§ 01Which Rule Do I Use? — Decision Guide
Before you differentiate anything, ask yourself what the function looks like. The shape of the function tells you which tool to reach for.
Rules can also combine, you may need the chain rule inside the product rule, or the chain rule applied to each factor in the quotient rule. This chapter covers each rule fully before showing you how to combine them.
A good strategy is always to name what you see before reaching for a formula. Write down the structure, is it a product? A composition? A fraction?, and then the correct rule follows naturally. Students who skip this step and go straight to algebra tend to make structural errors that are very hard to trace.
§ 02The Power Rule
The most used rule in all of calculus. Once you understand it, you can differentiate any polynomial in seconds.
In Plain English
If your function is x raised to any power n, then:
- Bring the power down in front as a multiplier.
- Reduce the power by 1.
That is all. Two steps. The rule works for any real number n, positive integers, negative integers, fractions, even irrational numbers like π.
Where Does This Come From?
This rule is derived from first principles using the limit definition of the derivative. If we apply the definition to f(x) = xⁿ and expand (x + h)ⁿ using the binomial theorem, all terms containing h² or higher vanish as h → 0, leaving exactly n·xⁿ⁻¹. The First Principles page carries out this derivation in full detail for several values of n.
f(x) = x½ → f′(x) = ½ · x⁻½ = 1 ÷ (2√x)
f(x) = x⁻² → f′(x) = −2x⁻³ = −2 ÷ x³
f(x) = 7 → f′(x) = 0 (constant rule)
Constants and Coefficients
Two additional facts make the power rule work on any polynomial:
Constant multiple rule: If there is a constant multiplier in front of x, it rides through the derivative unchanged. So d/dx[5x³] = 5 · d/dx[x³] = 5 · 3x² = 15x².
Sum/difference rule: When differentiating a sum or difference of terms, differentiate each term separately and add or subtract the results. So d/dx[4x³ − 7x + 2] = 12x² − 7 + 0 = 12x² − 7.
Wrong: d/dx[x³] = x² (just reduced the power, forgot the 3)
Correct: d/dx[x³] = 3x² ✓
Mistake 2 — Forgetting to reduce the power:
Wrong: d/dx[x³] = 3x³ (brought down the 3, kept the power as 3)
Correct: d/dx[x³] = 3x² ✓ (power goes from 3 down to 2)
§ 03The Product Rule
When two functions are multiplied together, you cannot just differentiate each one separately and multiply the results. The product rule tells you exactly what to do instead.
In Plain English
Call the two functions u and v. The derivative of their product is: differentiate the first and keep the second, then add the first kept times the second differentiated.
A helpful memory phrase: "d-first · second + first · d-second"
dy/dx = u′v + uv′
Why Can't You Just Multiply the Derivatives?
This is a very natural question and worth answering carefully. The sum rule works "termwise" because limits distribute over addition. But multiplication does not work the same way with limits. You can verify this with a simple counterexample: let u = x² and v = x³. Their product is u · v = x⁵, so d/dx[x⁵] = 5x⁴. Now try multiplying the individual derivatives: u′ = 2x and v′ = 3x², so u′ · v′ = 2x · 3x² = 6x³. That is clearly not 5x⁴. So simply multiplying derivatives gives the wrong answer. The product rule gives the correct one.
Set u = x², so u′ = 2x
Set v = sin x, so v′ = cos x
dy/dx = u′v + uv′
dy/dx = 2x · sin x + x² · cos x
Correct: d/dx[x² sin x] = 2x sin x + x² cos x ✓
The product rule produces a sum of two terms, not a product of two derivatives.
§ 04The Quotient Rule
When one function is divided by another, use the quotient rule. It looks intimidating at first, but the pattern becomes automatic with practice.
In Plain English
Call the top function u (the numerator) and the bottom function v (the denominator). The rule is: differentiate the top times the bottom, minus the top times the differentiated bottom, all divided by the bottom squared.
