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Product, Quotient, and Chain Rules · Tutorial 452 of 1495

Applying the Product Rule to Polynomial Factors

Identify each polynomial and its derivative, then combine the two terms of the product rule to differentiate their product.

Intermediate 9 min read

What You'll Learn

  • Identify the two factors in a product of polynomials.
  • Differentiate each factor using the power rule.
  • Apply the product rule in the correct order.
  • Expand and combine the resulting terms accurately.
  • Check a derivative by first expanding the original product.
  • Recognize and correct the error of multiplying the factor derivatives.

Why a Product Needs Its Own Rule

When a function is a product of two expressions that depend on \(x\), differentiating each expression separately and multiplying the results does not generally give the derivative of the product. Both factors change as \(x\) changes, so both contribute to the rate of change.

For polynomial factors, the calculation has a reliable structure: identify the factors, find the derivative of each factor, and combine them using the product rule. The power rule gives the derivatives of the individual polynomial terms; the product rule then handles how those factors work together.

Key idea: To differentiate a product, keep the original factors visible. The product rule uses each factor once undifferentiated and the other factor differentiated.

The Product Rule for Polynomial Factors

Suppose \(P\) and \(Q\) are polynomial functions. Since polynomials are differentiable for every real \(x\), their product is differentiable for every real \(x\), and the product rule gives

$$ \frac{d}{dx}\bigl(P(x)Q(x)\bigr) = P(x)Q'(x)+P'(x)Q(x). $$

The order of the two terms may be reversed, but each term must contain one original factor and the derivative of the other. The derivatives \(P'(x)\) and \(Q'(x)\) are found by differentiating the polynomial terms using the power rule.

For example, if \(P(x)=x^2+3x-2\) and \(Q(x)=2x^3-x+4\), the product is \(P(x)Q(x)\). The two product-rule terms are \(P(x)Q'(x)\) and \(P'(x)Q(x)\). They are not \(P'(x)Q'(x)\), and neither term differentiates both factors at once.

Why the Product Rule Works

The rule follows from the difference quotient. This short derivation also explains why both factors contribute. Let \(f\) and \(g\) be differentiable at \(x\). For nonzero \(h\), add and subtract \(f(x+h)g(x)\) in the numerator:

$$ \begin{aligned} \frac{f(x+h)g(x+h)-f(x)g(x)}{h} &= \frac{f(x+h)g(x+h)-f(x+h)g(x)}{h}\\ &\quad+ \frac{f(x+h)g(x)-f(x)g(x)}{h}\\ &= f(x+h)\frac{g(x+h)-g(x)}{h} + g(x)\frac{f(x+h)-f(x)}{h}. \end{aligned} $$

As \(h\) approaches \(0\), differentiability of \(f\) implies continuity of \(f\), so \(f(x+h)\) approaches \(f(x)\). The difference quotients approach \(g'(x)\) and \(f'(x)\), respectively. Taking the limit therefore gives

$$ \frac{d}{dx}\bigl(f(x)g(x)\bigr) = f(x)g'(x)+g(x)f'(x). $$

This argument applies in particular to polynomial functions. It shows why the rule has two terms: the change in the product can be separated into a change in one factor while the other is held at its current value, followed by a change in the other factor.

A Reliable Calculation Structure

For a product of two polynomial factors, write the factors as \(P(x)\) and \(Q(x)\) before differentiating. Keep the product rule in view as you calculate. This reduces the risk of losing a factor or differentiating the wrong expression.

1
Identify the factors.
Write the original product as \(P(x)Q(x)\), including all terms inside each factor.
2
Differentiate each factor.
Find \(P'(x)\) and \(Q'(x)\) using the power rule, including the derivative of each constant term.
3
Write both product-rule terms.
Form \(P(x)Q'(x)+P'(x)Q(x)\), retaining the original, undifferentiated factors.
4
Expand and simplify if useful.
Distribute carefully, then combine like terms. Keeping the factored product-rule form is also valid.

Worked Examples

Worked Example: Multiply a Quadratic by a Cubic

Differentiate \(y=(x^2+3x-2)(2x^3-x+4)\). Set \(P(x)=x^2+3x-2\) and \(Q(x)=2x^3-x+4\). Applying the power rule to each term gives

$$ P'(x)=2x+3, \qquad Q'(x)=6x^2-1. $$

Now apply the product rule, placing each original factor beside the derivative of the other:

$$ \begin{aligned} y' &=P(x)Q'(x)+P'(x)Q(x)\\ &=(x^2+3x-2)(6x^2-1)+(2x+3)(2x^3-x+4). \end{aligned} $$

Expand the first product term by term:

$$ \begin{aligned} (x^2+3x-2)(6x^2-1) &=6x^4+18x^3-12x^2-x^2-3x+2\\ &=6x^4+18x^3-13x^2-3x+2. \end{aligned} $$

Expand the second product:

$$ \begin{aligned} (2x+3)(2x^3-x+4) &=4x^4+6x^3-2x^2-3x+8x+12\\ &=4x^4+6x^3-2x^2+5x+12. \end{aligned} $$

Adding the two expressions and combining like terms gives

$$ \begin{aligned} y' &=(6x^4+18x^3-13x^2-3x+2) +(4x^4+6x^3-2x^2+5x+12)\\ &=10x^4+24x^3-15x^2+2x+14. \end{aligned} $$

Check the result by expanding the original product first:

$$ \begin{aligned} y &=2x^5+6x^4-4x^3-x^3-3x^2+2x+4x^2+12x-8\\ &=2x^5+6x^4-5x^3+x^2+14x-8,\\ \frac{dy}{dx} &=10x^4+24x^3-15x^2+2x+14. \end{aligned} $$

This matches the product-rule result.

