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Application of Derivatives

Chapter 6Notes

CBSE Class 12 Mathematics · NCERT Mathematics Part-I

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Shishya's notes

What this chapter is about

This chapter shows how the derivative, which you learned to compute in the previous chapter, becomes a powerful tool for analysing functions and solving practical problems. You move from the question "how do I find dy/dx?" to "what can dy/dx tell me about the behaviour of y?"

The main uses covered are: finding the rate at which one quantity changes when another changes, determining whether a function is increasing or decreasing on an interval, locating the tangent and normal to a curve at a point, using derivatives to approximate small changes in a quantity, and finding the maximum and minimum values a function can attain. These ideas appear constantly in physics (velocity, acceleration, optimisation of energy), economics (marginal cost, maximum profit) and geometry (shortest distance, largest area).

After working through this chapter you should be able to interpret the sign of the first and second derivatives, set up and solve optimisation problems from word descriptions, and write the equations of tangent and normal lines for curves defined by explicit functions.

Key ideas

  • Rate of change: If a quantity y depends on x, then dy/dx gives the instantaneous rate of change of y with respect to x. When x is time, this becomes velocity (for displacement) or the rate of growth (for population, cost, etc.).
  • Increasing and decreasing functions: On an interval where dy/dx > 0 the function rises as x increases; where dy/dx < 0 the function falls. At points where dy/dx = 0 or is undefined, the function may change its trend.
  • Tangent and normal: At a point (a, f(a)) on the curve y = f(x), the tangent has slope f′(a) and the normal has slope −1/f′(a) (provided f′(a) ≠ 0). The tangent touches the curve; the normal is perpendicular to it.
  • Approximation using differentials: For a small change δx in x, the change in y is approximately dy = f′(x) × δx. This lets you estimate values like √4.02 without a calculator.
  • Local maxima and minima: A function has a local maximum at c if f(c) ≥ f(x) for all x near c, and a local minimum if f(c) ≤ f(x) for all x near c. These occur where f′(c) = 0 or f′(c) does not exist.
  • First derivative test: If f′ changes from positive to negative at c, then c is a local maximum; if it changes from negative to positive, c is a local minimum; if it does not change sign, c is a point of inflection.
  • Second derivative test: At a critical point where f′(c) = 0, if f″(c) < 0 then c is a local maximum; if f″(c) > 0 then c is a local minimum; if f″(c) = 0 the test is inconclusive.
  • Absolute extrema on a closed interval: Evaluate f at every critical point inside [a, b] and at the endpoints a and b; the largest value is the absolute maximum, the smallest is the absolute minimum.

Formulas and facts to remember

  • Formula / Rule: Rate of change of y w.r.t. x = dy/dx · Meaning: How fast y changes per unit change in x
  • Formula / Rule: Slope of tangent at (a, f(a)) = f′(a) · Meaning: Gradient of the line that just touches the curve there
  • Formula / Rule: Equation of tangent: y − f(a) = f′(a)(x − a) · Meaning: Point-slope form using the derivative
  • Formula / Rule: Slope of normal = −1/f′(a) (when f′(a) ≠ 0) · Meaning: Perpendicular slope
  • Formula / Rule: Approximate change: δy ≈ f′(x) δx · Meaning: Linear approximation for small δx
  • Formula / Rule: Critical point: f′(c) = 0 or f′(c) undefined · Meaning: Candidates for local extrema
  • Formula / Rule: First derivative test: sign change of f′ around c decides max/min · Meaning: Uses behaviour on both sides of c
  • Formula / Rule: Second derivative test: f″(c) < 0 ⇒ max; f″(c) > 0 ⇒ min · Meaning: Concavity decides the type

Worked examples

### Example 1 – Rate of change

A circular oil spill expands so that its radius increases at 0.5 m/s. Find the rate at which the area is increasing when the radius is 4 m.

Solution

Area A = πr².

Differentiate with respect to time t: dA/dt = 2πr × dr/dt.

Substitute r = 4 m and dr/dt = 0.5 m/s: dA/dt = 2π × 4 × 0.5 = 4π m²/s ≈ 12.57 m²/s.

The area is increasing at 4π square metres per second when the radius is 4 m.

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### Example 2 – Equation of tangent and normal

Find the tangent and normal to the curve y = x³ − 2x + 1 at x = 1.

Solution

At x = 1: y = 1 − 2 + 1 = 0. Point is (1, 0).

Derivative: dy/dx = 3x² − 2. At x = 1: dy/dx = 3 − 2 = 1.

Tangent (slope 1): y − 0 = 1(x − 1) ⇒ y = x − 1.

Normal (slope −1): y − 0 = −1(x − 1) ⇒ y = −x + 1.

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### Example 3 – Finding maximum and minimum

A farmer has 60 m of fencing to enclose a rectangular plot against a long wall (no fence needed along the wall). What dimensions give the largest area?

Solution

Let the side perpendicular to the wall be x metres and the side parallel to the wall be y metres.

Fencing constraint: 2x + y = 60, so y = 60 − 2x.

Area A = x × y = x(60 − 2x) = 60x − 2x².

Differentiate: dA/dx = 60 − 4x.

Set dA/dx = 0: 60 − 4x = 0 ⇒ x = 15.

Second derivative: d²A/dx² = −4 < 0, confirming a maximum.

Dimensions: x = 15 m, y = 60 − 30 = 30 m.

Maximum area = 15 × 30 = 450 m².

Common mistakes

  • Forgetting to check endpoints when finding absolute extrema on a closed interval → always evaluate f at both endpoints and at every interior critical point.
  • Using the second derivative test when f″(c) = 0 and wrongly concluding there is no extremum → when f″(c) = 0 the test fails; apply the first derivative test instead.
  • Writing the normal slope as f′(a) instead of −1/f′(a) → remember the normal is perpendicular, so multiply the tangent slope by −1 and take the reciprocal.
  • Mixing up increasing/decreasing with concavity → f′ > 0 means increasing; f″ > 0 means concave upward; these are different properties.
  • Ignoring units in rate-of-change problems → carry units through (e.g., m/s, m²/s) to catch dimensional errors.

Quick revision

  • dy/dx > 0 ⇒ function increasing; dy/dx < 0 ⇒ function decreasing.
  • Tangent slope = f′(a); normal slope = −1/f′(a).
  • Critical points occur where f′ = 0 or f′ does not exist.
  • First derivative test: check sign change of f′ across the critical point.
  • Second derivative test: f″ < 0 gives max, f″ > 0 gives min, f″ = 0 is inconclusive.
  • For absolute extrema on [a, b], compare f at critical points and at a and b.

Written by Shishya's AI on 26 Sept 2026 from the chapter's title and class level, in Shishya's own words — not a copy or summary of the textbook. Read the official chapter for the book's own text, activities and exercises.