Difficulty: Intermediate | Prerequisites: Basic derivative rules (power rule, trig derivatives, product and quotient rules)
The chain rule is the single most-used differentiation technique in calculus. It handles every function built by plugging one function inside another, which is nearly every function you will meet after the first few weeks of the course. If you can only master one differentiation rule, this is the one. You should already be comfortable with the power rule, basic trig derivatives, and the idea of a composite function before working through these notes.
The chain rule tells you how to differentiate a composite function, one where a function sits inside another function. You take the derivative of the outer function (leaving the inner function untouched), then multiply by the derivative of the inner function. Once you can spot the "inner" and "outer" layers, the rest is mechanical.
Chain rule
A differentiation rule for composite functions. If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). In simple terms, differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.
Composite function
A function formed by plugging one function into another, written f(g(x)) or (f ∘ g)(x). Think of it as a two-step machine: x goes into g first, then the output of g feeds into f.
Inner function (u)
The function that sits inside the other one. In y = (x² + 3x)⁴, the inner function is u = x² + 3x. Students sometimes call this "the bit inside the brackets."
Outer function
The function that wraps around the inner function. In y = (x² + 3x)⁴, the outer function is u⁴. It is what you would see if you replaced the inner function with a single letter u.
Generalised power rule
A direct consequence of the chain rule: if y = [u(x)]ⁿ, then dy/dx = n[u(x)]ⁿ⁻¹ · u'(x). Think of it as the ordinary power rule, but with an extra factor for the derivative of whatever is inside the bracket.
Second derivative
The derivative of the derivative, written f''(x) or d²y/dx². When the first derivative itself involves a chain rule expression, finding the second derivative often requires the product rule as well.
Identify the composite structure. Ask: "Is there a function inside another function?" If yes, you need the chain rule.
Label the inner function u. Write u = (whatever is inside).
Differentiate the outer function with respect to u. Treat u as a single variable.
Multiply by du/dx, the derivative of the inner function.
Substitute u back in and simplify. The answer should contain no u, only x.
Any expression raised to a power where the base is not just x: (x² + 3x)⁴, (4x + 5)³.
Any trig function whose argument is not plain x: sin(3x), cos(2x³), tan(3x).
A radical (square root, cube root, fourth root) with something other than x under it: √(3x² − 8), ∛(x² + 4).
Nested compositions: cos⁵(2x³) requires the chain rule twice, once for the fifth power and once for the 2x³ inside the cosine.
Power of a function (generalised power rule)
y = [u(x)]ⁿ → dy/dx = n·[u(x)]ⁿ⁻¹ · u'(x)
Example: y = (x² + 3x)⁴. Here u = x² + 3x, n = 4. dy/dx = 4(x² + 3x)³ · (2x + 3)
Trig of a function
y = sin(u) → dy/dx = cos(u) · u' y = cos(u) → dy/dx = −sin(u) · u' y = tan(u) → dy/dx = sec²(u) · u'
Example: y = tan(3x). Here u = 3x. dy/dx = sec²(3x) · 3 = 3 sec²(3x)
Power of a trig function
y = sinⁿ(u) requires TWO chain rule layers: the power and the trig argument.
Example: y = sin³(x). Outer: u³, middle: sin(x). dy/dx = 3 sin²(x) · cos(x)
Example: y = cos⁵(2x³). Three layers: fifth power, cosine, 2x³. dy/dx = 5 cos⁴(2x³) · (−sin(2x³)) · 6x² = −30x² cos⁴(2x³) sin(2x³)
Radical of a function
Rewrite the radical as a fractional exponent first, then apply the generalised power rule.
Example: f(x) = ∛(x² + 4) = (x² + 4)^(1/3). f'(x) = (1/3)(x² + 4)^(−2/3) · 2x = 2x / [3(x² + 4)^(2/3)]
When a function involves both a product (or quotient) and a composition, apply the product rule or quotient rule for the overall structure and the chain rule for each piece that is itself a composite.
Example: g(x) = x³√(x² − 3). This is a product of x³ and (x² − 3)^(1/2).
Use the product rule: g'(x) = 3x²·(x² − 3)^(1/2) + x³ · (1/2)(x² − 3)^(−1/2) · 2x
Simplify: g'(x) = 3x²√(x² − 3) + x⁴/√(x² − 3)
Combine over a common denominator: g'(x) = [3x²(x² − 3) + x⁴] / √(x² − 3) = [4x⁴ − 9x²] / √(x² − 3)
For problems that ask for the equation of a tangent line:
Find f'(x) using the chain rule.
Evaluate f'(a) at the given x = a to get the slope m.
Evaluate f(a) to get the y-coordinate of the point.
Write the line: y − f(a) = m(x − a).
To find f''(x) when f'(x) was found using the chain rule, you will almost always need the product rule on f'(x), because f'(x) is typically a product of two functions of x. Then apply the chain rule again to each factor as needed.
\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\frac{d}{dx}[u^n] = n\,u^{n-1} \cdot \frac{du}{dx}\frac{d}{dx}[\sin(u)] = \cos(u) \cdot \frac{du}{dx}\frac{d}{dx}[\cos(u)] = -\sin(u) \cdot \frac{du}{dx}\frac{d}{dx}[\tan(u)] = \sec^2(u) \cdot \frac{du}{dx}\frac{d}{dx}[\sec(u)] = \sec(u)\tan(u) \cdot \frac{du}{dx}The chain rule is how you differentiate any quantity that depends on another changing quantity. In physics, if position depends on time through a composed relationship (say, the height of a point on a rotating wheel), the chain rule gives you velocity. In economics, if revenue depends on price and price depends on time, the chain rule connects the rate of change of revenue to the rate of change of price. Related-rates problems in later chapters rely almost entirely on the chain rule applied to implicit relationships.
