Proof of the Product Rule
When two functions are multiplied together, the derivative of their product is not simply the product of their derivatives. The product rule captures how changes in each factor contribute independently — and why there are exactly two terms, not one.
Statement
If f and g are both differentiable at x, then their product fg is also differentiable at x and:
Proof from the Limit Definition
Write h(x) = f(x)·g(x). The derivative of h at x is:
To split this into manageable pieces, introduce and subtract the term f(x)·g(x+ε) inside the numerator (adding zero in disguise):
Group the first two terms and the last two terms:
As ε → 0, the term g(x+ε) → g(x) (because g is continuous at x, which follows from differentiability). The two limits then evaluate to:
Intuition: Why Two Terms?
Think of the product f(x)·g(x) as the area of a rectangle with side lengths f(x) and g(x). When x changes by a small amount ε, both sides change: f changes by approximately f′ε and g changes by approximately g′ε.
The new area is (f + f′ε)·(g + g′ε) = fg + f·g′ε + f′ε·g + f′g′ε². Subtracting the original area fg and dividing by ε gives f·g′ + f′·g + f′g′ε. As ε → 0, the last term vanishes, leaving exactly the two terms of the product rule.
The key point is that the cross term f′g′ε² is of second order in ε — it is negligibly small compared to the first-order terms and disappears in the limit.
Worked Example
Differentiate h(x) = x³ · sin(x).
Set f(x) = x³ and g(x) = sin(x). Then:
- f′(x) = 3x² (power rule)
- g′(x) = cos(x) (standard result)
Applying the product rule:
Verification: for x = π/2, h(x) = (π/2)³·1 ≈ 3.876. h′(π/2) = 3(π/2)²·1 + (π/2)³·0 = 3(π/2)² ≈ 7.402. A numerical difference quotient at x = π/2 with ε = 10⁻⁶ confirms this result.
Frequently Asked Questions
What does the product rule state?
If f and g are differentiable functions, then (fg)′(x) = f′(x)·g(x) + f(x)·g′(x). The derivative of a product is not simply the product of the derivatives.
Why does the cross term df·dg vanish in the proof?
In the limit definition, the cross term df·dg is of order h² (since both df and dg are proportional to h). When divided by h and the limit h→0 is taken, this term goes to zero, leaving only the two first-order terms.
Can the product rule be extended to three functions?
Yes. For three functions f, g, h: (fgh)′ = f′gh + fg′h + fgh′. This follows from applying the two-function product rule twice.
How is the product rule used to derive the quotient rule?
Write f/g as f·(1/g). The product rule and chain rule together give (f·(1/g))′ = f′/g + f·(−g′/g²) = (f′g − fg′)/g². This is exactly the quotient rule.
Adapted and rewritten from Silas Maths source notes; formulas reviewed for CalxSolver publication.