Premed · Premed · Calculus 1
Lecture 6: The Chain Rule
Calculus I — Differential Calculus
Learning Objectives
By the end of this lecture, students will be able to:
- State and apply the chain rule for composite functions
- Identify the "inner" and "outer" functions in a composition
- Apply the chain rule in Leibniz notation
- Combine the chain rule with the product, quotient, and power rules
- Differentiate functions involving multiple nested compositions
Lecture Content
I. Motivation and Statement of the Chain Rule
Many functions encountered in practice are compositions of simpler functions. For example, h(x) = (3x^2 + 1)^5 is the composition of the outer function f(u) = u^5 with the inner function g(x) = 3x^2 + 1. The chain rule tells us how to differentiate such compositions.
The formal statement is as follows: if g is differentiable at x and f is differentiable at g(x), then the composite function h(x) = f(g(x)) is differentiable at x, and h'(x) = f'(g(x)) * g'(x). In words, this says "take the derivative of the outer function evaluated at the inner function, then multiply by the derivative of the inner function."
II. The Chain Rule in Leibniz Notation
If y = f(u) and u = g(x), the chain rule takes the elegant form dy/dx = (dy/du) * (du/dx). This notation looks like "cancellation" of the du terms. While this is not rigorous in terms of fractions, it provides excellent intuition and serves as a reliable mnemonic. The chain rule extends naturally to longer chains of composition: if y = f(u), u = g(v), and v = h(x), then dy/dx = (dy/du)(du/dv)(dv/dx).
III. Applying the Chain Rule — The General Power Rule
The General Power Rule is the chain rule applied to powers of functions: d/dx [g(x)]^n = n [g(x)]^{n-1} g'(x). This arises from composing f(u) = u^n with u = g(x).
Consider three examples. First, d/dx [(3x^2 + 1)^5]: the outer function is u^5 and the inner function is 3x^2 + 1. Applying the chain rule gives 5(3x^2 + 1)^4 6x = 30x(3x^2 + 1)^4. Second, d/dx [sqrt(x^2 + 4)] = d/dx [(x^2 + 4)^{1/2}] = (1/2)(x^2 + 4)^{-1/2} 2x = x / sqrt(x^2 + 4). Third, d/dx [1/(2x - 5)^3] = d/dx [(2x - 5)^{-3}] = -3(2x - 5)^{-4} * 2 = -6/(2x - 5)^4.
<image>A diagram illustrating the chain rule as a "function machine." Input x enters a box labeled "g (inner function)" producing u = g(x), which then enters a box labeled "f (outer function)" producing y = f(u) = f(g(x)). Below, the derivative chain is shown: dy/dx = (dy/du) times (du/dx), with arrows connecting the corresponding rates. Each rate is labeled: du/dx = g'(x) and dy/du = f'(u) = f'(g(x)). Title: "The chain rule as a pipeline of rates."</image>
IV. Combining the Chain Rule with Other Rules
In practice, the chain rule frequently appears alongside the product and quotient rules. When differentiating a product like f(x) g(h(x)), apply the product rule first: the result is f'(x) g(h(x)) + f(x) g'(h(x)) h'(x). Similarly, when the quotient rule produces terms involving composite functions, the chain rule must be applied to each one.
As an example, consider d/dx [x^2 (3x + 1)^4]. The product rule gives 2x(3x + 1)^4 + x^2 4(3x + 1)^3 * 3 = 2x(3x + 1)^4 + 12x^2(3x + 1)^3. Factoring yields 2x(3x + 1)^3 [(3x + 1) + 6x] = 2x(3x + 1)^3(9x + 1).
V. Multiple Applications of the Chain Rule
For deeply nested functions, the chain rule must be applied multiple times. Consider d/dx [sqrt(1 + (2x)^3)]. Setting u = 1 + (2x)^3, the function is u^{1/2}, so the derivative begins as (1/2)u^{-1/2} d/dx[1 + (2x)^3]. To differentiate (2x)^3 = 8x^3, we get 24x^2 (or equivalently, apply the chain rule to get 3(2x)^2 2 = 24x^2). The final result is (1/2)(1 + 8x^3)^{-1/2} * 24x^2 = 12x^2 / sqrt(1 + 8x^3).
VI. Common Mistakes with the Chain Rule
The most frequent error is forgetting the inner derivative. For instance, d/dx [(3x + 1)^5] is not simply 5(3x + 1)^4; it must be multiplied by 3, the derivative of the inner function. Another mistake is applying the chain rule when it is unnecessary: d/dx [x^5] = 5x^4 requires no chain rule because the inner function is just x, whose derivative is 1. Students should also be careful about confusing the order of operations: always differentiate the outer function first, then multiply by the derivative of the inner function. Finally, always simplify the result after applying the chain rule to catch errors and recognize patterns.
<image>A worked example shown step-by-step with color coding. The function y = (x^3 + 2x)^4 is differentiated. Step 1: Identify outer (u^4, colored blue) and inner (x^3 + 2x, colored red). Step 2: Apply chain rule — write 4(x^3 + 2x)^3 in blue, then multiply by (3x^2 + 2) in red. Step 3: Final answer y' = 4(3x^2 + 2)(x^3 + 2x)^3. A "common mistake" box below shows the incorrect answer y' = 4(x^3 + 2x)^3 with an X mark and the note "forgot the inner derivative."</image>
VII. Why the Chain Rule Works — Intuition
The chain rule has a natural interpretation in terms of rates. If u changes at a rate du/dx with respect to x, and y changes at a rate dy/du with respect to u, then y changes at a rate (dy/du)(du/dx) with respect to x. An analogy makes this concrete: if a car is moving at 60 km/h and the odometer converts kilometers to miles at 0.62 mi/km, then the mileage changes at 60 * 0.62 = 37.2 mi/h. The chain rule is simply the multiplication of rates along a chain of dependencies.
The formal proof relies on the fact that differentiability implies local linearity. Near any point, a differentiable function behaves approximately like a linear function, and the composition of two approximately linear functions is again approximately linear, with slope equal to the product of the individual slopes.
