9/15 Notes on Chain Rule, Inner/Outer Functions, and Applications (Exam Prep)

Key Concepts

  • Function composition and chain rule are central when a function has an inner function and an outer function.
  • Notation:
    • Inner function: g(x) (sometimes called u)
    • Outer function: f(u)
    • Composition: (f ∘ g)(x) = f(g(x))
  • Chain rule (two-function form): if y = f(u) and u = g(x), then
    • dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
    • More generally, if y = f(g(x)), then dydx=f′(g(x))⋅g′(x)\frac{dy}{dx} = f'(g(x)) \cdot g'(x)
  • Understanding inner vs outer function helps with complex derivatives. Often we substitute and differentiate piece by piece, then multiply the derivatives.
  • Real-life connection: rate of change in business models (e.g., simple vs compound interest) provides intuition for how outputs change when inputs are scaled or compounded.

Basic rules and notations (derivatives)

  • Power rule (basic): if y = x^n, then
    • dydx=nxn−1\frac{dy}{dx} = n x^{n-1}
  • Exponential rule: if y = e^{u(x)}, then
    • dydx=eu(x)⋅dudx\frac{dy}{dx} = e^{u(x)} \cdot \frac{du}{dx}
  • If y = a^{u(x)} for constant a > 0, then
    • dydx=au(x)ln⁡(a)⋅dudx\frac{dy}{dx} = a^{u(x)} \ln(a) \cdot \frac{du}{dx}
  • Natural logarithm: if y = \ln(u(x)), then
    • dydx=1u(x)⋅dudx\frac{dy}{dx} = \frac{1}{u(x)} \cdot \frac{du}{dx}
  • Square root: if y = \sqrt{u(x)}, then
    • dydx=12u(x)⋅dudx\frac{dy}{dx} = \frac{1}{2\sqrt{u(x)}} \cdot \frac{du}{dx}
  • If y = [u(x)]^n, then
    • dydx=n[u(x)]n−1⋅dudx\frac{dy}{dx} = n [u(x)]^{n-1} \cdot \frac{du}{dx}
  • Logarithmic properties (optional check method): if y = \ln(u(x)) and u(x) = 20x, you can also use properties of logs to simplify before differentiating.

Worked examples

  • Example 1: derivative of basic polynomial
    • y = x^3
    • dydx=3x2\frac{dy}{dx} = 3x^2
  • Example 2: derivative of an exponential with linear exponent
    • y = e^{3x}
    • Let u = 3x ⇒ du/dx = 3
    • dydx=e3x⋅3=3e3x\frac{dy}{dx} = e^{3x} \cdot 3 = 3e^{3x}
  • Example 3: derivative of a composed exponential with a quadratic inner
    • y = e^{3x^2}
    • Let u = 3x^2 ⇒ du/dx = 6x
    • dydx=e3x2⋅6x=6xe3x2\frac{dy}{dx} = e^{3x^2} \cdot 6x = 6x e^{3x^2}
  • Example 4: a composition with a general exponential outer function
    • Suppose f(u) = e^{u}, g(x) = 3x^2. Then y = f(g(x)) = e^{3x^2}
    • Using chain rule: dy/dx = f'(g(x)) \cdot g'(x) = e^{3x^2} \cdot 6x
  • Example 5: natural log of a linear function with inner substitution
    • y = \ln\left( \frac{1}{3}x + 1 \right)
    • Let u = \frac{1}{3}x + 1 ⇒ du/dx = \frac{1}{3}
    • Then dy/dx = \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{(\frac{1}{3}x + 1)} \cdot \frac{1}{3} = \frac{1}{x+3}
  • Example 6: derivative of a square root with inner function
    • y = \sqrt{u}, with u = g(x)
    • dy/dx = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}
  • Example 7: a three-level chain rule case
    • Consider y = e^{e^{2x+3}}
    • Let v = 2x + 3, dv/dx = 2
    • Let u = e^{v}, du/dv = e^{v}
    • Outer function: y = e^{u}, dy/du = e^{u}
    • Then dy/dx = (dy/du) (du/dv) (dv/dx) = e^{u} \cdot e^{v} \cdot 2 = 2 e^{e^{2x+3}} e^{2x+3}

