The Chapter in One Idea

Everything in this chapter is one statement wearing different costumes:

f(x)f'(x) is the instantaneous rate of change of ff — and geometrically, the slope of the tangent.

From this: rates of change of physical quantities (Section 1), where a function rises and falls (Section 2), where its peaks and valleys sit (Section 3), and how to optimise real quantities (Section 4).

Exam weight at a glance: CBSE Boards — typically one 2-mark rate/marginal question, one 3-mark monotonicity or local-extrema question, and a 5-mark optimisation problem. JEE Main — 1-2 questions, almost always monotonicity (often with a parameter) or maxima-minima. JEE Advanced adds tangents/normals and MVT (see JEE Corner).

Rate of Change — Formula Card

  • dydx\frac{dy}{dx} = rate of change of yy with respect to xx; dydxx=x0\left.\frac{dy}{dx}\right|_{x=x_0} = rate at the instant x=x0x = x_0.
  • dydx>0\frac{dy}{dx} > 0: yy increases as xx increases; dydx<0\frac{dy}{dx} < 0: yy decreases. A decreasing quantity enters equations with a negative rate.
  • Related rates (Chain Rule): if x=f(t)x = f(t), y=g(t)y = g(t), then dydx=dy/dtdx/dt\frac{dy}{dx} = \dfrac{dy/dt}{dx/dt} (for dxdt0\frac{dx}{dt} \neq 0). In practice: differentiate the connecting equation with respect to tt, substitute values after differentiating.
  • Marginal cost / revenue: MC =dCdx= \frac{dC}{dx}, MR =dRdx= \frac{dR}{dx} — differentiate, then substitute the output level.

Rates you should be able to write blindfolded:

Quantity Formula Rate
Circle area A=πr2A = \pi r^2 dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r\frac{dr}{dt}
Circumference C=2πrC = 2\pi r dCdt=2πdrdt\frac{dC}{dt} = 2\pi\frac{dr}{dt}
Sphere volume V=43πr3V = \frac{4}{3}\pi r^3 dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}
Cube volume / surface V=x3V = x^3, S=6x2S = 6x^2 3x2dxdt3x^2\frac{dx}{dt}, 12xdxdt12x\frac{dx}{dt}
Cone volume V=13πr2hV = \frac{1}{3}\pi r^2 h eliminate rr or hh via the given proportion first

Key Point: dVdr\frac{dV}{dr} for a sphere equals its surface area 4πr24\pi r^2 — growth happens by "adding a skin".

Monotonicity — Formula Card

Definitions (on an interval II): increasing: x1<x2f(x1)<f(x2)x_1 < x_2 \Rightarrow f(x_1) < f(x_2); decreasing: x1<x2f(x1)>f(x2)x_1 < x_2 \Rightarrow f(x_1) > f(x_2); constant: f(x)=cf(x) = c.

Theorem 1 (the workhorse). ff continuous on [a,b][a,b], differentiable on (a,b)(a,b):

  • f(x)>0f'(x) > 0 on (a,b)(a,b) \Rightarrow ff increasing on [a,b][a,b]
  • f(x)<0f'(x) < 0 on (a,b)(a,b) \Rightarrow ff decreasing on [a,b][a,b]
  • f(x)=0f'(x) = 0 on (a,b)(a,b) \Rightarrow ff constant

Working method: find ff', solve f(x)=0f'(x) = 0, split the domain, make a sign table, conclude interval-by-interval. Never union two increasing intervals into one claim.

Templates to remember:

  • ff' a perfect square (e.g. 3(x1)2+13(x-1)^2 + 1 or 6(x1)26(x-1)^2) \Rightarrow monotonic despite f=0f' = 0 at a point — isolated zeros of ff' don't break strict monotonicity (x3x^3 is the model).
  • Parameter template: px2+qx+r0 x    p>0, q24pr0px^2 + qx + r \geq 0\ \forall x \iff p > 0,\ q^2 - 4pr \leq 0.
  • Inequality template: to show g>hg > h on (0,)(0, \infty): set F=ghF = g - h, check F(0)=0F(0) = 0 and F>0F' > 0. Classics: x1+x<log(1+x)<x\frac{x}{1+x} < \log(1+x) < x; sinx<x<tanx\sin x < x < \tan x on (0,π2)\left(0, \frac{\pi}{2}\right); ex>1+xe^x > 1 + x.

