Math101 · Independent Ontario learning library
Math101learn.math101.caProduct Rule
The product rule differentiates two changing factors by adding the change from each factor while the other is held in place.
The derivative of a product is not the product of the derivatives; each factor contributes to the total change.
The rule
If $u(x)$ and $v(x)$ are differentiable, then
A common memory phrase is “first derivative times second, plus first times second derivative.” Clear function labels are safer than relying on wording alone.
Why two terms appear
When $x$ changes, both factors can change. The product's total change includes the effect of changing $u$ while $v$ is present and changing $v$ while $u$ is present.
The tiny overlap of both changes becomes negligible in the derivative limit, leaving the two main contributions.
Worked example
Expanding the original first and differentiating term by term gives the same result.
When expansion is easier
For polynomial factors, expanding may be efficient. However, products involving exponentials, roots, or complicated functions often become harder or impossible to expand meaningfully.
Choose the form that minimizes algebra while retaining domain information.
More than two factors
For three differentiable factors,
Each term differentiates one factor while leaving the others unchanged. The same pattern extends to more factors, though logarithmic differentiation can be more efficient in advanced work.
Product rule with the chain rule
Different factors may require other rules internally. For
the product rule handles the two factors and the chain rule differentiates $(3x+1)^5$:
Rule selection follows layers of structure.
Evaluating without fully simplifying
To find a slope at $x=a$, it may be fastest to substitute into
without expanding the derivative. A table can supply these four values even when formulas are unavailable.
For instance, if $u(2)=3$, $u'(2)=5$, $v(2)=-1$, and $v'(2)=4$, then $(uv)'(2)=5(-1)+3(4)=7$.
Geometric application
If a rectangle's sides $L(t)$ and $W(t)$ change with time, its area is $A(t)=L(t)W(t)$. Therefore
The two terms measure area change caused by each changing dimension. Units are area per unit time.
Factoring the derivative
After applying the rule, factor common terms when solving $f'(x)=0$ or analyzing signs. In the earlier chain-rule example,
reveals critical-number candidates more clearly than an expansion.
Common mistakes
Writing $(uv)'=u'v'$. The two contribution terms are required.
Differentiating only one factor. Both change with $x$.
Forgetting parentheses around a full factor. Preserve its structure.
Missing an inner chain-rule factor. Product rule handles the outside product only.
Expanding when it creates avoidable errors. Simplify strategically, not automatically.
Quick self-check
- Are there two nonconstant factors?
- Have I labelled $u,v,u',v'$ correctly?
- Does the result contain $u'v+uv'$?
- Do any factors need chain or other rules internally?
- Can the derivative be factored for later analysis?
- Do units or a second method confirm the result?
Related topics
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
Differentiate (x² + 1)(3x − 4).
- f′ = 2x(3x − 4) + 3(x² + 1).
- Expand and combine: 6x² − 8x + 3x² + 3.
- f′ = 9x² − 8x + 3.
End of lesson
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