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Basic Differentiation Rules · Tutorial 422 of 1495

Polynomial Derivatives Term By Term

Use the basic differentiation rules term by term to find a polynomial’s derivative, and check how constant terms contribute.

Beginner 9 min read

What You'll Learn

  • Identify a polynomial as a finite sum of constant multiples of nonnegative integer powers.
  • Apply the power rule to each nonconstant monomial.
  • Use the constant multiple and sum rules to combine term derivatives.
  • Explain why a constant term contributes zero to the derivative.
  • Check a polynomial derivative by differentiating each original term.
  • Recognize and correct a common error involving constant terms.

From Monomials to Polynomials

A polynomial is built by adding together a finite number of terms. Each term is a constant multiple of a nonnegative integer power of the variable. The earlier rules for differentiating monomials and sums therefore give a direct method for differentiating any polynomial: differentiate each term, then add the results.

This method matters because a polynomial may contain several different powers, positive or negative coefficients, and a constant term. Keeping the terms separate makes it easier to apply the right rule to each one and to see why constants disappear.

Definition: A polynomial in \(x\) is a function that can be written as a finite sum of terms \(c_kx^k\), where each coefficient \(c_k\) is a real number and each exponent \(k\) is a nonnegative integer. A constant term is the term with exponent zero, written \(c_0x^0=c_0\).

For example, \(4x^3-2x+7\) is a polynomial. Its terms are \(4x^3\), \(-2x\), and \(7\). The last term is constant: it does not change when \(x\) changes.

The Rules Used Term by Term

For a positive integer \(n\), the power rule states that the derivative of \(x^n\) is \(nx^{n-1}\). The constant multiple rule says that a constant factor can be kept outside the derivative. The sum and difference rules say that the derivative of a finite sum or difference is the sum or difference of the derivatives. These rules apply here because each polynomial term is differentiable for every real \(x\).

$$ \frac{d}{dx}\left(cx^n\right)=cnx^{n-1} \qquad (n\geq 1) $$

A constant term is the remaining case. If the term is \(c_0\), its value is the same for every input, so it is a constant function. The derivative of a constant function is zero. We can express the termwise rule, including the constant term, as follows:

$$ \frac{d}{dx}(c_0)=0. $$

Thus, when differentiating a polynomial, apply the power and constant multiple rules to every term with a positive exponent, and replace every constant term by zero. Then combine the resulting terms.

Term-by-term rule: If \(p(x)=c_0+c_1x+c_2x^2+\cdots+c_nx^n\), where \(n\geq 1\) is an integer and the coefficients are real numbers, then $$ p'(x)=c_1+2c_2x+3c_3x^2+\cdots+nc_nx^{n-1}. $$ If \(n=0\), then \(p(x)=c_0\) is constant and \(p'(x)=0\). The derivative of the constant term \(c_0\) is zero. Any coefficient that is zero contributes no term.

Why the Term-by-Term Method Works

The rule follows from the basic differentiation rules already established: the power rule differentiates each power, the constant multiple rule handles its coefficient, and the sum rule combines the results. The following argument makes explicit why those rules cover a polynomial with any finite number of terms.

1
Start with one term.
For one monomial \(c_kx^k\), the constant multiple rule and power rule give its derivative when \(k\geq 1\); if \(k=0\), it is constant and its derivative is zero.
2
Add one more term.
If the derivative of a sum of \(m\) terms is the sum of their derivatives, write a sum of \(m+1\) terms as the first \(m\) terms plus the last term.
3
Apply the sum rule.
The derivative of that sum is the derivative of the first \(m\) terms plus the derivative of the last term. By the assumption for \(m\) terms and the one-term rule, this is the sum of all \(m+1\) term derivatives.

The argument begins with a single term and shows that whenever it works for \(m\) terms, it also works for \(m+1\) terms. Therefore it works for every finite number of terms. This establishes the term-by-term rule for every polynomial, including polynomials with a constant term.

Worked Examples

Worked Example: A Polynomial with Four Terms

Differentiate \(f(x)=5x^4-3x^2+7x-9\) term by term.

First, differentiate each term separately. For \(5x^4\), the power rule gives \(4x^3\), and the constant multiple rule keeps the factor \(5\): \(5(4x^3)=20x^3\). For \(-3x^2\), the derivative is \(-3(2x)=-6x\). For \(7x\), use \(x=x^1\), so its derivative is \(7(1x^0)=7\). The last term, \(-9\), is constant and has derivative \(0\).

$$ \begin{aligned} f'(x) &=20x^3-6x+7+0\\ &=20x^3-6x+7. \end{aligned} $$

Check: Differentiating the original terms again gives \(\frac{d}{dx}(5x^4)=20x^3\), \(\frac{d}{dx}(-3x^2)=-6x\), \(\frac{d}{dx}(7x)=7\), and \(\frac{d}{dx}(-9)=0\). Their sum is \(20x^3-6x+7\), as found.

