Introduction
A polynomial derivative can be computed term by term, but it is easy to lose a sign, multiply by the wrong exponent, or change a power incorrectly. A small amount of organized bookkeeping makes these errors easier to prevent.
The method in this tutorial is to record two pieces of information for each nonconstant term: its coefficient and its exponent. The power rule then gives a clear update: multiply the coefficient by the original exponent, and reduce the exponent by one. Constant terms must be treated separately.
The Coefficient-and-Exponent Record
A nonconstant monomial has the form \(cx^n\), where \(c\) is its coefficient and \(n\) is a positive integer. The power rule says that its derivative is \(cnx^{n-1}\). Thus the original coefficient \(c\) and exponent \(n\) determine both parts of the differentiated term.
The exponent used to multiply the coefficient is the original exponent, before it decreases. The sign belongs to the coefficient: for example, in \(-4x^3\), the coefficient is \(-4\), not \(4\).
For a polynomial, the sum and difference rules allow us to differentiate each term separately. A constant is different: its derivative is zero by the constant rule. Do not apply the positive-integer power rule to a constant by lowering an exponent of zero.
Why the Record Gives the Right Result
Here is a justification for applying the record to an entire polynomial. Write a polynomial as a finite sum of nonconstant terms and, when present, a constant:
By the sum rule, the derivative of this finite sum is the sum of the derivatives of its terms. For each nonconstant term, the constant multiple rule and the positive-integer power rule give \(\frac{d}{dx}(c_ix^{n_i})=c_in_ix^{n_i-1}\). The constant rule gives \(\frac{d}{dx}(C)=0\). Therefore, combining those results gives
This proves the bookkeeping rule for every term in the polynomial: multiply each nonconstant term’s coefficient by its original exponent, lower that exponent by one, and differentiate any constant separately. The exponents \(n_i\) must be positive integers for the stated monomial rule here; the constant is not included among these terms.
A Reliable Table Layout
For each nonconstant term, make a row with its original coefficient and exponent, followed by the coefficient and exponent after differentiation. Keep the constant outside this calculation. The table is not a new differentiation rule; it is a way to make each use of the rule visible.
| Original term | Coefficient | Original exponent | New coefficient | New exponent | Differentiated term |
|---|---|---|---|---|---|
| \(cx^n\) | \(c\) | \(n\) | \(cn\) | \(n-1\) | \(cnx^{n-1}\) |
After completing the rows, write the resulting terms as a polynomial. Terms may be arranged by descending powers for readability. A power that did not occur in the original polynomial does not need to be inserted; the derivative is formed from the terms that are actually present.
Worked Examples
Worked Example: Tracking Positive and Negative Coefficients
Differentiate \(f(x)=6x^5-4x^3+7x-9\), keeping a separate record for each nonconstant term.
For \(6x^5\), the coefficient is \(6\) and the exponent is \(5\). Multiply \(6\) by \(5\), giving \(30\), and reduce the exponent from \(5\) to \(4\). This term contributes \(30x^4\).
For \(-4x^3\), the coefficient is \(-4\) and the exponent is \(3\). Multiply \(-4\) by \(3\), giving \(-12\), and reduce the exponent from \(3\) to \(2\). This term contributes \(-12x^2\).
For \(7x\), write the term as \(7x^1\). Its coefficient is \(7\) and its exponent is \(1\). Multiplying gives \(7\cdot1=7\), and the new exponent is \(1-1=0\). Since \(x^0=1\), this contribution is \(7x^0=7\).
The remaining term, \(-9\), is a constant, so its derivative is \(0\). It is handled by the constant rule, not by lowering an exponent.
| Original term | Coefficient calculation | Exponent calculation | Contribution |
|---|---|---|---|
| \(6x^5\) | \(6\cdot5=30\) | \(5-1=4\) | \(30x^4\) |
| \(-4x^3\) | \((-4)\cdot3=-12\) | \(3-1=2\) | \(-12x^2\) |
| \(7x\) | \(7\cdot1=7\) | \(1-1=0\) | \(7x^0=7\) |
| \(-9\) | Constant rule | Not lowered | \(0\) |
Putting the contributions together gives \(f'(x)=30x^4-12x^2+7\). Check the record: the coefficient \(6\) became \(30\) using exponent \(5\), the coefficient \(-4\) became \(-12\) using exponent \(3\), and \(7x\) contributed \(7\) because its original exponent was \(1\). The constant contributes nothing. Each term in the result matches its corresponding row.
Worked Example: Keeping Fractional Coefficients Exact
Differentiate \(g(x)=\frac{3}{2}x^4-\frac{5}{3}x^2+8\). Keep the fractional coefficients exact while recording the changes.
For \(\frac{3}{2}x^4\), multiply the coefficient \(\frac{3}{2}\) by the original exponent \(4\): \(\frac{3}{2}\cdot4=\frac{12}{2}=6\). The new exponent is \(4-1=3\), so the contribution is \(6x^3\).
