Chain Rule Product Rule and Quotient Rule
Chain Rule, Product Rule and Quotient Rule
These three differentiation rules extend your ability to differentiate far beyond simple polynomials. Together they allow you to find the derivative of virtually any function built from standard components.
The Chain Rule
The chain rule is used to differentiate a composite function — a function of a function. If y = f(g(x)), then:
dy/dx = f'(g(x)) × g'(x)
In Leibniz notation: if y is a function of u and u is a function of x, then dy/dx = (dy/du) × (du/dx).
Think of it as: "differentiate the outer function, keeping the inner function unchanged, then multiply by the derivative of the inner function."
Worked Example 1: Differentiate y = (3x + 1)⁵.
Let u = 3x + 1, so y = u⁵.
dy/du = 5u⁴ and du/dx = 3
dy/dx = 5u⁴ × 3 = 15(3x + 1)⁴
Worked Example 2: Differentiate y = sin(x²).
The outer function is sin(·) and the inner function is x².
dy/dx = cos(x²) × 2x = 2x·cos(x²)
Worked Example 3: Differentiate y = e^(4x³).
dy/dx = e^(4x³) × 12x² = 12x²·e^(4x³)
The Product Rule
The product rule differentiates two functions multiplied together. If y = u·v where u and v are both functions of x, then:
dy/dx = u·(dv/dx) + v·(du/dx)
Or more concisely: (uv)' = uv' + vu'
Worked Example: Differentiate y = x³·sin(x).
Let u = x³ and v = sinx.
u' = 3x² and v' = cosx
dy/dx = x³·cosx + sinx·3x² = x³cosx + 3x²sinx
This can be factorised as x²(xcosx + 3sinx).
The Quotient Rule
The quotient rule differentiates one function divided by another. If y = u/v, then:
dy/dx = (v·(du/dx) - u·(dv/dx)) / v²
Or: (u/v)' = (vu' - uv') / v²
Note the subtraction and the order — the denominator function v comes first in the numerator of the result. A common mnemonic is "low d-high minus high d-low, over the square of what's below".
Worked Example: Differentiate y = (2x + 1) / (x² + 3).
Let u = 2x + 1 and v = x² + 3.
u' = 2 and v' = 2x
dy/dx = ((x² + 3)(2) - (2x + 1)(2x)) / (x² + 3)²
= (2x² + 6 - 4x² - 2x) / (x² + 3)²
= (-2x² - 2x + 6) / (x² + 3)²
Combining the Rules
Many A-Level questions require you to apply two or more rules together.
Worked Example: Differentiate y = e²ˣ·cos(3x).
This is a product of two composite functions, so use the product rule with the chain rule applied to each factor.
u = e²ˣ, so u' = 2e²ˣ (chain rule)
v = cos(3x), so v' = -3sin(3x) (chain rule)
dy/dx = e²ˣ·(-3sin3x) + cos3x·(2e²ˣ)
= e²ˣ(2cos3x - 3sin3x)
Connected Rates of Change
The chain rule also applies to related rates problems. If two quantities both change with time, and you know how they are related, you can connect their rates:
dy/dt = (dy/dx) × (dx/dt)
Worked Example: A spherical balloon is inflated so that its volume increases at 10 cm³/s. Find the rate of increase of the radius when r = 5 cm.
V = (4/3)πr³, so dV/dr = 4πr²
dV/dt = (dV/dr) × (dr/dt)
10 = 4π(25) × dr/dt
dr/dt = 10/(100π) = 1/(10π) ≈ 0.0318 cm/s
Choosing the Right Rule
| Situation | Rule |
|---|---|
| Function of a function, e.g. sin(x²) | Chain rule |
| Two functions multiplied, e.g. x·eˣ | Product rule |
| One function divided by another, e.g. sinx/x | Quotient rule |
| Combination of the above | Apply rules in layers |
Exam Tips
- Always state u, v, u', v' clearly when using product or quotient rule — it prevents sign errors and earns method marks.
- The quotient rule can always be replaced by the product rule: write u/v as u·v⁻¹. Use whichever feels more natural.
- For the chain rule, practice identifying the "inner" and "outer" functions quickly.
- In connected rates problems, always state the chain rule relationship before substituting.
- Simplify your answer where possible — factorise common factors out.