Maxima and Minima — Formula Card

  • Critical point: f(c)=0f'(c) = 0 or ff not differentiable at cc (corners count!). Only critical points can host local extrema (Theorem 2) — but not every critical point does (x3x^3 at 0).
  • First Derivative Test: through the critical point cc, ff' changes (+)()(+) \to (-): local max; ()(+)(-) \to (+): local min; no change: point of inflexion.
  • Second Derivative Test: f(c)=0f'(c) = 0 and f(c)<0f''(c) < 0: local max; f(c)>0f''(c) > 0: local min; f(c)=0f''(c) = 0: test fails — go back to the first derivative test.
  • Absolute extrema on [a,b][a, b] (closed-interval method): evaluate ff at all critical points in (a,b)(a,b) AND at both endpoints; the largest value is the absolute maximum, the smallest the absolute minimum. A continuous function on a closed interval always attains both (Theorem 5).
  • Optimisation recipe: name variables (draw the figure) → write the target → use the constraint to reach ONE variable → state the domain → derivative test → answer the actual question with units.

Key Point: "Local maximum value" means f(c)f(c), not cc. Always report both the point and the value.

Standard Results and JEE Bounds — Quick Card

Optimisation punchlines (learn the setups in Sections 4-6):

  • Fixed sum kk: product maximised at k2,k2\frac{k}{2}, \frac{k}{2}; sum of squares minimised at k2,k2\frac{k}{2}, \frac{k}{2}; xmynx^m y^n maximised at mkm+n,nkm+n\frac{mk}{m+n}, \frac{nk}{m+n}.
  • Rectangle in a circle: square; rectangle of fixed perimeter: square.
  • Open box from an a×aa \times a sheet: cut x=a6x = \frac{a}{6}.
  • Cylinder of given surface, max volume: h=2rh = 2r. Cylinder in a sphere (radius RR): h=2R3h = \frac{2R}{\sqrt 3}.
  • Cylinder in a cone: greatest curved surface at x=r2x = \frac{r}{2}; greatest volume at x=2r3x = \frac{2r}{3} (height h3\frac{h}{3}).
  • Cone in a sphere: height 4r3\frac{4r}{3}, volume 827\frac{8}{27} of the sphere.
  • Cone of given slant height, max volume: α=tan12\alpha = \tan^{-1}\sqrt{2}; given total surface, max volume: α=sin113\alpha = \sin^{-1}\frac{1}{3}; given volume, least curved surface: h=2rh = \sqrt{2}\,r.
  • Wire of length LL → square + circle, minimum total area: square piece 4Lπ+4\frac{4L}{\pi+4}, circle piece πLπ+4\frac{\pi L}{\pi+4}.

JEE bounds to quote:

  • asinx+bcosx[a2+b2,a2+b2]a\sin x + b\cos x \in [-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]
  • x+a2x2ax + \frac{a^2}{x} \geq 2a (x>0x > 0)
  • logxx\frac{\log x}{x} peaks at ee (value 1e\frac{1}{e}); x1/xx^{1/x} peaks at ee (value e1/ee^{1/e}); xxx^x bottoms at 1e\frac{1}{e} (value e1/ee^{-1/e})
  • Minimum distance point-to-curve: minimise D2D^2; the shortest join is normal to the curve.

Syllabus reminder: JEE Main tests rate of change, monotonicity, maxima-minima only; tangents/normals and Rolle's/Lagrange's MVT are JEE Advanced territory (see JEE Corner).

Last-Minute Mistake Checklist

Before the exam, scan this list — each item is a real mark lost by thousands of students every year:

  1. Decreasing quantities need negative rates — "length decreasing at 3 cm/min" means dxdt=3\frac{dx}{dt} = -3.
  2. Differentiate first, substitute after. Plugging r=10r = 10 before differentiating freezes the variable and gives 0.
  3. Monotonic intervals are reported separately, never as a union with one verdict.
  4. Critical points include non-differentiable points — corners of x|x|-type functions.
  5. Second derivative test failing (f=0f'' = 0) is not a conclusion — switch to the first derivative test.
  6. Endpoints belong in every absolute-extrema table. The champion is an endpoint surprisingly often.
  7. Reject impossible roots with a reason (negative lengths, cuts larger than the sheet) — there's a mark for it.
  8. Answer the question asked — dimensions AND the max/min value, with units.
  9. In MCQs, verify with a cheap check: AM-GM for minima, a nearby test value for monotonicity, symmetry for two-answer distance problems.
  10. Budget time: a Board 5-marker deserves 8-9 minutes; a JEE question 2 minutes. If the setup isn't clear in 30 seconds, skip and return.

Done revising? Take the Section 8 mock drill under exam timing — that's the real test of readiness.