Memory phrase: "lo d-hi minus hi d-lo, over lo squared": where "hi" is the numerator and "lo" is the denominator.
dy/dx = (u′v − uv′) ÷ v²
Notice the minus sign in the middle, not a plus, as in the product rule. Getting this sign wrong is the most common error with this rule. The order matters critically: it is u′v − uv′, never uv′ − u′v. The denominator is always v², never v, never 2v.
u = sin x, u′ = cos x
v = x², v′ = 2x
Numerator: u′v − uv′ = cos x · x² − sin x · 2x
Denominator: v² = (x²)² = x⁴
dy/dx = (x² cos x − 2x sin x) ÷ x⁴
Factor x from numerator:
dy/dx = x(x cos x − 2 sin x) ÷ x⁴
dy/dx = (x cos x − 2 sin x) ÷ x³
Correct: dy/dx = (u′v − uv′) ÷ v² ✓
The minus sign is in the numerator, and it is u′v first, then subtract uv′. The denominator is always v squared.
§ 05The Chain Rule
The chain rule handles composite functions: a function living inside another function. This is the most powerful and most frequently needed rule in calculus.
In Plain English
If you have a function of a function, the derivative is: the derivative of the outer function (leaving the inner function untouched) multiplied by the derivative of the inner function.
Think of peeling an onion. You differentiate layer by layer, from outside to inside. Every layer contributes a factor to the final answer, none can be skipped.
dy/dx = f′(g(x)) · g′(x)
Or using substitution u = g(x):
dy/dx = (dy/du) · (du/dx)
How to Spot a Composite Function
Ask yourself: "Is there an expression inside something?" If yes, you need the chain rule.
- (3x + 1)⁵: the inside is 3x + 1, the outside is "something raised to the fifth power"
- sin(x²): the inside is x², the outside is "sine of something"
- e^(2x): the inside is 2x, the outside is "e to the something"
- √(x³ + 1), the inside is x³ + 1, the outside is "square root of something"
- ln(cos x), the inside is cos x, the outside is "natural log of something"
Step 1: Name the inner function: u = 3x + 1
Step 2: Rewrite: y = u⁵
Step 3: Differentiate the outer (leaving u intact): dy/du = 5u⁴
Step 4: Differentiate the inner: du/dx = 3
Step 5: Multiply (chain rule): dy/dx = 5u⁴ · 3
Step 6: Substitute back u = 3x + 1:
dy/dx = 15(3x + 1)⁴
Correct: d/dx[(3x+1)⁵] = 5(3x+1)⁴ · 3 = 15(3x+1)⁴ ✓
The inner derivative is never optional. Every layer of the composite function must be differentiated and multiplied in. This is the single most common error in calculus.
§ 06Practice Exercises — All Four Rules
Work through each problem yourself first. Click Show Solution to reveal the full step-by-step working. Every step is explained, no jumps, no shortcuts.
Power Rule (Exercises 1 – 4)
- Identify the exponent (power): n = 5. The function is xⁿ with n = 5.
- Apply the power rule, bring n down as a coefficient and subtract 1 from the exponent: dy/dx = 5 · x⁵⁻¹ We brought the 5 to the front, and the new exponent is 5 − 1 = 4.
- Simplify: dy/dx = 5x⁴
- Differentiate term by term: this uses the sum/difference rule, which says you can differentiate each term separately. There are three terms: 4x³, −7x, and 2.
- Differentiate the first term: 4x³. The constant 4 rides through; apply the power rule to x³. d/dx [4x³] = 4 · 3x² = 12x²
- Differentiate the second term: −7x. Recall that x = x¹, so n = 1. Bring down 1, reduce power to 0, and x⁰ = 1. d/dx [−7x] = −7 · 1 · x⁰ = −7 · 1 = −7 The −7 is a constant multiplier and carries through unchanged.
