Derivative Rules: Power, Product, Quotient and Chain
Calculus I / AP Calculus AB · 12 flashcards · 7 quiz questions · updated 2026-08-19
Most lost marks in Calculus I are not conceptual. They come from applying the product rule to a quotient, forgetting the inner derivative in a chain rule, or differentiating a constant multiple as though it were a product.
This guide is the four rules, a decision procedure for choosing between them, and the derivatives you should be able to write without thinking.
The four rules
The quotient rule's numerator order matters: it is f′g − fg′, not the other way round. A memory aid that survives exam pressure: 'low d-high minus high d-low, over the square of what's below'.
| Rule | Statement | Example |
|---|---|---|
| Power | d/dx [xⁿ] = n·xⁿ⁻¹ | d/dx [x⁵] = 5x⁴ |
| Product | (fg)' = f'g + fg' | d/dx [x²·sin x] = 2x·sin x + x²·cos x |
| Quotient | (f/g)' = (f'g − fg') / g² | d/dx [x / (x+1)] = 1/(x+1)² |
| Chain | d/dx [f(g(x))] = f'(g(x))·g'(x) | d/dx [sin(3x)] = 3·cos(3x) |
Choosing the right rule
- •Is the expression a single power of x? Power rule.
- •Is it a constant times a function? Constant multiple rule — pull the constant out. Do not use the product rule; the derivative of a constant is 0 and the extra term vanishes anyway.
- •Are two functions of x multiplied? Product rule.
- •Is one function divided by another? Quotient rule — or rewrite as a product with a negative exponent, which is often faster.
- •Is a function inside another function? Chain rule, and it composes with the others: expressions frequently need the product rule and the chain rule together.
Derivatives worth memorising
| f(x) | f′(x) | |
|---|---|---|
| c (a constant) | 0 | |
| eˣ | eˣ | |
| aˣ | aˣ · ln a | |
| ln x | 1/x | |
| logₐ x | 1 / (x · ln a) | |
| sin x | cos x | |
| cos x | −sin x | |
| tan x | sec² x | |
| arcsin x | 1 / √(1 − x²) | |
| arctan x | 1 / (1 + x²) |
Worked example: all three rules at once
Differentiate f(x) = x² · e^(3x).
This is a product, and the second factor is a composition — so product rule outside, chain rule inside.
f′(x) = 2x · e^(3x) + x² · e^(3x) · 3 = e^(3x)(2x + 3x²) = x·e^(3x)(2 + 3x).
The factored form is worth producing: questions that go on to ask for critical points want it anyway.
Implicit differentiation in one paragraph
When y is not isolated, differentiate both sides with respect to x and attach dy/dx every time you differentiate a y term — that is the chain rule doing its job. For x² + y² = 25: 2x + 2y·(dy/dx) = 0, so dy/dx = −x/y.
Common mistakes
- ✗Writing the quotient rule numerator backwards. It is f′g − fg′.
- ✗Dropping the inner derivative in the chain rule — d/dx[sin(3x)] is 3cos(3x), not cos(3x).
- ✗Using the product rule on a constant multiple, then losing a term in the algebra.
- ✗Treating d/dx[eˣ] and d/dx[xᵉ] as the same thing. The second is a power: e·xᵉ⁻¹.
- ✗Forgetting dy/dx in implicit differentiation when differentiating y terms.
Flashcards
Tap a card to reveal the answer.
Power rule⌄
d/dx [xⁿ] = n·xⁿ⁻¹, for any real n.
Product rule⌄
(fg)′ = f′g + fg′.
Quotient rule⌄
(f/g)′ = (f′g − fg′) / g².
Chain rule⌄
d/dx [f(g(x))] = f′(g(x)) · g′(x).
d/dx [eˣ]⌄
eˣ.
d/dx [ln x]⌄
1/x, for x > 0.
d/dx [tan x]⌄
sec² x.
d/dx [aˣ]⌄
aˣ · ln a.
d/dx [sin(5x)]⌄
5·cos(5x) — chain rule brings out the inner derivative.
d/dx [arctan x]⌄
1 / (1 + x²).
How do you differentiate x² · ln x?⌄
Product rule: 2x·ln x + x²·(1/x) = 2x·ln x + x.
What does implicit differentiation add?⌄
A factor of dy/dx whenever a y term is differentiated.
Practice quiz
Answer first, then open the explanation.
1. What is d/dx [x³ · cos x]?
- A. 3x² · (−sin x)
- B. 3x² cos x − x³ sin x
- C. 3x² cos x + x³ sin x
- D. −3x² sin x
Show answer
B. 3x² cos x − x³ sin x
Product rule: (3x²)(cos x) + (x³)(−sin x).2. What is d/dx [(2x + 1)⁵]?
- A. 5(2x + 1)⁴
- B. 10(2x + 1)⁴
- C. 2(2x + 1)⁴
- D. 10(2x + 1)⁵
Show answer
B. 10(2x + 1)⁴
Chain rule: 5(2x + 1)⁴ times the inner derivative 2.3. The quotient rule numerator is:
- A. fg′ − f′g
- B. f′g − fg′
- C. f′g + fg′
- D. f′g′ − fg
Show answer
B. f′g − fg′
Derivative of the top times the bottom, minus the top times the derivative of the bottom.4. What is d/dx [e^(x²)]?
- A. e^(x²)
- B. 2x · e^(x²)
- C. x² · e^(x²−1)
- D. 2x · e^(2x)
Show answer
B. 2x · e^(x²)
Chain rule: the outer derivative is e^(x²), the inner derivative of x² is 2x.5. What is d/dx [ln(3x)]?
- A. 3/x
- B. 1/(3x)
- C. 1/x
- D. 3 ln x
Show answer
C. 1/x
Chain rule gives (1/(3x))·3 = 1/x — which also follows from ln(3x) = ln 3 + ln x.6. Given x² + y² = 25, dy/dx equals:
- A. −x/y
- B. x/y
- C. −y/x
- D. 2x + 2y
Show answer
A. −x/y
Differentiating gives 2x + 2y·(dy/dx) = 0, so dy/dx = −x/y.7. Which rule alone handles d/dx [7x⁴]?
- A. Product rule
- B. Quotient rule
- C. Power rule with the constant multiple rule
- D. Chain rule
Show answer
C. Power rule with the constant multiple rule
Pull the 7 out and apply the power rule: 28x³.
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