The derivative toolkit: chain, product and quotient rules
What this lesson covers
- Every Level 3 derivative is built from a few standard results and three rules for combining them.
- A dropped factor or a missing bracket loses the whole question, even if later working is right.
A few results and three rules
- Standard results, from the booklet.
- Chain, product and quotient rules.
- Every factor, every bracket.
The results to know
- ln x gives 1/x.
- sec, cosec and cot are in the booklet.
- Check the signs there.
| f(x) | f′(x) |
|---|---|
| ln x | 1/x |
| sec x | sec x tan x |
| cosec x | −cosec x cot x |
| cot x | −cosec² x |
Surds: powers first
- Every root as a power.
- Then the power rule.
- y = 4√x + 2/√x
- = 4x^½ + 2x^−½roots as powers
- dy/dx = 4(½)x^−½ + 2(−½)x^−³ᐟ²
- = 2x^−½ − x^−³ᐟ²power rule
- x = 4: 2(4)^−½ − (4)^−³ᐟ²
- = 2(½) − 1/84^½ = 2, 4^³ᐟ² = 8
- = 1 − 1/8 = 7/8
The chain rule: y = √(x² + 9)
- Differentiate the outside.
- Times the inside derivative.
- y = (x² + 9)^½root as a power
- outside: ½(x² + 9)^−½inside left alone
- inside: d/dx(x² + 9) = 2x
- dy/dx = ½(x² + 9)^−½ × 2x
- = x(x² + 9)^−½½ × 2x = x
- = x/√(x² + 9)
Which rule?
- Inside another: chain.
- Multiplied: product.
- Divided: quotient.
| If | Rule | Example |
|---|---|---|
| function inside | chain | sec(x²) |
| two multiplied | product | x³e^−2x |
| one over another | quotient | ln x / x² |
The product rule: y = e²ˣ sin 3x
- A product: u′v + uv′.
- Chain rule inside each part.
- u = e²ˣ, u′ = 2e²ˣchain: × 2
- v = sin 3x, v′ = 3cos 3xchain: × 3
- dy/dx = u′v + uv′product rule
- = (2e²ˣ)(sin 3x) + (e²ˣ)(3cos 3x)
- = 2e²ˣ sin 3x + 3e²ˣ cos 3x
The 2025 error: a dropped factor
- Chain rule in each part.
- No trig identities needed.
- u′ = e²ˣthe 2 is dropped
- v′ = cos 3xthe 3 is dropped
- e²ˣ sin 3x + e²ˣ cos 3xscores nothing
- u′ = 2e²ˣ, v′ = 3cos 3xkeep both factors
- 2e²ˣ sin 3x + 3e²ˣ cos 3x
The quotient rule: y = (x² + 1)/(2x − 3)
- Top: u′v − uv′.
- Bottom: v².
- u = x² + 1, u′ = 2xthe top
- v = 2x − 3, v′ = 2the bottom
- dy/dx = (u′v − uv′)/v²quotient rule
- top: 2x(2x − 3) − 2(x² + 1)
- bottom: (2x − 3)²
The 2025 error: missing brackets
- The minus hits both terms.
- Then collect like terms.
- top: 2x(2x − 3) − 2(x² + 1)
- 2x(2x − 3) = 4x² − 6xexpand
- top: 4x² − 6x − 2x² + 1minus misses the 1
- −2(x² + 1) = −2x² − 2minus reaches both
- top: 4x² − 6x − 2x² − 2
- = 2x² − 6x − 2collect
- dy/dx = (2x² − 6x − 2)/(2x − 3)²
“You do not need to simplify”
- Take it literally: stop.
- A slip after that can cost the grade.
- Simplify only when the next step needs it.
Reciprocal trig and ln
- Booklet result, then × u′.
- ln u gives u′/u.
- d/du cosec u = −cosec u cot u
- y = cosec 3x: u = 3x, u′ = 3
- dy/dx = −cosec 3x cot 3x × 3
- = −3cosec 3x cot 3x
- ln(sin x): u = sin x, u′ = cos x
- dy/dx = (1/sin x) × cos xchain: × u′
- = cos x/sin x = cot x
Reciprocal trig: y = sec(x²)
- The inside is x².
- Its derivative 2x comes out.
- d/du sec u = sec u tan ubooklet
- u = x², u′ = 2xthe inside
- dy/dx = sec(x²) tan(x²) × 2x
- = 2x sec(x²) tan(x²)
Show every derivative
- No credit for a calculator answer alone.
- Unknown constants: eᵏˣ gives keᵏˣ.
- Radians for trig.
Where the marks are
- Achieved: the three rules, correctly.
- Merit: derivatives in a longer problem.
- Excellence: surds, trig, full algebra.
What to take away
- Choose the rule, keep every factor and bracket.
| Surds | fractional powers first |
| Inside | chain: × its derivative |
| Multiplied | product: u′v + uv′ |
| Divided | quotient, in brackets |
| Every factor | or no credit |
| Simplify? | not if told no need |
Your turn: four questions
- Pause here and answer each one.
- Question 1: surds to powers first.
- Question 3 simplifies neatly.
| Function | Find | |
|---|---|---|
| 1 | y = 6√x + 4/x | dy/dx |
| 2 | y = x³e^−2x | dy/dx |
| 3 | y = sin x / (1 + cos x) | dy/dx, simplified |
| 4 | y = cot 2x | gradient at x = π/8 |
Show answer
1. y = 6x^½ + 4x^−1, dy/dx = 3x^−½ − 4x^−2 = 3/√x − 4/x². 2. dy/dx = 3x²e^−2x − 2x³e^−2x. 3. Top cos x(1 + cos x) + sin² x = 1 + cos x, so dy/dx = 1/(1 + cos x). 4. dy/dx = −2cosec² 2x; cosec(π/4) = √2, so the gradient is −4.
The answers: questions 1 and 2
- 1. y = 6√x + 4/x
- = 6x^½ + 4x^−1roots as powers
- dy/dx = 3x^−½ − 4x^−2power rule
- = 3/√x − 4/x²
- 2. u = x³, u′ = 3x²
- v = e^−2x, v′ = −2e^−2xchain: × −2
- dy/dx = 3x²e^−2x + x³(−2e^−2x)u′v + uv′
- = 3x²e^−2x − 2x³e^−2x
The answers: question 3
- 3. u = sin x, u′ = cos x
- v = 1 + cos x, v′ = −sin x
- u′v = cos x(1 + cos x)
- uv′ = sin x(−sin x) = −sin² x
- top = cos x + cos² x + sin² xu′v − uv′
- = cos x + 1sin² x + cos² x = 1
- dy/dx = (1 + cos x)/(1 + cos x)²over v²
- = 1/(1 + cos x)cancel 1 + cos x
The answers: question 4
- 4. d/du cot u = −cosec² ubooklet
- dy/dx = −cosec² 2x × 2chain: u = 2x, u′ = 2
- = −2cosec² 2x
- x = π/8: cosec(π/4) = √22x = π/4
- −2(√2)² = −2 × 2 = −4the gradient
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