Applications of Derivatives

🟢 Lite — Quick Review

Rapid summary for last-minute revision before your exam.

Applications of Derivatives — Key Facts for JEE Main Equation of tangent at (x₁, y₁): (y − y₁) = f’(x₁)(x − x₁) Equation of normal: (y − y₁) = −1/f’(x₁) · (x − x₁) (perpendicular to tangent) Rate of change: dy/dx gives rate; related rates use dy/dt = (dy/dx) · (dx/dt) Maxima: f’(x) = 0 and f”(x) < 0; Minima: f’(x) = 0 and f”(x) > 0 Mean Value Theorem: f’(c) = [f(b)−f(a)]/(b−a) for some c ∈ (a, b) ⚡ Exam tip: JEE Main tests tangent/normal equations and maxima/minima regularly — practice with varied functions!

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🟡 Standard — Core Study

Standard content for students with a few days to months.

Applications of Derivatives — JEE Main Study Guide

Tangent and Normal:

• Slope of tangent at (x₁, y₁): m = f’(x₁)
• Equation: y − y₁ = m(x − x₁)
• Slope of normal = −1/m (perpendicular)
• If m = 0 (horizontal tangent): normal is vertical line x = x₁
• If m = ∞ (vertical tangent): normal is horizontal line y = y₁

Angle between curves: Angle θ between two curves at intersection satisfies tan θ = |m₁ − m₂|/(1 + m₁m₂)| Where m₁, m₂ are slopes of tangents at point of intersection

Monotonicity:

• f’(x) > 0 on (a, b) → f is strictly increasing
• f’(x) < 0 on (a, b) → f is strictly decreasing
• f’(x) ≥ 0 → f is non-decreasing (similarly for non-increasing)

Critical points: f’(x) = 0 or f’(x) does not exist; these are candidates for maxima/minima

First derivative test:

• If f’ changes from positive to negative at c: local maximum at c
• If f’ changes from negative to positive at c: local minimum at c
• If f’ does not change sign: neither max nor min

Second derivative test:

• f’(c) = 0 and f”(c) > 0 → local minimum
• f’(c) = 0 and f”(c) < 0 → local maximum
• f”(c) = 0 → test is inconclusive → use first derivative test

Global max/min on [a, b]: Evaluate f at critical points inside (a, b) and at endpoints a, b; highest is global max, lowest is global min

Rate problems:

• dy/dt = (dy/dx) · (dx/dt)
• Always identify what is changing and relate them through the function

Approximation: For small Δx, Δy ≈ f’(x) · Δx; useful for estimating function values near known points

• Key formula: Tangent: y − y₁ = f’(x₁)(x − x₁); Normal: y − y₁ = −(x − x₁)/f’(x₁)
• Common trap: For function with vertical tangent, f’(x) = ∞, so normal is horizontal line
• Exam weight: 2 questions per year (8 marks); tangent/normal and maxima/minima are most common

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🔴 Extended — Deep Dive

Comprehensive coverage for students on a longer study timeline.

Applications of Derivatives — Comprehensive JEE Main Notes

Rolle’s theorem: If f(a) = f(b) and f is continuous on [a,b] and differentiable on (a,b), then ∃ c ∈ (a,b) where f’(c) = 0 Geometrically: there is at least one stationary point where tangent is horizontal

Lagrange’s MVT: If f is continuous on [a,b] and differentiable on (a,b), then ∃ c ∈ (a,b) where f’(c) = [f(b)−f(a)]/(b−a) Interpretation: chord joining (a,f(a)) and (b,f(b)) is parallel to tangent at some point

Using MVT to prove inequalities: If f’(x) ≤ M on (a,b) and f is continuous, then |f(b) − f(a)| ≤ M|b − a|

Concavity and points of inflection:

• f”(x) > 0 on interval → curve is concave up (cup-shaped)
• f”(x) < 0 on interval → curve is concave down (cap-shaped)
• Point of inflection: f”(x) = 0 or undefined, and f” changes sign

