Maths · Class 12 · Limit, Continuity and Differentiability

Application of Derivatives for JEE

Everything you need for this topic on one page: the ideas, the formulas, the traps, and free lectures that explain it well.

Last updated: 26 October 2026

Direct answer

Rate of change, increasing and decreasing functions, tangents and normals, maxima and minima, and the mean value theorems.

Rate of change, increasing and decreasing functions, tangents and normals, maxima and minima, and the mean value theorems.

Key ideas

Application of Derivatives: what you must understand

  • f′(x) > 0 means increasing
  • At a local max f′ = 0 and f″ < 0
  • Check endpoints for absolute max/min on a closed interval
  • Rolle's and Lagrange's mean value theorems
Formula sheet

Must-know formulas for Application of Derivatives

  • Tangent: y − y₁ = m(x − x₁), m = f′(x₁)
  • Normal slope = −1/m
  • LMVT: f′(c) = (f(b) − f(a))/(b − a)
  • Second derivative test: f″(c) < 0 → max
Print

One-page formula sheet

Printable Application of Derivatives formula sheet: formulas, key ideas and traps on one page. Save it as a PDF for offline revision.

Avoid

Common mistakes students make

  • Forgetting endpoints
  • Using f′ = 0 points without checking sign change
  • Sign errors in normal slope
For the exam

How this topic is asked in JEE

Appears frequently. Maxima–minima word problems need a clean function first. Set it up carefully.

Prerequisites: Continuity and Differentiability.

NCERT map

Where Application of Derivatives is in NCERT

Chapter names follow the current rationalised NCERT textbooks. Read these sections before solving PYQs.

ClassBookChapterSections JEE asks from
Class 12Mathematics Part IApplication of DerivativesRate of Change of Quantities; Increasing and Decreasing Functions; Tangents and Normals; Maxima and Minima
JEE Advanced level

A JEE Advanced-level example on Application of Derivatives

Question. A curve is defined by y = x³ - 3x² + 2x. A tangent is drawn to this curve at a point P where the tangent makes an angle of 45° with the positive x-axis. If P lies in the first quadrant, find the equation of the normal at P.

Show solution
  1. Find dy/dx = 3x² - 6x + 2. Since tangent makes 45° with x-axis, slope = tan(45°) = 1.
  2. Solve 3x² - 6x + 2 = 1, giving 3x² - 6x + 1 = 0. Using quadratic formula: x = (3 ± √6)/3.
  3. Check which x gives a point in first quadrant. Write y = x(x-1)(x-2). For x = (3-√6)/3 ≈ 0.184, we get y > 0. For x = (3+√6)/3 ≈ 1.816, we get y < 0. So x = (3-√6)/3.
  4. Calculate y-coordinate: y = [(3-√6)/3] × [(-√6)/3] × [(-3-√6)/3] = √6/9. Point P = ((3-√6)/3, √6/9).
  5. Slope of normal = -1 (negative reciprocal of tangent slope). Equation: y - √6/9 = -1(x - (3-√6)/3). Simplify to get 9x + 9y = 9 - 2√6.

Answer: 9x + 9y = 9 - 2√6 (or equivalently x + y = 1 - (2√6)/9)

This example was solved twice independently and both solutions agree.

Free lectures

Best free lectures for Application of Derivatives

Full-chapter free lectures picked from YouTube search results by views, from established JEE channels (mostly in Hinglish). Use the subtopic list to jump straight to the part you need (timestamps come from each video's own chapter list). Watch one, then solve PYQs the same day.

FAQs

Frequently asked questions

Is Application of Derivatives important for JEE Main 2027?

Appears frequently. It is part of the official 'Limit, Continuity and Differentiability' unit in the JEE Main 2027 syllabus.

What are the most important formulas in Application of Derivatives?

Tangent: y − y₁ = m(x − x₁), m = f′(x₁); Normal slope = −1/m; LMVT: f′(c) = (f(b) − f(a))/(b − a).

How should I study Application of Derivatives for JEE?

Understand the key ideas from NCERT or one full lecture, learn the formula sheet, then solve 25–30 past JEE questions on the topic. Maxima–minima word problems need a clean function first. Set it up carefully.

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