- Differentiate the third term: 2. A plain constant has derivative zero: its graph is flat (zero slope). d/dx [2] = 0
- Combine all three results: f′(x) = 12x² − 7 + 0 f′(x) = 12x² − 7
- Rewrite the square root as a fractional exponent. This is essential, the power rule needs the function in xⁿ form. y = √x = x^(1/2) In general, the nth root of x equals x^(1/n). So √x = x^(1/2).
- Identify the exponent: n = 1/2. Now apply the power rule, bring 1/2 down as a coefficient and subtract 1 from the exponent. dy/dx = (1/2) · x^(1/2 − 1)
- Compute the new exponent: 1/2 − 1 = 1/2 − 2/2 = −1/2. dy/dx = (1/2) · x^(−1/2)
- A negative exponent means "one over". Rewrite in a cleaner form: x^(−1/2) = 1 ÷ x^(1/2) = 1 ÷ √x
- Final answer: dy/dx = 1 ÷ (2√x) This tells us the slope of y = √x at any point x. At x = 4, for example, the slope is 1 ÷ (2·2) = 1/4.
- Rewrite the fraction using a negative exponent. A term in the denominator becomes a negative power: y = 3 ÷ x² = 3x⁻² In general, 1/xⁿ = x⁻ⁿ. Here n = 2, so 1/x² = x⁻².
- Identify: there is a coefficient of 3 and a power of −2. The constant 3 carries through; apply the power rule to x⁻². dy/dx = 3 · (−2) · x^(−2 − 1)
- Multiply the constants: 3 · (−2) = −6. Compute the new exponent: −2 − 1 = −3. dy/dx = −6x⁻³
- Rewrite using fraction notation to present the answer cleanly: dy/dx = −6 ÷ x³ Both forms (−6x⁻³ and −6/x³) are equivalent and fully correct.
Product Rule (Exercises 5 – 8)
- Recognise the structure: two functions multiplied together → product rule is needed. Label them: u = x³ and v = eˣ.
- Differentiate each function separately: u′ = d/dx[x³] = 3x² v′ = d/dx[eˣ] = eˣ The derivative of eˣ is eˣ, it is its own derivative. This is a fundamental and important fact.
- Apply the product rule formula: dy/dx = u′v + uv′. Substitute in what we found: dy/dx = 3x² · eˣ + x³ · eˣ
- Both terms contain a factor of eˣ. Factorise it out to simplify: dy/dx = eˣ(3x² + x³)
- Both terms inside the bracket also share a factor of x². Factorise further: dy/dx = x²eˣ(3 + x) Fully factorised form is usually preferred. Both eˣ(3x² + x³) and x²eˣ(3 + x) are correct.
- Identify the structure: two functions multiplied → use the product rule. Label: u = x and v = ln x.
- Differentiate each: u′ = d/dx[x] = 1 v′ = d/dx[ln x] = 1/x The derivative of ln x is 1/x: a standard result you should know by heart.
- Apply the product rule: dy/dx = u′v + uv′: dy/dx = 1 · ln x + x · (1/x)
- Simplify: the second term is x · (1/x) = x/x = 1: dy/dx = ln x + 1 This is a result worth remembering: d/dx[x ln x] = ln x + 1. It appears frequently in integration by parts too.
- Two functions multiplied → product rule. Label: u = 2x + 1, v = sin x.
- Differentiate each: u′ = d/dx[2x + 1] = 2 v′ = d/dx[sin x] = cos x
- Apply the product rule: dy/dx = u′v + uv′. Substitute: dy/dx = 2 · sin x + (2x + 1) · cos x
- There is no further simplification here, the two terms do not share common factors. The final answer is: dy/dx = 2 sin x + (2x + 1) cos x It is acceptable to leave it in this expanded form. Do not try to combine sin and cos terms, they are not like terms.
- Two functions multiplied → product rule. Label: u = x², v = cos x.
