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Basic Differentiation Rules · Tutorial 440 of 1495

Negative And Fractional Exponents In One Expression

Apply the power rule to mixed powers while keeping the domain restrictions of every term in view.

Beginner 10 min read

What You'll Learn

  • Apply the power rule to positive, negative, and fractional powers in a sum
  • Determine where rational powers are real-valued
  • Intersect the domain restrictions imposed by all terms
  • Differentiate each term while retaining its original domain
  • Check a mixed-power derivative term by term

Why Mixed Powers Need Care

A single expression can combine ordinary polynomial terms, reciprocal powers, and roots. The power rule gives a consistent way to differentiate these terms, but the expression’s domain must be checked before calculating: each term can impose a different restriction on the allowed inputs.

For example, a square root requires a nonnegative input, while a negative power is undefined at zero. If both kinds of terms occur in one sum, the function is defined only where all of its terms are defined. The derivative formula must be interpreted on that shared domain.

Rational Powers and Their Real Domains

A fractional exponent is a rational number. Write it in lowest terms as \(p/q\), where \(p\) is an integer and \(q\) is a positive integer. When \(x\) is nonnegative, \(x^{p/q}\) is interpreted using the \(q\)-th root and integer powers. If \(x\) is negative, a real \(q\)-th root exists only when \(q\) is odd.

If \(p/q\) is negative, the power also involves division by a positive power of \(x\), so \(x=0\) must be excluded. Thus, for a reduced rational exponent, an even denominator restricts the base to \(x\geq 0\) when the exponent is positive, and to \(x>0\) when it is negative. An odd denominator permits negative bases, but a negative exponent still excludes zero.

Definition: The common domain of a sum is the intersection of the domains of all its terms. A real input belongs to the common domain only if every term in the expression is defined there.

For instance, \(x^{1/3}\) is real for every real \(x\), because cube roots of negative numbers are real. In contrast, \(x^{1/2}\) is real only for \(x\geq 0\). The term \(x^{-1}\) is defined for every real \(x\) except \(x=0\).

The Power Rule for a Sum of Powers

For a power \(x^r\), the power rule gives a derivative of \(r x^{r-1}\) at inputs where the power is defined and differentiable. In a finite sum, differentiate each term separately and preserve its coefficient. The resulting formula applies on open intervals within the common domain; a boundary point of the domain may require a separate derivative check.

Rule: If \(a_1,\ldots,a_n\) are constants and each \(x^{r_i}\) is differentiable on an open interval \(I\), then $$ \frac{d}{dx}\left(\sum_{i=1}^{n}a_i x^{r_i}\right) = \sum_{i=1}^{n}a_i r_i x^{r_i-1}, \qquad x\in I. $$ For mixed powers, take \(I\) to lie in the common domain of all terms.

The rule applies to positive integer powers, negative integer powers, and rational powers wherever the corresponding power is differentiable. A constant term has derivative zero. A nonzero constant coefficient stays in place while its power is differentiated.

Why Differentiating Term by Term Works

The sum rule follows directly from the definition of a derivative. Suppose \(f\) and \(g\) are differentiable at \(x\). The difference quotient for their sum can be separated into two difference quotients:

$$ \frac{(f+g)(x+h)-(f+g)(x)}{h} = \frac{f(x+h)-f(x)}{h} + \frac{g(x+h)-g(x)}{h}. $$

As \(h\) approaches zero, the first quotient approaches \(f'(x)\), and the second approaches \(g'(x)\). The limit of their sum is therefore \(f'(x)+g'(x)\). Repeating this argument gives the rule for any finite number of terms. Applying the power rule to each term then gives the displayed mixed-power formula.

A basic fractional-power example also shows why the rule has its familiar form. For \(x>0\), rationalizing the difference quotient for the square root gives

$$ \begin{aligned} \frac{d}{dx}x^{1/2} &=\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\\ &=\lim_{h\to 0} \frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})} {h(\sqrt{x+h}+\sqrt{x})}\\ &=\lim_{h\to 0}\frac{1}{\sqrt{x+h}+\sqrt{x}}\\ &=\frac{1}{2\sqrt{x}} =\frac{1}{2}x^{-1/2}. \end{aligned} $$

The cancellation in the second line uses \((\sqrt{x+h})^2-(\sqrt{x})^2=h\). The last limit is valid because \(x>0\), so the denominator approaches the nonzero value \(2\sqrt{x}\).

Worked Examples

Worked Example: Positive, Negative, and Square-Root Powers

Differentiate \(f(x)=3x^4-8x^{-2}+6x^{1/2}\), and state its domain.

First find where every term is real. The term \(3x^4\) is defined for all real \(x\). The term \(-8x^{-2}\) excludes \(x=0\). The square-root term \(6x^{1/2}\) requires \(x\geq 0\). Together these restrictions give the common domain \(x>0\).

