Difficulty: Intermediate | Prerequisites: Sections 10.1 through 11.3 study notes (limits, basic differentiation rules)
This block extends your differentiation toolkit and then puts it to work. The chain rule lets you differentiate compositions (functions inside functions), and the log/exponential rules open up a whole class of growth and decay models. Implicit differentiation handles curves that are not written as y = f(x). From 12.1 onward, you stop computing derivatives for their own sake and start using them to answer questions: where are the peaks and valleys of a function, how does a graph curve, what happens when two related quantities both change with time, and how sensitive is demand to price changes?
The chain rule differentiates composite functions by multiplying the outer derivative by the inner derivative. Log and exponential derivatives have their own formulas. With these tools in hand, you can find maxima and minima, sketch complete graphs, solve related-rates problems, and analyse elasticity of demand.
Chain rule
The rule for differentiating a composite function: d/dx [f(g(x))] = f'(g(x)) · g'(x). Think of it as "derivative of the outside, evaluated at the inside, times the derivative of the inside."
Implicit differentiation
A technique for finding dy/dx when y is not isolated on one side of the equation. Differentiate both sides with respect to x, treating y as a function of x (so every time you differentiate a y-term, you attach a dy/dx), then solve for dy/dx.
Absolute (global) maximum/minimum
The largest or smallest value a function takes on its entire domain (or on a closed interval).
Relative (local) maximum/minimum
A point where the function value is larger (or smaller) than at all nearby points, but not necessarily the largest (or smallest) overall.
Critical point
A value x = c in the domain of f where f'(c) = 0 or f'(c) is undefined. Every relative extremum occurs at a critical point (but not every critical point is an extremum).
Second derivative
The derivative of the derivative: f''(x) = d/dx [f'(x)]. It measures the rate at which the slope is changing.
Concavity
A function is concave up where f''(x) > 0 (the graph curves upward, like a bowl) and concave down where f''(x) < 0 (curves downward, like a hill).
Inflection point
A point where concavity changes, from up to down or down to up. At an inflection point, f''(x) = 0 or is undefined.
Related rates
Problems where two or more quantities change with respect to time, and you use implicit differentiation with respect to t to relate their rates of change.
Elasticity of demand
E = -(p/q) · (dq/dp), where q is the demand at price p. If E > 1, demand is elastic (price-sensitive). If E < 1, demand is inelastic. If E = 1, demand is unit elastic.
If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x).
Identify the outer function and the inner function. Differentiate the outer (leaving the inner untouched), then multiply by the derivative of the inner.
Example: d/dx [(3x + 1)⁵] = 5(3x + 1)⁴ · 3 = 15(3x + 1)⁴.
The chain rule combines freely with the product and quotient rules. When a problem has both a product and a composition, use the product rule first, then the chain rule on whichever factor is a composition.
d/dx [eˣ] = eˣ. The exponential function is its own derivative.
d/dx [e^(g(x))] = e^(g(x)) · g'(x). (Chain rule applied.)
d/dx [ln x] = 1/x, for x > 0.
d/dx [ln(g(x))] = g'(x) / g(x). (Chain rule applied.)
d/dx [aˣ] = aˣ · ln a, for any constant a > 0.
d/dx [log_a(x)] = 1 / (x · ln a).
These come up constantly in business models involving exponential growth, compound interest, and decay.
Used when the equation relating x and y is not (or cannot easily be) solved for y.
Differentiate every term with respect to x. Whenever you differentiate a term containing y, multiply by dy/dx (because y is a function of x).
Collect all dy/dx terms on one side, factor out dy/dx, and solve.
Example: given x² + y² = 25, differentiate to get 2x + 2y(dy/dx) = 0, so dy/dx = -x/y.
The result usually involves both x and y, which is normal. You need a specific point (x, y) on the curve to get a numerical slope.
To find relative extrema: compute f'(x), set it equal to zero, and solve for critical points. Also check where f'(x) is undefined.
First derivative test: check the sign of f' on each side of a critical point. If f' changes from + to -, you have a relative maximum. From - to +, a relative minimum. No sign change means no extremum.
On a closed interval [a, b], the absolute max and min occur either at a critical point or at an endpoint. Evaluate f at every critical point and both endpoints, then compare.
Optimisation word problems: identify the quantity to maximise or minimise, write it as a function of one variable (using a constraint equation if needed), differentiate, set equal to zero, solve, and verify.
Typical problems: maximising revenue, minimising cost, maximising area given a perimeter constraint, and maximising profit.
Always check that your answer is the type of extremum asked for (max vs. min) and that it lies within the problem's domain.
The second derivative f''(x) tells you about concavity.
f''(x) > 0: concave up (slope is increasing).
f''(x) < 0: concave down (slope is decreasing).
Second derivative test for extrema: at a critical point c where f'(c) = 0, if f''(c) > 0 the point is a relative minimum; if f''(c) < 0, a relative maximum; if f''(c) = 0, the test is inconclusive (fall back to the first derivative test).
Inflection points occur where f''(x) changes sign. Set f''(x) = 0 and check for a sign change.
A complete graph analysis combines everything: domain, intercepts, critical points, increasing/decreasing intervals, concavity, inflection points, and behaviour at the edges of the domain.
Steps: find f'(x) for critical points and monotonicity, find f''(x) for concavity and inflection points, evaluate f at key points, and sketch.
Horizontal asymptotes come from limits at ±∞. Vertical asymptotes come from values where the denominator is zero and the limit is ±∞.