- Differentiate each. Be careful with the sign for d/dx[cos x]: u′ = d/dx[x²] = 2x v′ = d/dx[cos x] = −sin x The derivative of cos x is NEGATIVE sin x. This negative sign is easy to drop by mistake.
- Apply the product rule: dy/dx = u′v + uv′. Substitute carefully: dy/dx = 2x · cos x + x² · (−sin x)
- Expand: the second term carries the negative sign from −sin x: dy/dx = 2x cos x − x² sin x
- Both terms share a factor of x. Factorise to get a neater form: dy/dx = x(2 cos x − x sin x) Factorised form is cleaner, but the expanded form 2x cos x − x² sin x is equally valid.
Quotient Rule (Exercises 9 – 11)
- Fraction with different functions on top and bottom → quotient rule. Label: u = sin x (top), v = x (bottom).
- Differentiate each: u′ = d/dx[sin x] = cos x v′ = d/dx[x] = 1
- Write out the quotient rule formula: dy/dx = (u′v − uv′) ÷ v². Substitute in every piece: dy/dx = (cos x · x − sin x · 1) ÷ x² Numerator = u′v − uv′ = cos x · x − sin x · 1. Denominator = v² = x².
- Write the numerator in standard order (x · cos x is cleaner than cos x · x, though they are equal): dy/dx = (x cos x − sin x) ÷ x² There is no further cancellation possible here. This is the final simplified answer.
- Fraction with functions top and bottom → quotient rule. Label: u = x² + 1 (top), v = x − 3 (bottom).
- Differentiate each: u′ = d/dx[x² + 1] = 2x v′ = d/dx[x − 3] = 1 The derivative of the constant −3 is 0, leaving just 1 from the x term.
- Apply the quotient rule formula dy/dx = (u′v − uv′) ÷ v²: dy/dx = (2x · (x − 3) − (x² + 1) · 1) ÷ (x − 3)²
- Expand the numerator. First expand 2x(x − 3): 2x(x − 3) = 2x² − 6x
- Now subtract (x² + 1) from that: 2x² − 6x − (x² + 1) = 2x² − 6x − x² − 1 = x² − 6x − 1 Be careful when distributing the minus sign: −(x² + 1) = −x² − 1, not −x² + 1.
- Write the final answer with the simplified numerator: dy/dx = (x² − 6x − 1) ÷ (x − 3)²
- Fraction → quotient rule. Label: u = eˣ (top), v = x² + 1 (bottom).
- Differentiate each: u′ = d/dx[eˣ] = eˣ v′ = d/dx[x² + 1] = 2x
- Apply the quotient rule formula dy/dx = (u′v − uv′) ÷ v²: dy/dx = (eˣ · (x² + 1) − eˣ · 2x) ÷ (x² + 1)²
- Both terms in the numerator share a factor of eˣ. Factorise it out: numerator = eˣ[(x² + 1) − 2x] = eˣ(x² − 2x + 1)
- Recognise that x² − 2x + 1 is a perfect square: it factors as (x − 1)²: numerator = eˣ(x − 1)² Check: (x − 1)² = x² − 2x + 1 ✓. Spotting this factorisation makes the answer look cleaner.
- Write the final answer: dy/dx = eˣ(x − 1)² ÷ (x² + 1)²
Chain Rule (Exercises 12 – 15)
- Identify the composite structure: there is a function (5x − 2) sitting inside a power function (something⁴). This is a composition → chain rule needed.
- Name the inner function: u = 5x − 2
- Rewrite y in terms of u: y = u⁴
- Differentiate the outer function with respect to u, leaving the inner function untouched (just write u, not 5x − 2, at this stage): dy/du = 4u³
- Differentiate the inner function with respect to x: du/dx = d/dx[5x − 2] = 5 The constant −2 disappears (derivative of a constant is 0). The 5x term gives 5.