Practice strategies and workflow

  • Step-by-step recipe for chain rule problems
    • Identify inner function: u = g(x)
    • Identify outer function: y = f(u)
    • Compute du/dx and dy/du separately
    • Multiply: dy/dx = (dy/du) (du/dx)
    • If multiple layers, apply substitutions in sequence (u-substitution, then v-substitution, etc.)
  • Three-step or more chain rule constructions
    • Break into consecutive substitutions: let v = something in x, let u = function of v, let y = function of u, then apply: dy/dx = (dy/du)(du/dv)(dv/dx)
  • Importance of parentheses
    • In products of derivatives, use parentheses to avoid ambiguity (e.g., (dy/du)(du/dx) rather than dy/dux)
  • Two methods to differentiate logarithmic expressions (illustrative)
    • Chain rule method: dy/dx = (1/u) * du/dx
    • Logarithmic property method: ln(ab) = ln(a) + ln(b) for a,b > 0; differentiate term-by-term to get the same result
  • Common pitfalls to watch for
    • Treating exponents as if they were the same x without substitution
    • Forgetting a derivative of the inner function (g'(x))
    • Skipping steps in multi-layer chains
    • Misplacing or miscounting constants when differentiating exponential functions

Business and real-world connections

  • Simple vs compound interest as intuition for composition
    • Simple interest: A = P(1 + r t)
    • Compound interest: A = P(1 + r)^t
    • Continuous compounding (alternative model): A = P e^{rt}
  • Sensitivity and rate of change concepts
    • If A = P(1 + r)^t, then dA/dt = P (1 + r)^t \ln(1 + r)
    • If continuously compounded: A = P e^{rt} ⇒ dA/dt = P r e^{rt}
  • When you change the rate or time, how the output changes is governed by the derivative, which is a direct application of the chain rule (different layers of the model correspond to inner/outer functions).
  • Preview of later topics (contextual relevance): elasticity, demand/supply curves, consumer/producer surplus, and how derivatives underpin these concepts.

Exam and course logistics mentioned in the session

  • Exam format and timing
    • Date: Monday, 22nd (as mentioned in the session)
    • Duration: 50 minutes (with a 25-minute window of extra time mentioned informally)
    • Format: Paper-based; no notes allowed
  • Study materials and practice
    • Review material and practice exam posted in announcements
    • Practice problems cover chain rule, inner/outer functions, and combinations like nested exponentials
  • Advice from the instructor
    • If you understand the core method, you can solve many problems quickly; practice helps automate the process
    • Two approaches to certain problems: direct chain-rule method vs algebraic/logarithmic manipulation; both yield the same result
    • Treat mathematics as a skill that benefits from practice; the instructor emphasizes practice to gain fluency

Quick reference formulas (summary)

  • Chain rule (two functions)
    • dydx=f′(g(x))⋅g′(x)\frac{dy}{dx} = f'(g(x)) \cdot g'(x)
  • General chain rule for inner/outer substitution
    • If y = f(u) and u = g(x), then
    • dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
  • Derivative examples
    • ddxe3x=3e3x\frac{d}{dx} e^{3x} = 3 e^{3x}
    • ddxe3x2=e3x2⋅6x\frac{d}{dx} e^{3x^2} = e^{3x^2} \cdot 6x
    • ddxu=12u⋅dudx\frac{d}{dx} \sqrt{u} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}
    • ddx[u(x)]n=n[u(x)]n−1⋅dudx\frac{d}{dx} [u(x)]^{n} = n [u(x)]^{n-1} \cdot \frac{du}{dx}
    • ddxln⁡(u)=1u⋅dudx\frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx}
  • Example: derivative of a logarithm with linear inner
    • If y = \ln\left( \frac{1}{3}x + 1 \right), then
    • dydx=1(13x+1)⋅13=1x+3\frac{dy}{dx} = \frac{1}{(\frac{1}{3}x + 1)} \cdot \frac{1}{3} = \frac{1}{x+3}
  • Three-level chain rule example
    • y = e^{e^{2x+3}}
    • dy/dx = 2 e^{e^{2x+3}} e^{2x+3}
  • Compound interest derivative (example)
    • For y = P (1 + r)^t, with r constant:
    • dydt=P(1+r)tln⁡(1+r)\frac{dy}{dt} = P (1 + r)^t \ln(1 + r)

Note: The notes above mirror the live lecture flow, including the chain-rule-based approach, inner/outer function distinction, and several worked examples. Where the lecturer’s explicit numeric steps were inconsistent (e.g., an additive misstatement in a reinforcement example), the standard correct forms and the intended learning points have been captured and clarified for exam preparation.