Learning Objectives
Identify the inner and outer functions in a composition.
Apply the chain rule to powers, trigonometric, exponential, and logarithmic expressions.
Combine chain and product or quotient rules without dropping factors.
Prerequisites: composite functions and basic derivative, product, and quotient rules. All trigonometric arguments are in radians. The chain rule tracks how a changing input affects an outer function.
1. Differentiate the Outer Function, Then Multiply
For y = f(g(x)), the chain rule is y′ = f′(g(x))g′(x), provided the component functions are differentiable at the relevant inputs. Differentiate the outer function while keeping its input unchanged, then multiply by the derivative of that input.
Worked Example: A Power of a Function
Differentiate y = (3x² + 1)⁵. The inner function is u = 3x² + 1 and the outer function is u⁵.
The outer derivative is 5u⁴, and the inner derivative is 6x. Therefore y′ = 5(3x² + 1)⁴(6x) = 30x(3x² + 1)⁴.
Leaving out 6x would treat the inner expression as if it changed at rate 1.
2. Standard Rules Extend to Inner Functions
Expression | Derivative |
|---|---|
[u(x)]ⁿ | n[u(x)]^(n − 1)u′(x), where defined and differentiable |
sin(u(x)) | cos(u(x))u′(x) |
cos(u(x)) | −sin(u(x))u′(x) |
e^(u(x)) | e^(u(x))u′(x) |
ln(u(x)) | u′(x)/u(x), for u(x) > 0 |
Worked Example: A Logarithm
For f(x) = ln(x² + 4), the logarithm is defined for every real x because x² + 4 is positive.
Differentiate the outer logarithm and multiply by 2x: f′(x) = 2x/(x² + 4).
The logarithm of a sum cannot be split into ln(x²) + ln 4. The chain rule applies directly to the actual input.
3. Work through More than Two Layers
Worked Example: Three Layers
Differentiate y = sin((x² + 1)³). The outermost operation is sine, then a cube, then x² + 1.
Following the layers gives y′ = cos((x² + 1)³) × 3(x² + 1)² × 2x.
Thus y′ = 6x(x² + 1)² cos((x² + 1)³). Each layer contributes exactly one derivative factor.
4. Use the Rule Matching the Outermost Operation
Worked Example: Product and Chain Together
Differentiate f(x) = x²e^(3x). The whole expression is a product, so start with the product rule.
The derivative is 2xe^(3x) + x²[e^(3x)·3]. The bracketed derivative uses the chain rule.
Hence f′(x) = e^(3x)(2x + 3x²). Applying only the chain rule to the exponential would omit the change in x².
When a table supplies function values, the same rule applies. To compute (f ∘ g)′(a), find g(a), then f′ evaluated at that output, and multiply by g′(a). The outer derivative is not generally evaluated at a itself.
Common Mistakes
Keep the inner expression unchanged while differentiating the outer layer.
Identify a product or quotient before treating its factors as compositions.
For a table-based composition, use f′(g(a)), not automatically f′(a).
5. Practice
1. Differentiate y = √(5x + 2) on x > −2/5.
2. Differentiate f(x) = cos(4x²).
3. Differentiate g(x) = x ln(2x + 1), on its domain x > −1/2.
4. If g(2) = 5, g′(2) = −3, and f′(5) = 4, find (f ∘ g)′(2).
Worked Solutions
1. y′ = (1/2)(5x + 2)^(−1/2)·5 = 5/[2√(5x + 2)]. The endpoint is excluded from this finite derivative formula.
2. f′(x) = −sin(4x²)·8x = −8x sin(4x²).
3. Use the product rule and then the chain rule: g′(x) = ln(2x + 1) + 2x/(2x + 1).
4. The derivative is f′(g(2))g′(2) = f′(5)(−3) = −12.
6. Summary
- Differentiate a composition from the outside inward.
- Multiply by the derivative of each inner layer.
- Choose the first rule from the expression's outermost operation.
- Evaluate outer derivatives at the inner output when using numerical data.