is Undefined
A tempting but incorrect calculation is to substitute \(x=0\) into the square-root derivative formula and say $$ \left.\frac{1}{2\sqrt{x}}\right|_{x=0} =\frac{1}{2\cdot 0} =\infty, $$ therefore the derivative at zero is “infinity.” This is not a valid derivative calculation: division by zero is undefined, and a derivative, when it exists, is a finite real number.
The correction is to use the difference quotient on the domain. For \(f(x)=\sqrt{x}\) at zero and \(h>0\), $$ \frac{\sqrt{0+h}-\sqrt{0}}{h} =\frac{\sqrt{h}}{h} =\frac{1}{\sqrt{h}}. $$ This grows without bound as \(h\to0^+\), so there is no finite right-hand derivative at zero. The formula \(1/(2\sqrt{x})\) is valid for \(x>0\), not at \(x=0\).
Why Domain Checks Matter
Fractional powers make the assumptions behind a differentiation rule visible. The power rule supplies an efficient derivative formula, but it does not erase the domain of the original function, nor does it guarantee that a derivative exists at a boundary point. Before using the formula, identify the real domain, check whether the input is nonzero, and treat zero separately when it belongs to the domain.
In particular, square-root powers require nonnegative inputs, cube-root powers allow negative inputs, and negative powers exclude zero. At nonzero inputs in the domain, the rational power rule applies. At zero, a positive exponent greater than \(1\) gives derivative \(0\), whereas a positive exponent less than \(1\) gives no finite derivative.
Check Your Understanding
For each question, identify the relevant domain before applying a derivative formula.
- For \(f(x)=x^{1/2}\), state the domain and the interval on which the derivative formula \(\frac{1}{2\sqrt{x}}\) is valid.
- Use the power rule to find the derivative of \(x^{2/3}\) at a nonzero input \(x\). Does the formula apply at \(x=0\)?
- For \(g(x)=x^{4/3}\), determine whether the derivative at zero exists and, if so, find it.
- Find the derivative of \(x^{-1/3}\) at nonzero real inputs, and state which input is excluded from its domain.
- Explain why the expression \(1/(2\sqrt{0})\) does not establish a derivative value for \(\sqrt{x}\) at zero.
