Core Concept
The derivative dy/dx represents the rate of change of y with respect to x. If y = f(x), then f′(x₀) gives the instantaneous rate of change at x = x₀. This concept applies to physical quantities like area, volume, speed, and marginal cost — all of which appear in AAI ATC exam problems.
• Chain Rule (two variables varying w.r.t. t):
dy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0
• Area of circle: A = πr² → dA/dt = 2πr · (dr/dt)
• Volume of sphere: V = (4/3)πr³ → dV/dt = 4πr² · (dr/dt)
• Volume of cone: V = (1/3)πr²h → use Chain Rule accordingly
• Marginal Cost (MC) = dC/dx | Marginal Revenue (MR) = dR/dx
Important Results to Remember
- For a circle expanding at rate dr/dt: dA/dt = 2πr · dr/dt
- For a cube with edge x: dV/dt = 3x² · dx/dt and dS/dt = 12x · dx/dt
- dy/dx > 0 when y increases with x; dy/dx < 0 when y decreases
- Marginal cost at x = 3 means evaluate dC/dx at x = 3
NCERT Figure — Rate of Change Diagram
Insert NCERT figure showing expanding circle / shadow problems here
🎯 Practice MCQs — Rate of Change
S = 6x² → dS/dt = 12x · dx/dt = 12×10×(3/100) = 3.6 cm²/s
At x = 7: MR = 26(7) + 26 = 182 + 26 = 208
Definition (Monotonicity)
A function f is increasing on interval I if x₁ < x₂ ⟹ f(x₁) < f(x₂) for all x₁, x₂ ∈ I.
A function f is decreasing on interval I if x₁ < x₂ ⟹ f(x₁) > f(x₂) for all x₁, x₂ ∈ I.
• f is increasing on [a,b] if f′(x) ≥ 0 for each x ∈ (a, b)
• f is decreasing on [a,b] if f′(x) ≤ 0 for each x ∈ (a, b)
• f is constant on [a,b] if f′(x) = 0 for each x ∈ (a, b)
Working Method — Finding Intervals
- Step 1: Find f′(x)
- Step 2: Set f′(x) = 0, find critical points
- Step 3: These points divide the real line into intervals
- Step 4: Check sign of f′(x) in each interval
- Step 5: f increasing where f′(x) > 0; decreasing where f′(x) < 0
NCERT Fig 6.1 & Fig 6.2 — Graph of x² showing increasing/decreasing regions
Insert NCERT figures showing strictly increasing, strictly decreasing, and constant function graphs
🎯 Practice MCQs — Increasing & Decreasing
f′(x) = 0 at x = −2, 3
Sign test: f′(x) < 0 for x ∈ (−2, 3)
∴ f is decreasing on (−2, 3)
dy/dx > 0 when 0 < x < 2 (both x > 0 and 2−x > 0)
∴ y is increasing on (0, 2)
cos x > 0 for x ∈ (0, π/2)
∴ sin x is increasing on (0, π/2)
Core Concept
The slope of the tangent to curve y = f(x) at point (x₀, y₀) is f′(x₀). The normal to the curve at the same point is perpendicular to the tangent, with slope −1/f′(x₀).
y − y₁ = f′(x₁)(x − x₁)
Normal at (x₁, y₁) [if f′(x₁) ≠ 0]:
y − y₁ = −[1/f′(x₁)](x − x₁)
Special cases:
• If f′(x₁) = 0 → tangent is horizontal (y = y₁)
• If f′(x₁) = ∞ → tangent is vertical (x = x₁)
NCERT Diagram — Tangent and Normal at a Point on a Curve
Insert diagram showing tangent line, normal line, angle of inclination at point P(x₀, y₀)
🎯 Practice MCQs — Tangents & Normals
At x = 2: slope = 3(4) − 1 = 12 − 1 = 11
Slope of normal = −1/2
Normal: y − 1 = −(1/2)(x − 1) → 2y − 2 = −x + 1 → x + 2y − 3 = 0
dy/dx = 2x = 8 → x = 4, y = 16
Point: (4, 16)
Local Maxima & Minima — Definitions
Local Maxima: f(c) ≥ f(x) for all x in (c−h, c+h) for some h > 0. The value f(c) is the local maximum value.
Local Minima: f(c) ≤ f(x) for all x in (c−h, c+h) for some h > 0. The value f(c) is the local minimum value.
A critical point is where f′(c) = 0 OR f is not differentiable at c.
• f′ changes + to − at c → Local Maxima at c
• f′ changes − to + at c → Local Minima at c
• f′ does not change sign → Point of Inflection
Second Derivative Test (easier for exams):
• f′(c) = 0 and f″(c) < 0 → Local Maxima at c
• f′(c) = 0 and f″(c) > 0 → Local Minima at c
• f′(c) = 0 and f″(c) = 0 → Test fails; use FDT
Absolute Maximum & Minimum on [a, b]
- Step 1: Find all critical points in (a, b) where f′(x) = 0
- Step 2: Evaluate f at critical points AND at endpoints a, b
- Step 3: Largest value = Absolute Maximum; Smallest = Absolute Minimum
NCERT Fig 6.11, 6.12, 6.14 — Local Maxima, Minima & Point of Inflection
Insert NCERT figures showing hills, valleys and the geometric interpretation of turning points
NCERT Fig 6.19 — Absolute Max and Min on Closed Interval [a, d]
Insert diagram showing absolute maximum at a, absolute minimum at d with local extrema at b, c
🎯 Practice MCQs — Maxima & Minima
f′(x) = 0 at x = 1, −1
f″(x) = 6x → f″(−1) = −6 < 0 → x = −1 is local maxima
f(−1) = −1 + 3 + 3 = 5
f(1) = 24, f(2) = 29, f(3) = 28, f(5) = 56
Absolute maximum = 56 at x = 5
Profit P = Revenue − Cost = 5x − x²/100 − x/5 − 500 = (24/5)x − x²/100 − 500
P′(x) = 24/5 − x/50 = 0 → x = 240
P″(x) = −1/50 < 0 → maximum at x = 240
Rate of Change
dy/dx = f′(x) is the instantaneous rate of change. Use Chain Rule when both variables depend on t.
Monotonicity
f increasing ↔ f′(x) ≥ 0. f decreasing ↔ f′(x) ≤ 0. Find critical points, then check sign.
Tangent & Normal
Slope of tangent = f′(x₁). Slope of normal = −1/f′(x₁). Use point-slope form for equations.
Maxima & Minima
SDT: f″(c) < 0 → max; f″(c) > 0 → min. For closed interval, also check endpoints.
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