The derivative of a product is NOT the product of derivatives. Each factor takes its turn being differentiated while the other stands still.
Differentiate
f(x) = x²(x³ + 1)
Product rule: 2x(x³ + 1) + x²(3x²) = 2x⁴ + 2x + 3x⁴ = 5x⁴ + 2x. C multiplies the two derivatives — the exact move the rule exists to forbid. Expanding first (x⁵ + x² → 5x⁴ + 2x) confirms it.
x · x = x² → derivative 2x
the true answer
1 · 1 = 1 ✗
derivatives multiplied — plainly wrong
One counterexample settles it forever: even the humblest product breaks the naive rule. Both factors are changing at once, and each factor's change matters while the other holds its value.
(uv)′ = u′v + uv′
differentiate one, freeze the other — both ways, added
u = x², v = x³ + 1
u′ = 2x, v′ = 3x²
2x(x³ + 1) + x²(3x²) = 5x⁴ + 2x
collect like terms at the end
x²(x³ + 1) = x⁵ + x²
just algebra
→ 5x⁴ + 2x ✓
the power rule agrees with the product rule
When both factors are polynomials, expanding is a legitimate alternative — and a perfect way to verify your product-rule work on exam day.
(uv)′ = u′v − uv′v²
"low d-high minus high d-low, over low squared"
x²x + 1 → 2x(x + 1) − x²(x + 1)² = x² + 2x(x + 1)²
the minus makes the order non-negotiable
Where is the tangent horizontal? Set f′ = 0. A fraction is zero only when its top is, so for this f: x² + 2x = x(x + 2) = 0, giving x = 0 or x = −2 (the bottom (x + 1)² is not zero at either).
A factor that decreases, at a negative x. Find f′(−1) for f(x) = (x² − 3)(4 − x).
u = x² − 3, u′ = 2x; v = 4 − x, v′ = −1
v′ is negative
f′(x) = 2x(4 − x) + (x² − 3)(−1)
u′v + uv′
f′(−1) = (−2)(5) + (−2)(−1) = −10 + 2 = −8
check by expanding: 8x − 3x² + 3 → −8 − 3 + 3 = −8 ✓
What is different: v′ = −1 and x = −1 put four signs in one line. Bracket every value as it goes in.
A constant on top. Find the slope of f(x) = 5x² + 1 at x = 1.
u = 5, u′ = 0; v = x² + 1, v′ = 2x
a constant's derivative is 0
f′(x) = [0 · (x² + 1) − 5 · 2x] ÷ (x² + 1)² = −10x(x² + 1)²
only the second term survives
f′(1) = −104 = −52
(1 + 1)² = 4
What is different: half the top vanishes, leaving a minus in front. The answer is not 52x.
A minus across a whole bracket. Differentiate f(x) = 3x − 2x + 4.
top: 3(x + 4) − (3x − 2)(1)
u′v − uv′
= 3x + 12 − 3x + 2 = 14
the minus reaches the −2 too
f′(x) = 14(x + 4)²
the x terms cancel completely
What is different: the top collapses to a number. For ax + bcx + d it is always ad − bc: here 3 × 4 − (−2) × 1 = 14.
Writing 3x + 12 − 3x − 2 subtracts only the 3x. Keep (3x − 2) in brackets until the minus has been multiplied through.
Besides formulas, the rules often come with only numbers: say f(1) = 4, f′(1) = 2, g(1) = −2, g′(1) = 3. Then (fg)′(1) = 2(−2) + 4(3) = 8, and (fg)′(1) = [2(−2) − 4(3)] ÷ (−2)² = −16 ÷ 4 = −4. Write the rule first, then drop the numbers in; the trap choices multiply or divide the two derivatives, or swap the order on top. If the function simplifies (it cancels, or it expands easily), do that first and use the power rule as a check.
Differentiate
f(x) = 3x + 1x − 2
Quotient rule: [3(x − 2) − (3x + 1)] / (x − 2)² = 3x − 6 − 3x − 1(x − 2)² = −7(x − 2)². A swaps the order in the numerator and loses the sign; C divides the derivatives — the forbidden shortcut, quotient edition.
