Why Factored Form Changes the Work
A polynomial may be presented as a product of factors rather than as a sum of powers. For example, \(f(x)=(x-2)(x+3)\) is already factored, while \(f(x)=x^2+x-6\) is its expanded form. Both expressions define the same function, but their forms suggest different ways to differentiate.
Expanding first lets you use the power rule term by term. Applying the product rule instead can avoid expansion and preserve information about the factors, such as where the original function is zero. Neither approach is automatically best in every problem. A useful habit is to choose a method, then compare the result with the other form when that comparison is manageable.
Choosing Between Expansion and the Product Rule
For two differentiable functions \(u\) and \(v\), the product rule is \((uv)'=u'v+uv'\). In this tutorial, the factors are polynomials, so each factor is differentiable for every real \(x\). You can expand a factored polynomial and differentiate its terms, or use the product rule directly. With several factors, apply the product rule repeatedly.
Expansion can be convenient when it is short and the resulting polynomial is easy to differentiate. The product rule can be convenient when expansion would create many terms or when the factors themselves are useful to keep visible. After differentiating, equivalent answers need not look alike: one may be factored and another expanded.
For three factors, this says to differentiate each factor in turn:
This formula follows from the product rule, not from a separate rule that multiplies all the derivatives. The proof also explains why the method extends to any finite number of factors.
Why the Repeated Product Rule Works
For \(P(x)=p_1(x)\), the derivative is \(P'(x)=p_1'(x)\), which is the stated sum with one term.
Let \(P_n(x)=p_1(x)\cdots p_n(x)\), and suppose its derivative is the sum formed by differentiating each of its \(n\) factors in turn.
Write \(P_{n+1}(x)=P_n(x)p_{n+1}(x)\). The product rule gives \(P_{n+1}'=P_n'p_{n+1}+P_np_{n+1}'\).
Substituting the assumed formula for \(P_n'\) into \(P_n'p_{n+1}\) produces the \(n\) terms in which one of the first \(n\) factors is differentiated. The term \(P_np_{n+1}'\) is the additional term in which the last factor is differentiated. Thus the formula holds for \(n+1\) factors.
The base case and the step from \(n\) factors to \(n+1\) factors establish the formula for every positive integer \(n\). This is why repeated use of the two-factor product rule gives exactly one term for each factor differentiated.
Worked Examples
Worked Example: A Product of Two Linear Factors
Differentiate \(f(x)=(x-2)(x+3)\) using the product rule, then check by expansion.
Let \(u(x)=x-2\) and \(v(x)=x+3\). Their derivatives are \(u'(x)=1\) and \(v'(x)=1\). Applying the product rule gives:
Now expand the original function to check:
The two methods agree. As a numerical check at \(x=0\), the product-rule expression gives \(1(3)+(-2)1=1\), and \(2(0)+1=1\).
Worked Example: A Quadratic Factor Times a Linear Factor
Find the derivative of \(g(x)=(2x^2-x+1)(x-2)\) in product-rule form and expanded form.
Differentiate each factor separately: \((2x^2-x+1)'=4x-1\), and \((x-2)'=1\). Therefore:
For an independent check, expand \(g\) first:
At \(x=2\), the product-rule form gives \((8-1)(0)+(8-2+1)=7\). The expanded derivative gives \(6(2^2)-10(2)+3=24-20+3=7\), confirming agreement at that input as well.
Worked Example: Three Factors with a Repeated Factor
Differentiate \(h(x)=(x-1)^2(x+2)\) by treating it as three factors, and compare with expansion. Write the repeated factor explicitly as \((x-1)(x-1)(x+2)\).
In each product-rule term, differentiate one factor and leave the other two unchanged. Each linear factor has derivative \(1\):
Now expand the original function:
The forms agree. Also, \(h'(1)=3(1)^2-3=0\); the product-rule expression gives \((1-1)(3(1)+3)=0\). The repeated factor makes the zero at \(x=1\) visible in the factored derivative.
Equivalent Forms Reveal Different Information
A derivative can be correct in more than one algebraic form. The product-rule result may retain factors, while expansion combines terms into a polynomial in standard form. For instance, \(3(x-1)(x+1)\) and \(3x^2-3\) are equal because \((x-1)(x+1)=x^2-1\).
The factored form can make zeros of the derivative apparent: \(3(x-1)(x+1)=0\) at \(x=1\) and \(x=-1\). The expanded form can make it easier to compare coefficients or substitute a value using a familiar polynomial expression. Choose the form that best serves the next step; do not assume that a different-looking answer is wrong.
Common Mistake: Multiplying the Derivatives
A frequent error is to differentiate each factor and then multiply the results. For example, for \(q(x)=(x^2+1)(x-3)\), an incorrect calculation is:
This omits the changes in each factor while the other factor remains present. The product rule requires two terms:
Check at \(x=0\). The correct derivative gives \(1\), while the incorrect answer \(2x\) gives \(0\). This concrete mismatch exposes the error. In general, whenever differentiating a product, check that there is one term for each factor and that each term leaves all the other factors in place.
A Practical Method
For a factored polynomial, use this sequence of decisions:
- Identify each factor and its derivative, using the power rule and the sum rule.
- Estimate whether expansion is short or likely to create many terms.
- If using the product rule, write one term for each factor differentiated in turn.
- Simplify carefully, keeping a factored result if it is useful.
- When practical, expand the original expression or the derivative to verify that the forms agree.
These steps apply to products of polynomial factors, including repeated factors. Because every polynomial is differentiable for every real input, no additional domain restriction is needed for these examples.
Check Your Understanding
For each question, decide which form or rule is useful and show enough work to justify the result.
- Differentiate \(r(x)=(x+4)(x-1)\) using the product rule, then check by expansion.
- For \(s(x)=(x^2+2)(x+1)\), write the product-rule derivative and simplify it.
- Differentiate \(t(x)=(x-3)^2(x+1)\) by treating it as three factors. How many product-rule terms should appear before simplification?
- Explain why differentiating \(u(x)v(x)\) as \(u'(x)v'(x)\) does not follow from the product rule.
- Give one reason to keep a derivative factored and one reason to expand it.
