Phase 2 - Lesson 2.3

Differentiation Rules, Implicit Differentiation, and Related Rates

The algebra of derivatives - product, quotient and chain rules - plus differentiating relations implicitly and linking rates that move together.

⏱ 55 min● Intermediate🔗 Prereqs: 2.2
↖ Phase 2 hub
Builds on: 2.2 defined the derivative; here we build the toolkit that makes computing it fast.
Leads to: The chain rule reappears as Itô’s lemma (Phase 9); implicit differentiation underlies option greeks and yield curves.

Learning Objectives

Click a status chip to cycle: Not started → In progress → Studied → Practiced → Needs review → Mastered.

Key Vocabulary

Power rule
d/dx x^n = n x^{n-1} for any real n.
Product rule
(fg)′ = f′g + fg′.
Quotient rule
(f/g)′ = (f′g − fg′)/g².
Chain rule
(f∘g)′(x) = f′(g(x))·g′(x); differentiate outer, keep inner, times inner′.
Implicit differentiation
Differentiating both sides of F(x,y)=0 treating y as a function of x, then solving for y′.
Related rates
Using a constraint to relate the time-derivatives of interdependent quantities.

The four rules you compose everything from

With four rules you can differentiate essentially any elementary formula. The subtle one is the chain rule: a composition’s rate is the outer rate times the inner rate.

\[(fg)'=f'g+fg',\qquad \Big(\tfrac{f}{g}\Big)'=\frac{f'g-fg'}{g^2},\qquad (f\circ g)'(x)=f'(g(x))\,g'(x)\] (2.4)
Worked Example - Chain + product together
1
Differentiate \(h(x)=x^2 e^{3x}\). Product rule with \(f=x^2,\ g=e^{3x}\).
2
\(f'=2x\); for \(g=e^{3x}\) the chain rule gives \(g'=3e^{3x}\).
3
So \(h'(x)=2x\,e^{3x}+x^2\cdot 3e^{3x}=e^{3x}(2x+3x^2)\).

Implicit differentiation

When a curve is given by a relation you cannot (or would rather not) solve for \(y\), differentiate both sides treating \(y=y(x)\) and using the chain rule on every \(y\) term, then solve for \(y'\).

Worked Example - Slope on the circle x²+y²=25
1
Differentiate both sides in \(x\): \(2x+2y\,y'=0\).
2
Solve: \(y'=-x/y\).
3
At \((3,4)\): \(y'=-3/4\) - the tangent slope, obtained without ever solving for \(y\).

Related rates

If two quantities satisfy a relation and both change with time, differentiating the relation with respect to \(t\) links their rates. Example: a spherical balloon with \(V=\tfrac43\pi r^3\) gives \(\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\), so a known inflation rate \(dV/dt\) determines how fast the radius grows.

Interactive: check a derivative numerically

Common Mistakes to Avoid
  • Forgetting the inner derivative in the chain rule - writing (e^{3x})′=e^{3x} instead of 3e^{3x}.
  • Dropping the y′ factor when differentiating a y term implicitly (that factor is the chain rule).
  • Mixing up the quotient-rule sign: the numerator is f′g − fg′, in that order.
  • In related rates, plugging in specific numbers BEFORE differentiating, which freezes the variables.
Quant Practitioner Tips
  • Differentiate the general relation first; substitute the instantaneous values only at the end.
  • For nested compositions, peel from the outside in, multiplying each layer’s derivative.
  • A central difference with h≈1e−5 is a cheap, reliable check on any hand derivative.
  • Implicit differentiation gives greeks: differentiating a pricing relation yields sensitivities like delta and gamma.

Knowledge Check

Q1 Easy
The derivative of sin(x²) is:
cos(x²)
2x cos(x²)
2x sin(x²)
cos(2x)
Q2 Medium
Differentiating x²+y²=25 implicitly gives 2x+2yy′=0, so y′ equals:
−x/y
x/y
−y/x
−2x
Q3 Medium
A 13 ft ladder slides down a wall; the base moves out. Which relation should you differentiate w.r.t. t for a related-rates problem?
x+y=13
x²+y²=13²
xy=13
x²−y²=13

Practical Exercise

A company’s revenue is \(R(q)=q\,p(q)\) where price depends on quantity via \(p(q)=100-2q\). (a) Find marginal revenue \(R'(q)\). (b) At what \(q\) is marginal revenue zero, and what does that mean?

▶ Show full solution

(a) \(R(q)=q(100-2q)=100q-2q^2\), so by the power rule \(R'(q)=100-4q\). (You can also use the product rule on \(q\cdot p(q)\): \(R'=p(q)+q p'(q)=(100-2q)+q(-2)=100-4q\).)

(b) Set \(R'(q)=0\): \(100-4q=0\Rightarrow q=25\). Marginal revenue zero means selling one more unit adds no revenue - the revenue curve peaks here (a maximum, since \(R''=-4\lt 0\)). Beyond \(q=25\) the price cut outweighs the extra volume.

After the reveal, answer for yourself: Marginal revenue = 0 is the profit-relevant boundary studied in 2.5. Notice the product rule and expand-then-differentiate agree - a good self-check.

Lesson Summary

Four rules - power, product, quotient, chain - differentiate any elementary formula, with the chain rule (outer times inner) the one to master. Implicit differentiation handles relations you cannot solve for y, and related rates link time-derivatives through a shared constraint. A central difference verifies any of it numerically.

Formula Sheet Additions

Chain rule
\[(f\circ g)'(x)=f'(g(x))\,g'(x)\]
The composition rule that becomes Itô’s lemma in stochastic calculus.

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers
Q: State the chain rule and the single most common error in applying it.
A: (f∘g)′(x)=f′(g(x))g′(x): differentiate the outer function at the inner value, then multiply by the inner derivative. The common error is omitting the inner derivative g′(x).
Q: Outline the steps of a related-rates problem.
A: Write a relation among the quantities, differentiate it with respect to time (chain rule on each variable), then substitute the given instantaneous values and solve for the unknown rate - substituting numbers only after differentiating.

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