Worked Example: Track Negative Terms Carefully

Differentiate \(y=(3x^2-2)(x^3+2x^2-x+1)\). Let \(P(x)=3x^2-2\) and \(Q(x)=x^3+2x^2-x+1\). Their derivatives are

$$ P'(x)=6x, \qquad Q'(x)=3x^2+4x-1. $$

Using the product rule gives

$$ \begin{aligned} y' &=P(x)Q'(x)+P'(x)Q(x)\\ &=(3x^2-2)(3x^2+4x-1) +6x(x^3+2x^2-x+1). \end{aligned} $$

First expand the term containing \(P(x)\):

$$ \begin{aligned} (3x^2-2)(3x^2+4x-1) &=9x^4+12x^3-3x^2-6x^2-8x+2\\ &=9x^4+12x^3-9x^2-8x+2. \end{aligned} $$

Then expand the term containing \(P'(x)\):

$$ 6x(x^3+2x^2-x+1)=6x^4+12x^3-6x^2+6x. $$

Adding the terms gives

$$ \begin{aligned} y' &=(9x^4+12x^3-9x^2-8x+2) +(6x^4+12x^3-6x^2+6x)\\ &=15x^4+24x^3-15x^2-2x+2. \end{aligned} $$

To check, expand the original product:

$$ \begin{aligned} y &=3x^5+6x^4-3x^3+3x^2-2x^3-4x^2+2x-2\\ &=3x^5+6x^4-5x^3-x^2+2x-2,\\ \frac{dy}{dx} &=15x^4+24x^3-15x^2-2x+2. \end{aligned} $$

The independently differentiated expansion agrees, including the negative signs.

Worked Example: Evaluate the Derivative at a Point

Let \(y=(x^2-4x+5)(2x^2+x-3)\). Find \(y'(1)\). Define \(P(x)=x^2-4x+5\) and \(Q(x)=2x^2+x-3\). Differentiate each factor:

$$ P'(x)=2x-4, \qquad Q'(x)=4x+1. $$

The product rule gives

$$ y'=(x^2-4x+5)(4x+1)+(2x-4)(2x^2+x-3). $$

At \(x=1\), evaluate all four factor values:

$$ \begin{aligned} P(1)&=1-4+5=2, & Q(1)&=2+1-3=0,\\ P'(1)&=2-4=-2, & Q'(1)&=4+1=5. \end{aligned} $$

Substitute these values into the product rule:

$$ y'(1)=P(1)Q'(1)+P'(1)Q(1) =2(5)+(-2)(0)=10. $$

For a check, expand the original product:

$$ \begin{aligned} y&=2x^4+x^3-3x^2-8x^3-4x^2+12x+10x^2+5x-15\\ &=2x^4-7x^3+3x^2+17x-15,\\ y'&=8x^3-21x^2+6x+17,\\ y'(1)&=8-21+6+17=10. \end{aligned} $$

The value agrees. In particular, \(Q(1)=0\) makes the term \(P'(1)Q(1)\) vanish, but the other term \(P(1)Q'(1)\) remains.

A Common Error: Multiplying the Derivatives

A frequent mistake is to differentiate both factors and then multiply those derivatives. For instance, for \(y=x^2x^3\), this incorrect calculation gives

$$ y' \ne (2x)(3x^2)=6x^3. $$

The product rule instead gives one term for each factor’s contribution:

$$ \begin{aligned} y' &=x^2(3x^2)+(2x)x^3\\ &=3x^4+2x^4\\ &=5x^4. \end{aligned} $$

Check by simplifying the original product: \(y=x^5\), so the power rule gives \(y'=5x^4\). The incorrect answer \(6x^3\) does not match. The key correction is structural: in \(P Q'+P'Q\), each term keeps one original factor unchanged.

Choosing a Useful Form and Checking Your Work

The derivative can be left in product-rule form or expanded. Both are correct when they are algebraically equivalent. The factored form often makes the origin of each term easy to see; an expanded polynomial can be convenient when combining like powers or evaluating at a specific input. Use the form that suits the task, and preserve parentheses until each multiplication is complete.

Expanding the original product before differentiating provides a useful independent check, as in the examples. It may involve more algebra, so it is not always the most efficient route, but it can expose a missing term or sign error. When checking, differentiate the complete expanded polynomial using the power rule and compare the result with the product-rule expression after simplification.

  • Confirm that both original factors appear in the product-rule expression.
  • Confirm that each factor has been differentiated correctly.
  • Check the signs when distributing negative terms.
  • Combine only terms with the same power of \(x\).
  • If practical, expand the original product and differentiate it as a check.

Check Your Understanding

For each question, identify the two factors before applying the product rule. Leave answers factored when the question does not request expansion.

  1. Differentiate \(y=(x^2+1)(x^3-2)\) using the product rule. What are the two terms before simplifying?
  2. For \(y=(2x^2-x)(x^2+3)\), find each factor’s derivative and write \(y'\) in product-rule form.
  3. Explain why \(P'(x)Q'(x)\) is not the product rule for \(P(x)Q(x)\).
  4. Differentiate \(y=(x^2-1)(x^3+x)\), then check your result by expanding the original product.
  5. If \(P(2)=3\), \(P'(2)=4\), \(Q(2)=0\), and \(Q'(2)=5\), find the derivative of \(P(x)Q(x)\) at \(x=2\).

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