- Apply the chain rule: multiply the two derivatives together: dy/dx = (dy/du) · (du/dx) = 4u³ · 5 = 20u³
- Substitute back u = 5x − 2 to express the answer in terms of x: dy/dx = 20(5x − 2)³
- Identify the composite structure: there is 3x² sitting inside a sine function. Composition → chain rule.
- Name the inner function: u = 3x²
- Rewrite in terms of u: y = sin(u)
- Differentiate the outer function with respect to u: dy/du = cos(u) The derivative of sin(u) with respect to u is cos(u). Keep the inside as u for now.
- Now substitute u back to express this in terms of x: dy/du = cos(3x²)
- Differentiate the inner function with respect to x: du/dx = d/dx[3x²] = 6x Apply the power rule to x²: bring down 2, reduce power to 1, giving 3 · 2x = 6x.
- Apply the chain rule: multiply the two derivatives: dy/dx = cos(3x²) · 6x
- Write in conventional order (coefficient first): dy/dx = 6x cos(3x²)
- Identify the composite structure: there is x² + 3x inside the exponential function e^(something). Composition → chain rule.
- Name the inner function: u = x² + 3x
- Rewrite in terms of u: y = eᵘ
- Differentiate the outer function with respect to u. The derivative of eᵘ is eᵘ, the exponential function is its own derivative: dy/du = eᵘ = e^(x² + 3x) We immediately substitute u back: eᵘ = e^(x² + 3x). This outer derivative is unchanged by differentiation.
- Differentiate the inner function with respect to x. Use the power rule on each term: du/dx = d/dx[x² + 3x] = 2x + 3
- Apply the chain rule, multiply the outer and inner derivatives: dy/dx = e^(x² + 3x) · (2x + 3)
- Write in conventional order: dy/dx = (2x + 3) e^(x² + 3x) Placing the polynomial factor (2x + 3) first is the conventional style, but the order of multiplication does not affect the answer.
- Identify the outer structure: two functions multiplied together → product rule is needed first. Label the two factors: u = x² and v = (2x + 1)³.
- Differentiate u using the power rule (straightforward): u′ = d/dx[x²] = 2x
- Differentiate v using the chain rule, there is a function (2x + 1) inside a power (something³): Inner function: w = 2x + 1 → dw/dx = 2 Outer function: v = w³ → dv/dw = 3w² v′ = 3(2x + 1)² · 2 = 6(2x + 1)² We brought down the power 3, reduced the power by 1 to get (2x+1)², then multiplied by the inner derivative 2.
- Now apply the product rule: dy/dx = u′v + uv′. Substitute everything in: dy/dx = 2x · (2x + 1)³ + x² · 6(2x + 1)²
- Identify common factors in both terms. Both contain x and (2x + 1)²: = 2x(2x + 1)² · (2x + 1) + 6x² · (2x + 1)² Rewriting the first term makes the common factor visible: 2x(2x+1)³ = 2x · (2x+1)² · (2x+1).
- Factorise out the common factors 2x and (2x + 1)²: = 2x(2x + 1)²[(2x + 1) + 3x] Inside the bracket: we divided the first group by 2x(2x+1)² to get (2x+1), and the second group 6x²(2x+1)² ÷ [2x(2x+1)²] = 3x.
- Simplify the bracket: (2x + 1) + 3x = 5x + 1: dy/dx = 2x(2x + 1)²(5x + 1) This fully factorised form is the cleanest answer. You can verify it is correct by expanding, but that is rarely necessary in an exam.
§ 07Combining Rules — A Summary
Real exam questions rarely use just one rule. Here is how to identify what is needed at a glance.
f(x) · g(x) → Product Rule
f(x) ÷ g(x) → Quotient Rule
f(g(x)) → Chain Rule
f(x) · g(h(x)) → Product + Chain
f(g(x)) ÷ h(x) → Quotient + Chain
When in doubt, write out each component, label it, differentiate it individually, then assemble. Never try to hold all of it in your head at once, the algebra will slip.