Learning Objectives
Differentiate products of two changing functions.
Differentiate quotients while preserving subtraction order and domain restrictions.
Choose simplification or a derivative rule based on the expression's structure.
Prerequisites: basic derivatives, algebraic expansion, and factoring. Let u and v denote differentiable functions of x. Primes denote derivatives with respect to x.
1. Both Factors Can Contribute to Change
The product rule is (uv)′ = u′v + uv′. Differentiate one factor at a time while leaving the other unchanged, then add the two contributions. The derivative of a product is generally not u′v′.
Worked Example: A Polynomial Times an Exponential
Differentiate f(x) = x²eˣ. Choose u = x² and v = eˣ, so u′ = 2x and v′ = eˣ.
The product rule gives f′(x) = 2xeˣ + x²eˣ = eˣ(2x + x²).
At x = 1, f′(1) = 3e. Multiplying the two derivatives alone would give 2e and miss one contribution.
2. Expansion Can Be a Useful Alternative
Worked Example: Check with a Second Method
Let g(x) = x²(x + 3). The product rule gives g′(x) = 2x(x + 3) + x² = 3x² + 6x.
Alternatively, expand g(x) = x³ + 3x² and differentiate term by term to obtain the same result.
For a short polynomial product, expansion may be convenient. For x²eˣ, expansion does not remove the need for the product rule.
3. The Quotient Rule Includes a Subtraction
For v(x) ≠ 0, the quotient rule is (u/v)′ = (u′v − uv′)/v². The original denominator is squared in the new denominator. In the numerator, the first term differentiates the original numerator, and the second term differentiates the original denominator.
Worked Example: A Rational Function
Differentiate h(x) = (x² + 1)/(x − 2), where x ≠ 2. Set u = x² + 1 and v = x − 2.
Then h′(x) = [2x(x − 2) − (x² + 1)(1)]/(x − 2)².
Simplify the numerator: 2x² − 4x − x² − 1 = x² − 4x − 1. Thus h′(x) = (x² − 4x − 1)/(x − 2)², still with x ≠ 2.
4. Simplify First When It Reduces the Work
For f(x) = (x² + 3x)/x with x ≠ 0, simplification gives f(x) = x + 3 on its original domain. Therefore f′(x) = 1 for x ≠ 0. The original function remains undefined at zero, so no derivative at zero has been established.
Example: Recover a Known Trigonometric Rule
Since tan x = sin x/cos x where cos x ≠ 0, the quotient rule gives (tan x)′ = [cos²x + sin²x]/cos²x.
Use cos²x + sin²x = 1 to obtain 1/cos²x = sec²x. The denominator restriction matches the domain of tan x.
Common Mistakes
Keep the original factor unchanged in each product-rule term.
Use parentheses around the full numerator of a quotient-rule result.
Reversing the subtraction in the quotient rule negates the answer.
Simplification does not restore excluded inputs.
5. Practice
1. Differentiate f(x) = x sin x.
2. Differentiate g(x) = (3x + 1)/(x + 2).
3. Differentiate h(x) = x³ ln x for x > 0.
4. Simplify before differentiating q(x) = (x³ − x)/x, stating its domain restriction.
Worked Solutions
1. f′(x) = sin x + x cos x by the product rule.
2. g′(x) = [3(x + 2) − (3x + 1)]/(x + 2)² = 5/(x + 2)², with x ≠ −2.
3. h′(x) = 3x² ln x + x³(1/x) = 3x² ln x + x², with x > 0.
4. For x ≠ 0, q(x) = x² − 1, so q′(x) = 2x on that domain. The original q has no value or derivative at zero.
6. Summary
- Product derivatives add two contributions.
- Quotient derivatives require the correct subtraction order and a squared denominator.
- Simplify when it helps, while preserving the original domain.
- Use expansion or a known identity as an independent check when possible.