Class 12 Maths Chapter 6: Application of Derivatives

Application of Derivatives is Chapter 6 of NCERT Class 12 Maths. It shows how derivatives are used to measure the rate of change of quantities, to find where a function is increasing or decreasing, and to locate maximum and minimum values in real-life problems. These Class 12 Maths Chapter 6 notes give you the key tests, formulas and solved examples you need for the CBSE board exam.

What you will learn

Using derivatives to find the rate of change of one quantity with respect to another
Finding intervals where a function is increasing or decreasing
The meaning of critical points, local maxima and local minima
The first derivative test and the second derivative test
Finding absolute maximum and minimum values on a closed interval and solving word problems

Key Concepts and Formulas

Rate of Change of Quantities

If y = f(x), then dy/dx tells us how fast y changes when x changes. If both depend on time t, the chain rule connects their rates.

dy/dx = rate of change of y w.r.t. x
dy/dt = (dy/dx) × (dx/dt)
Marginal cost:    MC = dC/dx
Marginal revenue: MR = dR/dx

In marginal cost and revenue, x is the number of units produced or sold.
A positive rate means the quantity is increasing; a negative rate means it is decreasing.

Increasing and Decreasing Functions

Let f be continuous on [a, b] and differentiable on (a, b). The sign of f′(x) decides whether the graph rises or falls.

f′(x) > 0 on (a, b) ⇒ f strictly increasing
f′(x) < 0 on (a, b) ⇒ f strictly decreasing
f′(x) = 0 on (a, b) ⇒ f is constant

To find the intervals, solve f′(x) = 0, mark the points on the number line and test the sign of f′(x) in each part.

Maxima, Minima and Critical Points

A point c where f′(c) = 0 or f′(c) does not exist is called a critical point. Local maximum and minimum values can occur only at critical points.

First derivative test at c:
f′ changes + to −  ⇒ local maximum
f′ changes − to +  ⇒ local minimum
no sign change     ⇒ point of inflection
Second derivative test (f′(c) = 0):
f″(c) < 0 ⇒ local maximum
f″(c) > 0 ⇒ local minimum
f″(c) = 0 ⇒ test fails

If the second derivative test fails, go back to the first derivative test.

Absolute Maximum and Minimum

A continuous function on a closed interval [a, b] always has an absolute maximum and an absolute minimum value.

Find all critical points of f in (a, b).
Evaluate f at these critical points and at the end points a and b.
The largest value is the absolute maximum and the smallest is the absolute minimum.
In word problems, first write the quantity to be maximised or minimised as a function of one variable.

Solved Examples

Example 1: The radius of a circle increases at 0.3 cm/s. Find the rate of increase of its area when the radius is 5 cm.

A = πr²
dA/dt = 2πr × dr/dt
At r = 5, dr/dt = 0.3:
dA/dt = 2π × 5 × 0.3
∴ dA/dt = 3π cm²/s

Example 2: The volume of a cube increases at 12 cm³/s. How fast is the edge increasing when the edge is 4 cm?

V = a³
dV/dt = 3a² × da/dt
12 = 3 × 4² × da/dt
12 = 48 × da/dt
∴ da/dt = 0.25 cm/s

Example 3: The total cost of producing x items is C(x) = 0.02x³ + 5x + 40 rupees. Find the marginal cost when 10 items are produced.

MC = dC/dx = 0.06x² + 5
At x = 10: MC = 0.06 × 100 + 5
         = 6 + 5
∴ Marginal cost = ₹11

Example 4: Find the intervals in which f(x) = x³ − 6x² + 9x + 2 is increasing or decreasing.

f′(x) = 3x² − 12x + 9
      = 3(x − 1)(x − 3)
f′(x) = 0 ⇒ x = 1, 3
x < 1:      f′(x) > 0
1 < x < 3:  f′(x) < 0
x > 3:      f′(x) > 0
∴ Increasing on (−∞, 1) and (3, ∞),
  decreasing on (1, 3)

Example 5: Find the local maximum and local minimum values of f(x) = x³ − 6x² + 9x + 2.

f′(x) = 3(x − 1)(x − 3) = 0 ⇒ x = 1, 3
f″(x) = 6x − 12
f″(1) = −6 < 0 ⇒ local max at x = 1
f(1) = 1 − 6 + 9 + 2 = 6
f″(3) = 6 > 0 ⇒ local min at x = 3
f(3) = 27 − 54 + 27 + 2 = 2
∴ Local max = 6 at x = 1, local min = 2 at x = 3

Graph of a cubic function showing increasing and decreasing intervals with local maximum and minimum, Class 12 Application of Derivatives — The sign of f′(x) shows rises, falls, max and min

Example 6: Find the absolute maximum and minimum values of f(x) = 2x³ − 3x² − 12x + 5 on [−2, 3].

f′(x) = 6x² − 6x − 12 = 6(x − 2)(x + 1)
Critical points: x = −1, x = 2
f(−2) = −16 − 12 + 24 + 5 = 1
f(−1) = −2 − 3 + 12 + 5 = 12
f(2) = 16 − 12 − 24 + 5 = −15
f(3) = 54 − 27 − 36 + 5 = −4
∴ Absolute max = 12 at x = −1,
  absolute min = −15 at x = 2

Example 7: An open box is made from a square sheet of side 30 cm by cutting equal squares of side x cm from each corner and folding up the sides. Find x for maximum volume and the maximum volume.

Base side = 30 − 2x, height = x
V = x(30 − 2x)² = 4x³ − 120x² + 900x
V′ = 12x² − 240x + 900
   = 12(x − 5)(x − 15)
V′ = 0 ⇒ x = 5 or x = 15
x = 15 makes the base 0, so reject it
V″ = 24x − 240, V″(5) = −120 < 0
V(5) = 5 × 20 × 20
∴ Max volume = 2000 cm³ when x = 5 cm

Square sheet with corner squares cut and folded into an open box of maximum volume, Class 12 Maths — Cutting 5 cm squares gives the largest box, 2000 cm³

Common Mistakes to Avoid

In rate of change questions, substitute the given value (like r = 5) only after differentiating, never before.
Always write the sign of f′(x) in each interval clearly; just finding the roots of f′(x) = 0 does not earn full marks.
For absolute extrema on a closed interval, do not forget to check the end points.
In word problems, reject values that do not make physical sense, such as a negative length or a side that makes the volume zero.

Quick Revision

dy/dt = (dy/dx)(dx/dt)
Marginal cost = dC/dx
f′(x) > 0 ⇒ increasing
f′(x) < 0 ⇒ decreasing
Critical point: f′(c) = 0 or undefined
f″(c) < 0 ⇒ local maximum
f″(c) > 0 ⇒ local minimum

Absolute extrema: compare f at critical points and end points
Always reject answers that make no physical sense in word problems