When two or more quantities depend on time t, their rates of change (dx/dt, dy/dt, etc.) are related through an equation connecting x, y, etc.
Strategy:
Draw a picture and label all variables.
Write an equation relating the variables (geometry, area, volume, Pythagorean theorem, etc.).
Differentiate both sides with respect to t using the chain rule.
Substitute known values and solve for the unknown rate.
Common setups: expanding/contracting circles or spheres, ladders sliding down walls, filling tanks, shadows lengthening.
Always note units. Rates are typically "per unit time."
Elasticity measures how sensitive quantity demanded is to a change in price: E = -(p/q) · (dq/dp).
If E > 1 (elastic): a price increase causes revenue to fall; lowering the price increases revenue.
If E < 1 (inelastic): a price increase causes revenue to rise; the demand is not very price-sensitive.
If E = 1 (unit elastic): revenue is at a maximum with respect to price.
To find the price that maximises revenue, set E = 1 and solve for p.
Chain rule: d/dx [f(g(x))] = f'(g(x)) · g'(x)
Exponential: d/dx [eˣ] = eˣ; d/dx [e^(g(x))] = e^(g(x)) · g'(x)
Natural log: d/dx [ln x] = 1/x; d/dx [ln(g(x))] = g'(x) / g(x)
General exponential: d/dx [aˣ] = aˣ · ln a
General log: d/dx [log_a(x)] = 1 / (x · ln a)
Second derivative test: at f'(c) = 0, f''(c) > 0 means relative min, f''(c) < 0 means relative max
Elasticity: E = -(p/q) · (dq/dp); revenue maximised when E = 1
Optimisation is everywhere. Businesses use it to set prices that maximise profit, engineers use it to minimise material in packaging (e.g. finding the dimensions of a box that minimise surface area for a given volume). Related rates appear whenever two measurements change simultaneously: a balloon inflating, a shadow moving, or a stock price and a derivative contract both shifting with time. Elasticity is the standard tool economists use to decide whether raising or lowering a price will bring in more revenue.
Students often forget the inner derivative when applying the chain rule. d/dx [e^(3x)] is 3e^(3x), not e^(3x). Missing that factor of 3 is one of the most common errors on exams.
Students assume every critical point is an extremum. A critical point where f' does not change sign (e.g. f(x) = x³ at x = 0) is neither a max nor a min.
In related-rates problems, students substitute numerical values too early. You must differentiate with respect to t before plugging in specific numbers.
Students confuse concave up with increasing. A function can be increasing and concave down (its slope is positive but shrinking), or decreasing and concave up (its slope is negative but getting less steep).
⚠️ Chain rule problems are guaranteed on Exam 2. Practise nested compositions (e.g. e raised to a polynomial, ln of a product).
⚠️ Optimisation word problems are high-value questions. The setup (writing the function from the word problem) is usually the hardest part.
⚠️ Know both the first and second derivative tests for classifying critical points. The exam may specify which test to use.
⚠️ Related rates: expect one multi-step problem. Draw the picture, label every variable, write the equation, then differentiate.
⚠️ Elasticity: know the formula and be able to classify demand as elastic, inelastic, or unit elastic at a given price.
Fill in the blank: d/dx [ln(5x)] = ____.
True or false: If f'(c) = 0 and f''(c) < 0, then x = c is a relative minimum.
Fill in the blank: In a related-rates problem, you differentiate with respect to ____ before substituting known values.
True or false: If elasticity E = 0.6, demand is elastic.
Fill in the blank: d/dx [e^(4x)] = ____.
Answers: 1. 1/x (the 5s cancel: 5/(5x) = 1/x). 2. False (f''(c) < 0 means relative maximum). 3. Time, t. 4. False (E < 1 means inelastic). 5. 4e^(4x).
Q: Differentiate f(x) = (2x³ + 1)⁴.
A: Using the chain rule: f'(x) = 4(2x³ + 1)³ · 6x² = 24x²(2x³ + 1)³.
Q: Find the critical points of f(x) = x³ - 12x + 5 and classify each.
A: f'(x) = 3x² - 12 = 3(x² - 4) = 3(x - 2)(x + 2). Critical points at x = 2 and x = -2. f''(x) = 6x. f''(2) = 12 > 0, so x = 2 is a relative minimum. f''(-2) = -12 < 0, so x = -2 is a relative maximum.
Q: A circular oil spill has radius r increasing at 3 m/min. How fast is the area increasing when r = 10 m?
A: A = πr². dA/dt = 2πr · dr/dt = 2π(10)(3) = 60π m²/min.
Q: Demand is q = 200 - 4p. Find the elasticity at p = 25 and classify.
A: dq/dp = -4. At p = 25, q = 200 - 100 = 100. E = -(25/100)(-4) = 1. Demand is unit elastic at p = 25, so revenue is maximised there.
Q: Given x² + xy = 10, find dy/dx.
A: Differentiate: 2x + y + x(dy/dx) = 0. Solve: dy/dx = -(2x + y)/x.
The chain rule is essential for integration by substitution (Section 13.2), which is essentially the chain rule run in reverse. Optimisation techniques reappear whenever you need to find a maximum or minimum of an integral expression. Concavity and inflection points connect to the shape of cumulative functions you will meet in the integration chapters. Elasticity ties back to marginal revenue and provides a pricing-side complement to the marginal analysis from 11.2.
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