Students often forget to multiply by the derivative of the inner function. They write d/dx[(x² + 1)⁴] = 4(x² + 1)³ and stop. The missing factor of 2x is the most common chain rule error on exams.
Students sometimes differentiate the inner function and replace it inside the outer function, rather than leaving it untouched. For sin(3x), the derivative is cos(3x) · 3, not cos(3).
When there are multiple nested layers, students apply the chain rule once and forget the remaining layers. For cos⁵(2x³), there are three layers (the fifth power, the cosine, and 2x³), and each one produces a factor in the derivative.
Students confuse the chain rule with the product rule. The chain rule applies to f(g(x)), a function inside a function. The product rule applies to f(x) · g(x), two functions multiplied together. If both structures are present (as in x³√(x² − 3)), you need both rules.
⚠️ Chain rule problems appear on virtually every calculus exam. Expect at least 3 to 5 questions that require it.
⚠️ Tangent line problems at a given point are a classic exam format: find f'(x) with the chain rule, evaluate, then write the line equation.
⚠️ Problems that combine the chain rule with the product or quotient rule are considered higher difficulty and often carry more marks.
⚠️ Second derivative questions where the first derivative involves the chain rule test whether you can apply the product rule to a chain rule result.
⚠️ Table-based problems (like problems 15 to 17 and 33 to 36 in this worksheet) test whether you understand f'(g(x)) · g'(x) conceptually, not just mechanically. You read values from a table or graph instead of computing derivatives from a formula.
True or False: The derivative of (3x + 1)⁵ is 5(3x + 1)⁴. (False: missing the factor of 3 from the inner derivative.)
Fill in the blank: d/dx[sin(5x)] = ___ · 5. (cos(5x))
True or False: To differentiate cos³(x), you need the chain rule. (True: it is [cos(x)]³, a power of a function.)
Fill in the blank: If y = √(x² + 4), rewrite as y = (x² + 4)^(___) before differentiating. (1/2)
True or False: For g(x) = x²·sin(x), you should use the chain rule. (False: this is a product, not a composition. Use the product rule.)
Q: Find dy/dx for y = (x² + 3x)⁴.
A: dy/dx = 4(x² + 3x)³ · (2x + 3)
Q: Find f'(x) for f(x) = ∛(x² + 4) = (x² + 4)^(1/3).
A: f'(x) = (1/3)(x² + 4)^(−2/3) · 2x = 2x / [3(x² + 4)^(2/3)]
Q: Find dy/dx for y = 2/(4x + 5)³.
A: Rewrite as y = 2(4x + 5)^(−3). Then dy/dx = 2 · (−3)(4x + 5)^(−4) · 4 = −24/(4x + 5)⁴.
Q: Find dy/dx for y = tan(3x).
A: dy/dx = sec²(3x) · 3 = 3 sec²(3x)
Q: Find f'(x) for f(x) = sin³(x).
A: f'(x) = 3 sin²(x) · cos(x)
Q: Find dy/dx for y = cos⁵(2x³).
A: dy/dx = 5 cos⁴(2x³) · (−sin(2x³)) · 6x² = −30x² cos⁴(2x³) sin(2x³)
Q: Find g'(x) for g(x) = 5 sec²(πx).
A: g'(x) = 5 · 2 sec(πx) · sec(πx)tan(πx) · π = 10π sec²(πx) tan(πx)
Q: Find f'(x) and the tangent line equation for f(x) = √(3x² − 8) at x = 2.
A: f'(x) = (1/2)(3x² − 8)^(−1/2) · 6x = 3x/√(3x² − 8). At x = 2: f(2) = √(12 − 8) = 2, f'(2) = 6/2 = 3. Tangent line: y − 2 = 3(x − 2), or y = 3x − 4.
Q: Find g'(x) and the tangent line for g(x) = (2x³ + 1)⁴ at x = 1.
A: g'(x) = 4(2x³ + 1)³ · 6x² = 24x²(2x³ + 1)³. At x = 1: g(1) = 3⁴ = 81, g'(1) = 24 · 27 = 648. Tangent line: y − 81 = 648(x − 1).
Q: Find dy/dx and the tangent line for y = sin(4x) at x = π/3.
A: dy/dx = 4 cos(4x). At x = π/3: y = sin(4π/3) = −√3/2, dy/dx = 4 cos(4π/3) = 4(−1/2) = −2. Tangent line: y + √3/2 = −2(x − π/3).
Q: Find the second derivative of f(x) = (3 − x²)⁴.
A: f'(x) = 4(3 − x²)³ · (−2x) = −8x(3 − x²)³. Then f''(x) uses the product rule on −8x and (3 − x²)³: f''(x) = −8(3 − x²)³ + (−8x) · 3(3 − x²)² · (−2x) = −8(3 − x²)³ + 48x²(3 − x²)². Factor: f''(x) = 8(3 − x²)²[−(3 − x²) + 6x²] = 8(3 − x²)²(7x² − 3).
The chain rule connects directly to implicit differentiation, which treats y as a function of x and applies the chain rule to every y term. Related rates problems in later chapters are essentially the chain rule applied with respect to time. The chain rule also reappears in integration as u-substitution, which is the chain rule in reverse.
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