Curvature: Radius of curvature R at (x, y) = [(1 + (dy/dx)²)^{3/2}] / |d²y/dx²| For parametric (x(t), y(t)): R = [(ẋ² + ẏ²)^{3/2}] / |ẋÿ − ÿẋ|

Envelope of family of curves: Given y = f(x, a), eliminate a between y = f(x,a) and ∂f/∂a = 0 Result gives the envelope curve that is tangent to each member of the family

Orthogonal trajectories: For family of curves F(x, y, c) = 0, first find differential equation by eliminating c Replace dy/dx by −dx/dy to get OT differential equation, then solve

Approximation formula: √(25.3) = √(25 + 0.3) ≈ 5 + 0.3/(2·5) = 5 + 0.03 = 5.03 General: √(a + δ) ≈ √a + δ/(2√a) for small δ

Differential of arc length: ds = √[(dx)² + (dy)²] For y = f(x): ds/dx = √[1 + (dy/dx)²] For x = f(t), y = g(t): ds/dt = √[(dx/dt)² + (dy/dt)²]

Applications in physics:

• Velocity = dy/dt, Acceleration = d²y/dt²
• For motion along line: v = dx/dt, a = dv/dt = d²x/dt²
• Related rates: for circular ripple (radius r expanding at dr/dt), area A = πr² → dA/dt = 2πr·dr/dt

Optimization problems: Steps: identify variable to optimize, express in terms of single variable, find domain, differentiate and set = 0, check endpoints and second derivative

Mean Value Theorem for Cauchy: If f and g are continuous on [a,b] and differentiable on (a,b) and g’(x) ≠ 0, then ∃ c where: [f(b)−f(a)]/[g(b)−g(a)] = f’(c)/g’(c)

Advanced maxima/minima:

• For two variables with constraint, use Lagrange multipliers

• If z = f(x,y) subject to g(x,y) = 0, then ∇f = λ∇g

• Remember: f’(x) > 0 → increasing; f’(x) < 0 → decreasing; f”(x) > 0 → concave up; Max: f’=0, f”<0; Min: f’=0, f”>0; Tangent: y−y₁ = f’(x₁)(x−x₁); Normal: slope = −1/f’(x₁)

• Previous years: “Find equation of tangent to y² = x³ at (1,1)” [2023]; “Find maximum value of x(1−x) for x∈[0,1]” [2024]; “If radius of sphere is increasing at 2 cm/s, find rate of change of volume when r=5 cm” [2024]

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📊 JEE Main Exam Essentials

Detail	Value
Questions	90 (30 per subject)
Time	3 hours
Marks	300 (90 per subject)
Section	Physics (30), Chemistry (30), Mathematics (30)
Negative	−1 for wrong answer
Mode	Computer-based

🎯 High-Yield Topics for JEE Main Mathematics

• Calculus (Differentiation + Integration) — ~35 marks combined
• Coordinate Geometry (straight lines, circles, conics) — ~20 marks
• Algebra (Complex Numbers, Quadratics, P&C, Probability) — ~25 marks
• Trigonometry + Inverse Trigonometry — ~15 marks
• Vector + 3D — ~15 marks

📝 Previous Year Question Patterns

• Applications of Derivatives: 2 questions per year, 8 marks
• Common patterns: equation of tangent/normal, find local/global maxima and minima, rate problems, monotonicity
• Weight: very high frequency, highly scoring

💡 Pro Tips

• For tangent/normal equations, always find the slope first
• In rate problems, always state what dy/dx equals and relate through the given relation
• For optimization, express everything in terms of one variable before differentiating
• Check endpoints for global max/min — critical points alone don’t guarantee global extremes
• The first derivative test is more reliable than the second when f”(c) = 0
• For approximation using differentials: Δy ≈ dy = f’(x)·Δx

🔗 Official Resources

• NTA Official JEE Main
• JEE Main Syllabus PDF

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