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Basic Differentiation Rules · Tutorial 437 of 1495

Rational Expressions As Sums Of Integer Powers

Rewrite each term of a rational expression as an integer power of the variable, then use the power and sum rules to find its derivative.

Beginner 8 min read

What You'll Learn

  • Convert a polynomial divided by a monomial into a sum of integer powers.
  • Find the exponent of each term by subtracting the denominator exponent from the numerator exponent.
  • Apply the power rule to positive, zero, and negative integer powers.
  • Differentiate the rewritten expression term by term.
  • Preserve the original domain when simplifying a rational expression.
  • Check a rewrite by combining its terms over a common denominator.

Why rewrite before differentiating?

A polynomial divided by a power of its variable can look like a single complicated fraction. But the numerator is a sum, and each numerator term can be divided by the denominator separately. Rewriting the expression this way turns it into a sum of powers, which can be differentiated term by term.

The resulting powers may be negative integers. For example, \(x^{-2}\) means \(1/x^2\), so it is another way to write a rational term. This notation makes the power rule useful for expressions whose original form has a monomial denominator.

Key idea: Divide each numerator term by the monomial denominator, simplify its coefficient, and subtract the denominator’s exponent from the numerator term’s exponent. Then apply the power rule to each resulting term.

From a rational expression to integer powers

A monomial denominator has the form \(c x^m\), where \(c\ne 0\) and \(m\) is a nonnegative integer. When \(m>0\), the original expression is defined only for \(x\ne 0\). When \(m=0\), the denominator is the nonzero constant \(c\), so it imposes no restriction on \(x\).

Write a polynomial numerator as \(a_0+a_1x+\cdots+a_nx^n\), allowing some coefficients \(a_k\) to be zero. For each term \(a_kx^k\), division by \(cx^m\) gives a coefficient \(a_k/c\) and a power \(x^{k-m}\).

$$ \frac{a_0+a_1x+\cdots+a_nx^n}{cx^m} = \frac{a_0}{c}x^{-m} +\frac{a_1}{c}x^{1-m} +\cdots+ \frac{a_n}{c}x^{n-m}. $$

If a term’s exponent \(k-m\) is negative, its power is defined for \(x\ne 0\). If \(k=m\), that term becomes a constant because \(x^0=1\). A zero numerator coefficient simply contributes no term.

Definition: An expression is written as a sum of integer powers when it is expressed as a finite sum of constant multiples of \(x^r\), where each exponent \(r\) is an integer. Negative powers are interpreted by \(x^{-q}=1/x^q\) for positive integers \(q\), and require \(x\ne 0\).

Differentiate the resulting powers

The power rule states that the derivative of \(x^r\) is \(r x^{r-1}\) for a positive integer \(r\), and also for a negative integer \(r\) when \(x\ne 0\). For \(r=0\), the function is the constant \(1\), whose derivative is \(0\). The sum rule says that a finite sum can be differentiated by differentiating each term, and constant factors remain as factors.

$$ \frac{d}{dx}\left(Cx^r\right)=Crx^{r-1}. $$

Thus, once the rational expression is rewritten, differentiate every term using its own exponent. The resulting derivative is valid on the domain of the original expression. In particular, if the original denominator contains \(x^m\) with \(m>0\), keep \(x\ne 0\) as a domain restriction even if the rewritten expression contains a constant term.

Why the rewriting method works

The conversion is an identity on the domain of the original fraction. It follows from distributing division across a finite sum and applying the exponent law \(x^k/x^m=x^{k-m}\), which is valid when \(x\ne 0\). Differentiation then follows from the power rule and sum rule.

1
Split the numerator.
For a polynomial numerator, divide each term by the common denominator separately.
2
Simplify each quotient.
Divide the numerical coefficients and subtract the denominator’s exponent from the numerator term’s exponent.
3
Differentiate term by term.
Use the power rule on each integer power and preserve the domain of the original fraction.

More explicitly, for \(x\ne 0\), the exponent law gives \(x^k/x^m=x^{k-m}\) for every nonnegative integer \(k\) and positive integer \(m\). Therefore each term in the numerator satisfies

$$ \frac{a_kx^k}{cx^m} = \frac{a_k}{c}x^{k-m}. $$

Adding these equalities gives the rewritten expression. Applying the power rule and sum rule to that finite sum gives its derivative. This explains both why the conversion is valid and why it leads directly to a term-by-term differentiation method.

Worked examples

Worked Example: Differentiate a fraction with several powers

Differentiate \(f(x)=\dfrac{6x^4-9x^2+3x}{3x^3}\), stating its domain.