Apply the power rule to each term, keeping each coefficient:

$$ \begin{aligned} \frac{d}{dx}(3x^4)&=3(4)x^{4-1}=12x^3,\\ \frac{d}{dx}(-8x^{-2})&=(-8)(-2)x^{-2-1}=16x^{-3},\\ \frac{d}{dx}(6x^{1/2})&=6\left(\frac12\right)x^{1/2-1}=3x^{-1/2}. \end{aligned} $$

Therefore, for \(x>0\), $$ f'(x)=12x^3+16x^{-3}+3x^{-1/2}. $$

Check at \(x=4\). The derivative’s three terms evaluate to \(12(4^3)=768\), \(16(4^{-3})=16/64=1/4\), and \(3(4^{-1/2})=3/2\). Their sum is \(768+1/4+3/2=3079/4\). Each term in the answer matches the power rule applied to its corresponding original term, and \(4\) lies in the stated domain.

Worked Example: Odd-Denominator Powers and a Negative Input

Differentiate \(g(x)=5x^{4/3}-2x^{-2/3}+7x^{1/3}\), and determine its common domain.

All the fractional exponents have denominator \(3\) in lowest terms. Since \(3\) is odd, these powers are real for negative as well as positive inputs. However, \(x^{-2/3}\) is undefined at zero. Thus the common domain is \(\mathbb{R}\setminus\{0\}\).

Differentiate the terms one at a time:

$$ \begin{aligned} \frac{d}{dx}(5x^{4/3})&=5\left(\frac43\right)x^{4/3-1} =\frac{20}{3}x^{1/3},\\ \frac{d}{dx}(-2x^{-2/3})&=(-2)\left(-\frac23\right)x^{-2/3-1} =\frac43x^{-5/3},\\ \frac{d}{dx}(7x^{1/3})&=7\left(\frac13\right)x^{1/3-1} =\frac73x^{-2/3}. \end{aligned} $$

Hence $$ g'(x)=\frac{20}{3}x^{1/3}+\frac43x^{-5/3}+\frac73x^{-2/3}, \qquad x\ne 0. $$

Check the formula at \(x=-1\), which is in the domain. The powers have values \((-1)^{1/3}=-1\), \((-1)^{-5/3}=-1\), and \((-1)^{-2/3}=1\). Therefore $$ g'(-1)=-\frac{20}{3}-\frac43+\frac73=-\frac{17}{3}. $$ The signs follow from the odd cube-root denominator: the first two powers are negative at \(-1\), while the last is positive.

Worked Example: Intersecting Restrictions from Several Terms

Differentiate \(q(x)=2x^{7/2}-3x^{1/2}+4x^{-1}\), and state where the formula applies.

The two fractional powers with denominator \(2\) require \(x\geq 0\). The negative integer power \(x^{-1}\) requires \(x\ne 0\). The common domain is therefore \(x>0\). In particular, the square-root terms alone would permit \(x=0\), but the reciprocal term does not.

Now apply the power rule:

$$ \begin{aligned} \frac{d}{dx}(2x^{7/2})&=2\left(\frac72\right)x^{7/2-1}=7x^{5/2},\\ \frac{d}{dx}(-3x^{1/2})&=(-3)\left(\frac12\right)x^{1/2-1} =-\frac32x^{-1/2},\\ \frac{d}{dx}(4x^{-1})&=4(-1)x^{-1-1}=-4x^{-2}. \end{aligned} $$

Thus $$ q'(x)=7x^{5/2}-\frac32x^{-1/2}-4x^{-2}, \qquad x>0. $$

At \(x=1\), the three derivative terms are \(7\), \(-3/2\), and \(-4\). Their sum is \(7-3/2-4=3/2\). This input is in the common domain, and the evaluated result includes all three differentiated terms.

Common Domain: Check Every Term

A frequent source of error is to find the domain from only the most conspicuous term. Instead, list each restriction and intersect them. For example, a square root may require nonnegative inputs, while a negative power may exclude zero; together, these restrictions can reduce the allowable inputs to positive values only.

Do not assume that an algebraic simplification changes the domain of the expression as originally given. If a term is undefined at a point, the original expression is not defined there, even if a simplified formula has a value at that point. Keep the original domain restrictions attached to the derivative formula.

Takeaway: Differentiate each power on the common domain, but determine that domain from the original terms before differentiating. A derivative formula does not make an excluded input valid.

A Reliable Calculation Process

1
Find each term’s domain.
Check even-denominator roots, negative exponents, and any other restrictions term by term.
2
Intersect the restrictions.
State the common domain of the full expression, not just the domain of one term.
3
Apply the power rule to every term.
Multiply by the exponent and reduce the exponent by one, retaining the original coefficient.
4
State where the derivative formula applies.
Use the open intervals in the common domain; check a boundary separately if it is relevant.

Check Your Understanding

For each question, consider both the derivative calculation and the domain of the original expression.

  1. What is the common domain of \(x^{3/2}+x^{-2}\)?
  2. Differentiate \(4x^5-3x^{-1}\), and state where the formula applies.
  3. Why is \(x^{2/3}\) real at negative inputs, while \(x^{1/2}\) is not?
  4. Differentiate \(2x^{5/3}-6x^{-1/3}\), and identify the excluded input.
  5. If a sum contains a square-root term and a negative integer power, why must you check both restrictions before giving the domain?

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