Worked Example: Fractional Coefficients

Differentiate \(g(x)=\frac{3}{2}x^5-\frac{2}{3}x^3+4\).

For the first term, apply the power rule to \(x^5\), giving \(5x^4\), then multiply by \(\frac{3}{2}\): \(\frac{3}{2}(5x^4)=\frac{15}{2}x^4\). For the second term, the derivative of \(x^3\) is \(3x^2\); multiplying by \(-\frac{2}{3}\) gives \(-\frac{2}{3}(3x^2)=-2x^2\). The constant \(4\) contributes \(0\).

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

Check: Applying the rules to the original terms yields \(\frac{d}{dx}(\frac{3}{2}x^5)=\frac{3}{2}(5x^4)=\frac{15}{2}x^4\), \(\frac{d}{dx}(-\frac{2}{3}x^3)=-\frac{2}{3}(3x^2)=-2x^2\), and \(\frac{d}{dx}(4)=0\). Adding these gives the stated derivative.

Worked Example: Terms in an Unusual Order

Differentiate \(h(x)=\frac{2}{5}x^6-7+\ 6x-\frac{7}{2}x^2-x^3\). The terms do not need to be written in order of exponent before applying the rule.

Differentiate the first term: \(\frac{2}{5}(6x^5)=\frac{12}{5}x^5\). The constant \(-7\) has derivative \(0\). The term \(6x\) has derivative \(6\). For \(-\frac{7}{2}x^2\), the derivative is \(-\frac{7}{2}(2x)=-7x\). Finally, \(-x^3\) has derivative \(-1(3x^2)=-3x^2\).

$$ \begin{aligned} h'(x) &=\frac{12}{5}x^5+0+6-7x-3x^2\\ &=\frac{12}{5}x^5-3x^2-7x+6. \end{aligned} $$

Check: The five original terms produce, in their original order, \(\frac{12}{5}x^5\), \(0\), \(6\), \(-7x\), and \(-3x^2\). Adding them gives \(\frac{12}{5}x^5-3x^2-7x+6\), matching the result. Reordering the derivative terms changes neither their values nor their sum.

Constant Terms and Common Errors

A constant term contributes no change as the input changes, so its derivative is zero. It is important not to confuse the constant term with a coefficient attached to a variable term. For example, in \(8x+5\), the \(8\) is the coefficient of \(x\), and the derivative of \(8x\) is \(8\). The \(5\) is a separate constant term, and its derivative is \(0\).

Common mistake: Treating a constant term as if it stays unchanged in the derivative. For example, the incorrect calculation \(\frac{d}{dx}(3x^2+5)=6x+5\) leaves the constant \(5\) in place. The correction is \(\frac{d}{dx}(3x^2+5)=6x+0=6x\), because \(\frac{d}{dx}(5)=0\).

Another useful check is to keep track of each original term before combining anything. A missing term or sign error is easier to spot when the derivative of each term is written separately. In particular, a negative coefficient stays negative when multiplied by the positive exponent in the power rule.

The zero polynomial, whose value is zero for every input, also fits the same idea: it has no nonzero terms, and its derivative is zero. More generally, if a polynomial has only a constant term, its derivative is zero. These are not exceptions to the method; they follow from the constant-function rule.

Why This Method Is Useful

Term-by-term differentiation turns the derivative of a polynomial into a sequence of small applications of familiar rules. Each term can be handled independently, and the sum rule assembles those results. The method works for polynomials of any finite degree, whether their terms are ordered or not, and whether their coefficients are integers, fractions, or zero.

As a practical routine, identify the terms first, differentiate each one, and then combine the results. Keep the constant term visible long enough to record its derivative as zero. This avoids the common error of carrying it into the answer unchanged and provides a straightforward way to check the final expression.

Check Your Understanding

For each question, use the power, constant multiple, and sum rules. Keep constant terms distinct from coefficients of variable terms.

  1. Differentiate \(p(x)=4x^3-2x+11\), showing the derivative of each term.
  2. What is the derivative of the constant term \(-6\), and why?
  3. Differentiate \(q(x)=\frac{1}{3}x^4-5x^2+2\).
  4. A student writes \(\frac{d}{dx}(9x+4)=9x+4\). Identify the error and give the correct derivative.
  5. Explain why the sum rule allows the derivative of a polynomial to be found by adding the derivatives of its terms.

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