For \(-\frac{5}{3}x^2\), multiply \(-\frac{5}{3}\) by \(2\): \(-\frac{5}{3}\cdot2=-\frac{10}{3}\). The new exponent is \(2-1=1\), so the contribution is \(-\frac{10}{3}x\).
The term \(8\) is constant and differentiates to \(0\). Therefore, \(g'(x)=6x^3-\frac{10}{3}x\).
Check the two nonconstant rows: the original exponents \(4\) and \(2\) were used in the coefficient multiplications, and the resulting exponents are \(3\) and \(1\). The fractional coefficient was multiplied before any change to its sign or value, so both differentiated terms agree with the recorded calculations.
Worked Example: Handling Terms Written Out of Order
Differentiate \(h(x)=11-2x^5+3x^2-\frac{1}{2}x\). The terms are not written in descending powers, so identify each coefficient and exponent before arranging the derivative.
The term \(-2x^5\) has coefficient \(-2\) and exponent \(5\). The new coefficient is \((-2)\cdot5=-10\), and the new exponent is \(5-1=4\). Its contribution is \(-10x^4\).
The term \(3x^2\) has coefficient \(3\) and exponent \(2\). The new coefficient is \(3\cdot2=6\), and the new exponent is \(2-1=1\). Its contribution is \(6x\).
The term \(-\frac{1}{2}x\) has coefficient \(-\frac{1}{2}\) and exponent \(1\). The new coefficient is \(-\frac{1}{2}\cdot1=-\frac{1}{2}\), and the new exponent is \(1-1=0\). Its contribution is \(-\frac{1}{2}x^0=-\frac{1}{2}\).
The constant \(11\) differentiates to \(0\). There are no other terms to differentiate. Thus, arranging the contributions by descending powers gives \(h'(x)=-10x^4+6x-\frac{1}{2}\).
Check each contribution against the original: the exponent \(5\) produces coefficient \(-10\) and power \(4\); exponent \(2\) produces coefficient \(6\) and power \(1\); and exponent \(1\) produces coefficient \(-\frac{1}{2}\) and power \(0\). The constant contributes zero. The order in which the original terms were written does not affect these checks.
Common Errors and Their Corrections
One common error is to use the reduced exponent as the multiplier. For example, the incorrect calculation \(\frac{d}{dx}(5x^4)=5\cdot3x^3=15x^3\) uses \(3\), the new exponent, to multiply the coefficient. The correct multiplier is the original exponent \(4\):
A second error is to lower the power in the coefficient calculation as well as in the exponent, or to forget that a negative sign is part of the coefficient. For \(-3x^2\), the coefficient calculation is \((-3)\cdot2=-6\), and the new power is \(2-1=1\). The correct derivative is \(-6x\), not \(6x\) and not \(-3x\).
A particularly important mistake is to treat a constant \(-9\) as though it were a nonconstant monomial with exponent zero, then write \(0x^{-1}\). The expression \(x^{-1}\) is undefined at \(x=0\), so that is not a valid way to state the derivative of the constant on the real line. The correction is direct: \(\frac{d}{dx}(-9)=0\) by the constant rule. Do not lower a constant’s exponent to \(-1\).
Why This Bookkeeping Matters
In a short expression, it may seem easy to differentiate by inspection. As a polynomial contains more terms, however, the same two changes must be made repeatedly and independently. A record makes it possible to check each term rather than relying on a general impression that the final expression looks plausible.
It also makes the structure of the rule visible. The coefficient does not simply remain unchanged: it is multiplied by the original exponent. The power does not determine that multiplier after it has been lowered; it is the original power that supplies the multiplier. Keeping those roles distinct prevents many routine errors.
Finally, a table helps distinguish nonconstant terms from constants. This matters because the positive-integer power rule applies to terms with positive integer exponents, while constants are differentiated using the constant rule. Keeping that boundary clear avoids creating an undefined negative power when the original function is defined for all real inputs.
Summary
To organize the derivative of a polynomial, process each nonconstant term \(cx^n\) separately. Record \(c\) and \(n\), calculate the new coefficient \(cn\), and calculate the new exponent \(n-1\). Then combine the resulting terms, preserving signs and exact coefficients. Differentiate every constant separately as zero.
Check Your Understanding
For each question, use the coefficient-and-exponent record and remember to handle constants separately.
- For the term \(-7x^4\), what are the original coefficient and exponent, and what term results after differentiation?
- Differentiate \(3x^3-5x+4\), showing the coefficient and exponent update for each nonconstant term.
- In differentiating \(\frac{2}{5}x^2\), what number multiplies the coefficient: the original exponent or the new exponent?
- Why is it not appropriate to differentiate a constant by writing \(0x^{-1}\)?
- A student differentiates \(8x^3\) as \(16x^2\). Identify the bookkeeping error and give the correct derivative.
