Product Rule for Three Functions
A product of three variable factors produces three derivative terms.
Differentiate one factor in each term while preserving the others.
Factor common pieces only after all three terms are present.
For two functions, the product rule is \((uv)'=u'v+uv'\). For three functions, apply that same rule to \((uv)w\), then expand. The result is
Extended product rule
\[(uvw)'=u'vw+uv'w+uvw'.\]
Each term differentiates exactly one factor. The other two factors remain unchanged. This pattern generalizes: a product of four factors produces four terms, with a different factor differentiated in each term.
Example 1: polynomial, exponential, and sine
Let \(f(x)=x^2e^x\sin x\). Use \(u=x^2\), \(v=e^x\), and \(w=\sin x\). Their derivatives are \(u'=2x\), \(v'=e^x\), and \(w'=\cos x\). Substitute into the three-factor rule:
\[f'(x)=2xe^x\sin x+x^2e^x\sin x+x^2e^x\cos x.\]
Every term contains \(xe^x\), so an optional factored form is \(xe^x[2\sin x+x\sin x+x\cos x]\). Both forms are correct.
Example 2: simplify before using the rule
Suppose \(g(x)=x\cdot x^2\cdot x^3\). The extended rule works, but multiplying first gives \(g(x)=x^6\), so \(g'(x)=6x^5\). If you use the three-factor rule, you obtain \(x^2x^3+2x^2x^3+3x^3x^2=6x^5\). The comparison shows why simplifying first can save time when factors combine cleanly.
Example 3: one factor is constant
For \(h(x)=4x^2\cos x\), treat 4 as a constant multiplier rather than a third variable function. Pull it outside and apply the two-factor product rule: \(h'(x)=4[2x\cos x-x^2\sin x]\). A constant factor contributes no extra product-rule term because its derivative is zero.
Why there are three terms
A small change in the product can come from a change in the first factor, the second factor, or the third factor. To first order, those three contributions add. Terms involving simultaneous changes in two or three factors are too small to remain after division by the input change and taking the limit. This is the same reasoning behind the ordinary two-factor product rule.
Common mistakes
- Writing only two terms: every variable factor needs one turn being differentiated.
- Differentiating all factors in one term: \(u'v'w'\) is not the product rule.
- Changing untouched factors: when differentiating \(u\), copy \(v\) and \(w\) exactly.
- Applying the rule to addition: the extended formula is for multiplication; sums are differentiated term by term.
- Expanding too early: write the complete rule first, then factor or combine terms.
Practice problems with solutions
- Differentiate \(xe^x\cos x\).
Show answer
\(e^x\cos x+xe^x\cos x-xe^x\sin x\).
- Differentiate \(x^2\sin x\ln x\), for \(x>0\).
Show answer
\(2x\sin x\ln x+x^2\cos x\ln x+x\sin x\).
- Differentiate \((x+1)(x-1)e^x\).
Show answer
\((x-1)e^x+(x+1)e^x+(x+1)(x-1)e^x\), which simplifies to \(e^x(x^2+2x-1)\).
For the two-factor foundation, study product and quotient rules and the complete x·e^x walkthrough.
Reference standard: Rule statements and notation follow OpenStax Calculus Volume 1, Chapter 3. DerivCalc's explanations and examples are independently written.