The denominator is zero when \(x=0\), so the original expression is defined for \(x\ne 0\). Divide each numerator term by \(3x^3\):

$$ \begin{aligned} f(x) &=\frac{6x^4}{3x^3}-\frac{9x^2}{3x^3}+\frac{3x}{3x^3}\\ &=2x^{4-3}-3x^{2-3}+x^{1-3}\\ &=2x-3x^{-1}+x^{-2}. \end{aligned} $$

Apply the power rule to each term:

$$ \begin{aligned} f'(x) &=2(1)x^{1-1}-3(-1)x^{-1-1}+(-2)x^{-2-1}\\ &=2+3x^{-2}-2x^{-3}. \end{aligned} $$

Check: Combining the rewritten function over \(x^2\) gives \(2x-3x^{-1}+x^{-2}=(2x^3-3x+1)/x^2\). The original fraction also simplifies to \((6x^4-9x^2+3x)/(3x^3)=(2x^3-3x+1)/x^2\), so the rewrite agrees with the original wherever \(x\ne 0\). The derivative was obtained by applying the power rule separately to all three terms, and its domain remains \(x\ne 0\).

Worked Example: Include a term that becomes a constant

Differentiate \(g(x)=\dfrac{5x^4-10x^2+15}{5x^4}\).

Since the denominator is zero at \(x=0\), the domain is \(x\ne 0\). Divide each term by \(5x^4\):

$$ \begin{aligned} g(x) &=\frac{5x^4}{5x^4}-\frac{10x^2}{5x^4}+\frac{15}{5x^4}\\ &=x^{4-4}-2x^{2-4}+3x^{0-4}\\ &=1-2x^{-2}+3x^{-4}. \end{aligned} $$

The first term is constant, so its derivative is zero. Differentiate the other terms with the power rule:

$$ \begin{aligned} g'(x) &=0-2(-2)x^{-2-1}+3(-4)x^{-4-1}\\ &=4x^{-3}-12x^{-5}. \end{aligned} $$

Check: Putting the rewritten function over \(x^4\) gives \((x^4-2x^2+3)/x^4\). Dividing the original numerator and denominator by \(5\) gives the same expression, so the simplification is correct. The constant contributes zero to the derivative, as required.

Worked Example: Handle a negative leading power

Differentiate \(h(x)=\dfrac{8x^5+4x^3-12x}{4x^2}\).

The original denominator requires \(x\ne 0\). Dividing term by term gives

$$ \begin{aligned} h(x) &=\frac{8x^5}{4x^2}+\frac{4x^3}{4x^2}-\frac{12x}{4x^2}\\ &=2x^{5-2}+x^{3-2}-3x^{1-2}\\ &=2x^3+x-3x^{-1}. \end{aligned} $$

Now differentiate each term:

$$ \begin{aligned} h'(x) &=2(3)x^{3-1}+1(1)x^{1-1}-3(-1)x^{-1-1}\\ &=6x^2+1+3x^{-2}. \end{aligned} $$

Check: Recombining the function gives \(2x^3+x-3x^{-1}=(8x^5+4x^3-12x)/(4x^2)\). The power-rule calculation gives \(6x^2+1+3x^{-2}\), with the negative-power term contributing positively because its coefficient is multiplied by the exponent \(-1\). Both expressions are considered on \(x\ne 0\).

A common exponent error

When dividing powers with the same base, subtract the denominator exponent from the numerator exponent. Reversing the order can turn a negative power into a positive one and change the function.

Worked Example: Correct the order of subtraction

Consider the term \(x^2/x^5\). A wrong calculation is

$$ \frac{x^2}{x^5}=x^{5-2}=x^3. $$

This subtracts in the wrong order. The exponent law gives numerator exponent minus denominator exponent:

$$ \frac{x^2}{x^5}=x^{2-5}=x^{-3}=\frac{1}{x^3},\qquad x\ne 0. $$

To check, multiply \(x^{-3}\) by \(x^5\): the result is \(x^2\), as required. Multiplying the incorrect answer \(x^3\) by \(x^5\) gives \(x^8\), not \(x^2\). Recombining a rewritten expression over its original denominator is a useful way to catch this type of mistake.

Keep the original domain in view

Negative powers make the restriction at zero visible, but the restriction comes from the original denominator. For example, \(x^4/x^4\) rewrites as \(1\) for \(x\ne 0\), yet the original fraction is still undefined at zero. Do not use a simplified expression to add inputs that were excluded by the original denominator.

The method applies when the denominator is a nonzero constant times a single power of the variable. It does not turn a denominator such as \(x+2\) into integer powers by termwise division; that denominator is not a monomial. For the monomial case, the reliable sequence is: record the domain, divide each term, subtract exponents in the correct order, and differentiate the resulting sum.

Check Your Understanding

For each question, first simplify the expression into integer powers. Keep track of any restriction imposed by its original denominator.

  1. Rewrite \(\dfrac{12x^5-6x^2}{3x^3}\) as a sum of integer powers.
  2. Differentiate \(\dfrac{4x^3+8}{2x^4}\), stating the domain.
  3. What exponent results when \(x^2\) is divided by \(x^6\)?
  4. Why does rewriting \(x^3/x^3\) as \(1\) not make the original fraction defined at \(x=0\)?
  5. Does the term-by-term method in this tutorial directly apply to a denominator \(x+1\)? Explain briefly.

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