Chain Rule & Generalized Power Rule — Comprehensive Lecture Notes

Revisiting Derivative Rules & Their Limitations

• Opening remark: previously-covered rules (sum, product, quotient, basic power) all fail when the expression looks like “a function inside a function raised to a power.”
• The new scenario: expressions such as $(3x+1)^{0.1}$ cannot be tackled by naïve application of earlier rules.
• Motivation: need a systematic way to differentiate composite functions (i.e.
  one function plugged into another).

Composite‐Function Language

• Let $u(x)=3x+1$ (the “inside” or inner function).
• Let $h(x)=x^{0.1}$ (the “outside” or outer power function).
• Then $f(x)=h\big(u(x)\big)=(3x+1)^{0.1}.$
  • Terminology: “$f$ is the composite of $h$ and $u$.”
• Key insight: Differentiating $f$ requires accounting for both layers of variation—how $h$ changes and how $u$ changes.

Chain Rule (Formal Statement)

• For any composition $f(x)=h\big(u(x)\big)$:
  $\displaystyle f'(x)=h'\big(u(x)\big)\,u'(x).$
• Words: “Differentiate the outside, keep the inside unchanged, then multiply by the derivative of the inside.”
• Significance: Unifies the treatment of all nested expressions, is the backbone for more sophisticated techniques (implicit, inverse, substitutions in integration, etc.).

Example 1 — $f(x)=(3x+1)^{0.1}$

• Identify components:
  • $u(x)=3x+1\;\Rightarrow\;u'(x)=3$
  • $h(x)=x^{0.1}\;\Rightarrow\;h'(x)=0.1x^{-0.9}$ (power rule)
• Apply chain rule:
  $\displaystyle f'(x)=h'\big(u(x)\big)\,u'(x)=0.1(3x+1)^{-0.9}\,\cdot 3=0.3\,(3x+1)^{-0.9}.$
• Interpretation: outer power term produces the fractional exponent and negative power; inner linear term supplies the factor $3$.

Generalized Power Rule (Quick-Use Version)

• Derived directly from chain rule.
• For any differentiable $u(x)$ and real $n$:
  $\displaystyle \frac{d}{dx}\Big[u(x)^n\Big]=n\,u(x)^{n-1}\,u'(x).$
• Memorization recommended; drastically reduces algebraic overhead.

Example 2 — $y=(x^3+x)^{100}$

• Using generalized power rule:
  • $u(x)=x^3+x,\;u'(x)=3x^2+1,\;n=100.$
  • $\displaystyle y'=100\,(x^3+x)^{99}\,(3x^2+1).$
• Emphasized caution: order matters (outer power, inner base). Interchanging would misidentify $u,h$ and yield wrong derivative.

Example 3 — Square-Root Form: $f(x)=\sqrt{3x+1}$

• Rewrite as $(3x+1)^{1/2}$ so $n=1/2,\;u(x)=3x+1,\;u'(x)=3.$
• Derivative:
  $\displaystyle f'(x)=\tfrac12\,(3x+1)^{-1/2}\,\cdot3=\tfrac32\,(3x+1)^{-1/2}.$

Example 4 — Nested Sum & Negative Power: $g(x)=\Big[(x+1)^{-2.5}+3x\Big]^{-3}$

1. Outer structure: $n=-3,\;u(x)=(x+1)^{-2.5}+3x.$
2. Outer derivative via generalized power rule: $-3\,u(x)^{-4}\,u'(x).$
3. Compute $u'(x):$
   • Two-term sum ⇒ use sum rule.
   • First term: inner $v(x)=x+1,\;v'(x)=1,\;m=-2.5\Rightarrow -2.5\,(x+1)^{-3.5}.$
   • Second term derivative is $3.$
   • Thus $u'(x)=-2.5\,(x+1)^{-3.5}+3.$
4. Assemble final answer:
   $\displaystyle g'(x)=-3\Big[(x+1)^{-2.5}+3x\Big]^{-4}\Big[-2.5\,(x+1)^{-3.5}+3\Big].$

• Highlight: multi-layer compositions may involve repeated chain-rule invocations inside sums/products.

Business Application — Marginal Product (Profit vs. Workers)

• Profit function (consultant): $P(q)=4000q-0.46q^2-0.00001q^3.$ (Coefficients combined in lecture to $4000-0.92q-0.00003q^2$ when differentiated.)
• Output–labor link: $q(n)=100n.$ (Each worker produces $100$ lasers.)
• Goal: $\frac{dP}{dn}$ — “marginal product” of an additional worker.
• Chain-rule setup: $\displaystyle \frac{dP}{dn}=\frac{dP}{dq}\cdot\frac{dq}{dn}=P'(q)\,q'(n).$
  1. $P'(q)=4000-0.92q-0.00003q^2.$
  2. $q'(n)=100.$
• Substitute $q=100n$ afterward to express in labor units:
  $\displaystyle \frac{dP}{dn}=100\big(4000-0.92\cdot100n-0.00003\cdot100^2n^2\big).$
• Interpretation: predicts incremental profit for each additional assembly-line worker; vital for hiring decisions and cost-benefit analysis.

Related-Rates Example — Expanding Oil Slick

• Physical setup: Oil spill forms a circle with radius $r(t)$ growing at $\displaystyle \frac{dr}{dt}=2\ \text{mi/h}.$
• Area function: $A=\pi r^2.$
• Differentiate implicitly with respect to $t$:
  $\displaystyle \frac{dA}{dt}=2\pi r\frac{dr}{dt}.$
• Plug known rate and evaluate at $r=3\ \text{mi}$:
  $\displaystyle \frac{dA}{dt}=2\pi(3)(2)=12\pi\ \text{mi}^2/\text{h}.$
• Significance: Chain rule links geometric growth (area) to directly measurable linear growth (radius), a common pattern in physics & engineering.

Conceptual Connections & Practical Notes

• Chain rule unifies all previous derivative techniques; the product, quotient, and basic power rules are special-case outer layers.
• Evident across disciplines:
  • Economics (marginal analysis, elasticity).
  • Physics (kinematics with multi-stage dependencies).
  • Biology (population models with layered environmental factors).
• Ethical/Philosophical aside: In modeling real-world systems, ensure each composed function is valid in its domain; incorrect nesting may misrepresent realities (e.g., negative production levels or radii).

Numerical & Notational Reminders

• Always keep LaTeX negative exponents in parentheses—e.g.
  $(3x+1)^{-0.9}$ vs. $3x+1^{-0.9}$ (latter is ambiguous).
• Avoid swapping inner/outer roles; $u(h(x))\neq h(u(x)).$
• For fractional/negative powers, rewriting radicals as powers makes chain rule mechanical.

Study Checklist

• Memorize generalized power rule $n\,u^{\,n-1}u'.$
• Practice layered compositions where more than two functions nest.
• Reproduce the business & oil-slick examples without notes; these integrate conceptual recognition, rule application, and contextual interpretation.
• Cross-verify results via alternative methods (e.g., expand simple powers once, differentiate, match with chain-rule output).
• Stay mindful of domain constraints and units when applying to real data.