Why a Product of Three Factors Needs Three Terms
A product of three functions changes when any of its factors changes. Its derivative must therefore account for the change in the first factor, the change in the second factor, and the change in the third factor. The two-factor product rule gives a precise way to combine these contributions.
The goal here is to derive the formula rather than simply memorize it. The derivation uses the two-factor product rule twice: first for the product of the first two functions, and then for that product multiplied by the third function.
The Two-Factor Rule and the Required Conditions
We use the two-factor product rule: if \(u\) and \(v\) are differentiable on an interval \(I\), then their product \(uv\) is differentiable on \(I\), and
Here, the prime denotes differentiation with respect to \(x\). The rule applies to functions, not only to particular algebraic expressions, so we can use it with a product such as \(fg\) as one of the factors.
It is important that the functions be differentiable on the interval under consideration. The formula does not assert that a derivative exists at a point where one of the factors is not differentiable.
Deriving the Three-Factor Formula
Group the first two factors together. This lets us view the full product \(fgh\) as the product \((fg)h\). Apply the two-factor rule to those two factors:
The expression still contains \((fg)'\). Since \(f\) and \(g\) are differentiable, apply the two-factor rule to their product:
Substitute this expression into the derivative of \((fg)h\), then distribute the factor \(h\):
Since \((fg)h=fgh\), this establishes the three-factor product rule.
This is a complete derivation from the two-factor product rule. The first application differentiates the grouped product \(fg\) or the factor \(h\); the second application expands the derivative of \(fg\). Together, the applications produce the three contributions.
Why Each Factor Is Differentiated Once
Read the terms from left to right. In \(f'gh\), the derivative is applied to \(f\), while \(g\) and \(h\) remain unchanged. In \(fg'h\), it is applied to \(g\). In \(fgh'\), it is applied to \(h\). Thus every term represents one factor changing while the other two are held in their original, undifferentiated forms.
This pattern is not a matter of writing down every possible combination of primes. There are exactly three terms because there are three factors, and each term contains exactly one differentiated factor. For example, \(f'g'h\) would differentiate two factors at once; it is not one of the contributions given by the product rule.
Write \(f'gh\), keeping \(g\) and \(h\) unchanged.
Write \(fg'h\), keeping \(f\) and \(h\) unchanged.
Write \(fgh'\), keeping \(f\) and \(g\) unchanged.
The complete derivative is \(f'gh+fg'h+fgh'\).
Worked Examples
Worked Example: Differentiating Three Polynomial Factors
Find the derivative of \(y=(x^2+1)(2x-1)(x+3)\).
Set \(f=x^2+1\), \(g=2x-1\), and \(h=x+3\). Their derivatives are \(f'=2x\), \(g'=2\), and \(h'=1\). Apply the three-factor formula:
Expand each term. First, \((2x-1)(x+3)=2x^2+5x-3\), so the first term is \(4x^3+10x^2-6x\). For the second term, \((x^2+1)(x+3)=x^3+3x^2+x+3\), giving \(2x^3+6x^2+2x+6\). The third term is \(2x^3-x^2+2x-1\). Adding like terms gives
Check by expanding the original product. First, \((2x-1)(x+3)=2x^2+5x-3\). Therefore,
and direct differentiation gives \(y'=8x^3+15x^2-2x+5\), agreeing with the product-rule result.
Worked Example: Evaluating the Derivative at a Point
Let \(y=(x^2-2x+3)(x+1)(3x-1)\). Find \(y'(1)\) using the three-factor rule.
Write \(f=x^2-2x+3\), \(g=x+1\), and \(h=3x-1\). At \(x=1\), the factor values and derivative values are
Substitute these values into \(y'=f'gh+fg'h+fgh'\):
To check, multiply the factors: \((x^2-2x+3)(x+1)=x^3-x^2+x+3\), and then
Differentiate this polynomial: \(y'=12x^3-12x^2+8x+8\). At \(x=1\), \(y'(1)=12-12+8+8=16\), as before.
Worked Example: Repeated Factors
Find the derivative of \(y=x^2\cdot x^2\cdot x^2\). Treat the three occurrences as the factors \(f=x^2\), \(g=x^2\), and \(h=x^2\). Each has derivative \(2x\). The rule gives
Since the original product is \(x^2x^2x^2=x^6\), direct differentiation gives \(y'=6x^5\). The check confirms the result. The repeated factors are still treated as three factor positions: the rule contributes one term for differentiating each position.
A Common Mistake: Differentiating All Factors at Once
A tempting but incorrect calculation is to differentiate each factor and multiply the results. For \(y=(x^2+1)(2x-1)(x+3)\), that mistake would give
This expression differentiates all three factors simultaneously and omits the cases in which only one factor changes. For example, at \(x=0\), the incorrect expression gives \(0\). But the correct formula gives
The correction is to make three separate terms, differentiating a different factor in each one, and then add them. The expansion check in the first example also gives \(y'(0)=5\), since \(8(0)^3+15(0)^2-2(0)+5=5\).
Organizing and Checking the Formula
When applying the rule, it helps to label the factors before differentiating. This makes it easier to see whether a factor has been omitted or differentiated in the wrong terms. In the expression \(fgh\), the three intended terms are \(f'gh\), \(fg'h\), and \(fgh'\).
After writing the three terms, check each one separately:
- Does the first term differentiate \(f\) and leave \(g\) and \(h\) unchanged?
- Does the second term differentiate \(g\) and leave \(f\) and \(h\) unchanged?
- Does the third term differentiate \(h\) and leave \(f\) and \(g\) unchanged?
- Are all three terms added?
If the factors are polynomials, expanding the original product and differentiating it directly can provide an independent check, as in the examples. That check is useful for detecting algebra errors, but it does not replace the derivation of the rule. For more complicated functions, the product-rule expression may be clearer than a fully expanded result.
Check Your Understanding
Use the three-factor product rule to answer the following questions. In each derivative, identify which factor is differentiated in each term.
- Write the derivative formula for \(y=fgh\), assuming all three functions are differentiable on an interval.
- For \(y=(x+2)(x^2+1)(2x-3)\), write the three product-rule terms without expanding or simplifying them.
- Suppose \(f(1)=2\), \(g(1)=3\), \(h(1)=4\), \(f'(1)=1\), \(g'(1)=0\), and \(h'(1)=-2\). Find \((fgh)'(1)\).
- What is wrong with the calculation \((fgh)'=f'g'h'\)? State the corrected formula.
- Apply the formula to \(y=x^2\cdot x\cdot(x+1)\), then check your result by expanding the product first.
