Lesson 12 of 12 · 11 min

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Must-know facts

10 facts

1dy/dx is the rate of change of y with x; with time, dy/dt = (dy/dx)(dx/dt).
2Related rates: write the relation, differentiate in t, then substitute.
3Marginal cost = dC/dx, marginal revenue = dR/dx.
4f′ > 0 on (a, b) ⇒ increasing on [a, b]; f′ < 0 ⇒ decreasing; f′ = 0 ⇒ constant.
5Intervals of monotonicity come from the sign of f′ between its zeros.
6Critical point: f′(c) = 0 or f′(c) undefined.
7First test: + to − is a local max, − to + a local min, no change means inflection.
8Second test: f′(c) = 0 with f″(c) < 0 is a max, f″(c) > 0 a min, f″(c) = 0 no verdict.
9Closed interval: compare f at the critical points and at both ends.
10Tangent slope f′(x₀); normal slope −1/f′(x₀) (JEE).

Common traps

Where marks are lost

Substituting the given values before differentiating.

Differentiate the general relation first; a quantity frozen early has rate 0.

Forgetting the end points on a closed interval.

f = 2x³ − 15x² + 36x + 1 on [1, 5] has its absolute maximum 56 at the end x = 5, not at a critical point.

Calling every critical point an extremum.

x³ and 2x³ − 6x² + 6x + 5 have f′ = 0 without a sign change; those are inflections.

Reading f″(c) = 0 as an inflection or as a minimum.

The second test is silent then. Use the sign of f′ on both sides.

Ignoring points where f′ does not exist.

3 + |x| has its minimum at 0 where there is no derivative; x^(1/3)-type terms give such points too.

Stating the local maximum as the answer when the absolute maximum is asked.

Local means only nearby. Compare all candidates when the question says greatest or least.

Keeping an optimisation root that the physical set-up forbids.

For the 3 × 8 box, x must be below 3/2, so x = 3 is rejected.

Writing an increasing interval as a union and calling f increasing on it.

4x³ − 6x² − 72x + 30 increases on (−∞, −2) and on (3, ∞) separately, not on their union.

Formulas

7 to know

Rate chain

dy/dt = (dy/dx)·(dx/dt)

Also dy/dx = (dy/dt)/(dx/dt) where dx/dt ≠ 0.

Circle and sphere

dA/dt = 2πr dr/dt, dV/dt = 4πr² dr/dt

From A = πr² and V = 4πr³/3.

Marginal cost, revenue

MC = dC/dx, MR = dR/dx

Profit peaks where MR = MC and P″ < 0.

Monotonicity

f′ > 0 ⇒ increasing, f′ < 0 ⇒ decreasing

f continuous on [a, b], differentiable on (a, b).

First derivative test

f′: + → − max, − → + min

No sign change: point of inflection.

Second derivative test

f′(c) = 0: f″(c) < 0 max, f″(c) > 0 min

f″(c) = 0 gives no verdict.

Tangent and normal (JEE)

y − y₀ = m(x − x₀), m = f′(x₀); normal slope −1/m

Needs m ≠ 0 for the normal slope.

Key terms

6 terms

Marginal cost: The rate of change of total cost with output, dC/dx.
Strictly increasing: x₁ < x₂ always gives f(x₁) < f(x₂) on the interval.
Critical point: A point of the domain where f′ is zero or does not exist.
Local maximum: A value not exceeded by f anywhere in some open interval around the point.
Point of inflection: A point where f′ = 0 but f′ keeps its sign, so f has no extremum.
Absolute maximum: The largest value of f on the whole interval.